Research Note

Vertical Flexibility in Supply Chains

Wallace J. Hopp, Seyed M.R. Iravani, Wendy Lu Xu

Department of Industrial Engineering and Management SciencesNorthwestern University, Evanston, IL 60208, USA

Abstract

Jordan and Graves (1995) initiated a stream of research on supply chain flexibility, which wasfurthered by Graves and Tomlin (2003), that examined various structures for achieving horizontalflexibility within a single level of a supply chain. In this note, we extend the theory of supplychain flexibility by considering placement of vertical flexibility across multiple stages in a supplychain. Specifically, we consider two types of flexibility - logistics flexibility and process flexibility- and examine how demand, production and supply variability at a single stage impacts the beststage in the supply chain for each type of flexibility. Under the assumptions that margins are thesame regardless of flexibility location, capacity investment costs are the same within and acrossstages, and flexibility is limited to a single stage of logistics (process) flexibility accompanied withnecessary process (logistics) flexibility, we show that both types of flexibility are most effectivewhen positioned directly at the source of variability. However, while expected profit increases aslogistics flexibility is positioned closer to the source of variability (i.e., downstream for demandvariability and upstream for supply variability), locating process flexibility anywhere except atthe stage with variability leads to the same decrease in expected profit.

Keywords: Supply Chain, Flexibility, Capacity Investment.

1 Introduction

The fundamental problem in any supply chain system is efficiently matching supply with demand.

Because supply and demand are uncertain, we must make use of various buffers, including safety

stock, safety lead time and safety capacity, to facilitate this matching problem. A well-known

principle of factory physics is that flexibility can reduce the amount of buffering needed to mitigate

the effects of variability (Hopp and Spearman 2008). Examples of flexible capacity in a supply chain

include: (a) Dell sourcing multiple mother boards from a single supplier, (b) Hewlett-Packard (HP)

assembling voltage adaptors to printers in their European distribution center before shipping them

to countries with different AC voltage standards, and (c) General Motors tooling stamping plants

to produce body parts for more than one model.

In each of these cases, by using capacity that can be shifted from one product type to another,

the firm enhances its ability to adjust to fluctuations in either the supply of materials or demand

for products. However, as these examples highlight, the flexibility can be positioned at different

levels of the supply chain, including suppliers (Dell), component plants (GM) or distribution (HP).

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Several authors have studied the problem of how to use process flexibility. Jordan and Graves

(1995) showed that most of the benefits of full flexibility (ability to produce and ship all products

from all plants) can be achieved by partial flexibility. Iravani et al. (2005) introduced the concept

of structural flexibility to capture the ability of a flexibility structure to respond to demand or

supply variability. Graves and Tomlin (2003) presented a framework for analyzing the benefits

of flexibility in a multi-stage supply chain and developed a flexibility measure and guidelines for

flexibility investment. Their paper addressed the question of which flexibility structure is most

efficient provided all stages of the supply chain make use of the same flexibility structure. Other

studies include Fine and Freund (1990), Gupta et al. (1992), and Van Mieghem (1998), among

others. Taken as a whole, this stream of research has provided a number of useful insights that

describe the impact of flexible technology in a supply chain. However, all of these studies have

focused on process flexibility within a single stage of the supply chain.

A multi-echelon supply chain also presents the question of which stage to target for flexibil-

ity investment. We term this the vertical flexibility problem because of the analogy to vertical

integration. Consequently, we refer to flexibility within a stage as horizontal flexibility. Figure

1 contrasts sample horizontal flexibility structures (Figure 1-left) with sample vertical flexibility

structures (Figure 1-right). These structures contain two types of flexibility, process flexibility (the

ability to produce different types of products) and logistics flexibility (the ability to ship products to

different locations). Aprile et al. (2005) used numerical studies to compare lost sales resulting from

different process and logistics flexibility configurations in a fixed-capacity, five-product, two-stage

supply chain. They observed that, given some degree of logistics flexibility, process flexibility in

the supply stage enables the system to cope with demand variability. They also noted that process

flexibility in the assembler stage is more beneficial when there is capacity variability in both supplier

and assembler stages.

Our note goes beyond the results of Aprile et al. (2005) toward a theory of vertical flexibility by

providing analytical results of where to locate flexibility within a supply chain. We do this by proving

a principle that describes how the optimal location for full logistics and process flexibility in a multi-

echelon multi-product supply chain is affected by variability in supply, demand and intermediate

processing. Our main insight is that, if (a) margins are the same regardless of flexibility location,

(b) capacity investment costs are the same within and across stages, (c) only one stage in the

supply chain has variability, and (d) flexibility decisions are limited to locating a single stage of full

logistics (process) flexibility accompanied with necessary process (logistics) flexibility, then logistics

(process) flexibility is most effective when positioned directly at the source of variability. However,

2

ABC

Demand

Demand

Demand

Plant 1

Plant 2

Plant 3

(a) Full Flexibility

(b) Chained Flexibility

A

B

C

ABC

ABC

Demand

Demand

Demand

Plant 1

Plant 2

Plant 3

A

B

C

AB

BC

CA

Demand

Demand

Demand

Plant 1

Plant 2

Plant 3

Plant 1

Plant 2

Plant 3

Stage 0Stage 1Stage 2

(c) Single-Stage Full Logistics Flexibility

Plant 1

Plant 2

Plant 3

Plant 1

Plant 2

Plant 3

Stage 4 Stage 3Plant 1

Plant 2

Plant 3

Stage 5

(d) Single-Stage Full Process Flexibility

A

B

C

ABC

ABC

ABC

ABC

ABC

ABC

ABC

ABC

ABC

A

B

C

A

B

C

Stage 0Stage 1Stage 2Stage 4 Stage 3Stage 5Demand

Demand

Demand

Plant 1

Plant 2

Plant 3

Plant 1

Plant 2

Plant 3

Stage 0Stage 1Stage 2Plant 1

Plant 2

Plant 3

Plant 1

Plant 2

Plant 3

Stage 4 Stage 3Plant 1

Plant 2

Plant 3

Stage 5

A

B

C

A

B

C

A

B

C

ABC

ABC

ABC

A

B

C

A

B

C

Figure 1: Left: Examples of horizontal flexibility (how to use flexibility in a given stage); Right: Examples ofvertical flexibility (what stage to make flexible). (Unshaded boxes: specialized plants, Shaded boxes: flexibleplants)

while the effectiveness of logistics flexibility increases with proximity to the source of variability, the

effectiveness of process flexibility is equally suboptimal when located at any stage other than the

stage with variability.

2 Model Formulation

We consider a multi-echelon, multi-product supply chain, that produces I different products, indexed

by n = 1, 2, . . . , I, and has K + 1 stages, indexed by k = 0, 1, . . . ,K, and I plants per stage. Note

that the number of plants at each stage is assumed to be the same as the number of products so

that, for a supply chain with no flexibility, each product has a dedicated plant at each stage of the

supply chain. Stage 0 denotes demand, so that node n of stage 0 represents the retail outlet for

product n, while stage K represents the initial (supply) stage. We assume that there are always

sufficient raw materials at stage K and that the cost of these materials is included in how much the

company earns for selling one unit of the product.

Shipping routes between plants in stage k and stage k − 1, or between plants in stage 1 and

demand nodes at stage 0, are represented by an arc set Ak, k = 1, 2, . . . , K, where plant i at

stage k can supply plant j at stage k − 1 (or demand node j if k = 1) iff (i, j) ∈ Ak. Plant i at

stage k can produce product n iff (i, n) ∈ Bk, k = 1, 2, . . . , K. Set A = {A1, . . . , AK} and set

B = {B1, . . . , BK}, respectively, represent the logistics and process flexibility configurations of the

supply chain.

We assume both demands and production capacities can be random. However, to focus on the

3

effect of variability on the optimal location of flexibility, we restrict our attention to cases where

only a single stage has variability. For a given flexibility configuration (i.e., fixed A and B), we

formulate the problem of maximizing expected profit as a two-step sequential decision process:

1. Capacity Investment Decision: First, before demand is observed, the firm chooses production

capacity levels for all plants, µki , by taking into account demand distributions and unit capacity

investment costs cki , i = 1, 2, . . . , I, k = 1, 2, . . . , K. For a plant with yield loss and machine failures

and other sources of variability, we consider the production capacity to be a random variable, Qki (µ

ki ),

which follows distribution f(Qki (µ

ki )) that depends on the level of µk

i . We use qki to represent

a realization of capacity Qki (µ

ki ) and let µ =

(µk

i

)and c =

(cki

)be corresponding matrices of

capacity and capacity investment cost, with µki and ck

i representing the entries at the ith row and

kth column.

2. Production Flow Decision: After all uncertain demand and/or production capacities have been

observed, the firm chooses a matrix of production and shipping flows, X = (Xkijn), where Xk

ijn

represents the quantity of product n produced in plant i of stage k for plant j of stage k − 1. Note

that this flow matrix is constrained by the flexibility configuration of the supply chain. A flow Xkijn

can be non-zero only if (i, j) ∈ Ak, (i, n) ∈ Bk, and (j, n) ∈ Bk−1. In other words, a flow of product

n from plant i to plant j can only exist if there is a shipping route from i to j, and both plants are

able to process the product. Taking into account the fact that distribution center i at stage 0 is

for product type i, to make our model concise, we define a set B0 for the demand nodes, such that

(j, n) ∈ B0 iff j = n. To simplify the notation, we define set F (k) as the set of all triples (i, j, n) for

all feasible flows Xkijn that satisfy (i, j) ∈ Ak, (i, n) ∈ Bk, and (j, n) ∈ Bk−1. Let rn denote the unit

selling price of product n minus the cost of raw materials, pkin the unit production cost of product n

in plant i of stage k, tkijn the unit transportation cost of product n from plant i at stage k to plant

j of stage k− 1, and r, p and t represent the corresponding vectors (matrices). To maximize profit,

the firm observes demand vector d = (d1, d2, . . . , dI) and production capacity matrix q =(qki

), and

then chooses its production flow matrix X as the optimal solution to the following linear program,

which we call problem P2(A,B,d,q), where π(·) represents the maximum profit:

π(A,B,d,q) = maxX

{ ∑

(i,j,n)∈F (1)

rnX1ijn −

K∑

k=1

∑

(i,j,n)∈F (k)

(pk

in + tkijn

)Xk

ijn

}(1)

subject to:∑

u:(u,i,n)∈F (k+1)

Xk+1uin =

∑

j:(i,j,n)∈F (k)

Xkijn i, n = 1, 2, . . . , I, k = 1, 2, . . . , K − 1, (2)

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∑

j,n:(i,j,n)∈F (k)

Xkijn ≤ qk

i i = 1, 2, . . . , I, k = 1, 2, . . . , K, (3)

∑

i,j:(i,j,n)∈F (1)

X1ijn ≤ dn n = 1, 2, . . . , I, (4)

Xkijn ≥ 0 i, j, n = 1, 2, . . . , I, k = 1, 2, . . . , K, (5)

Constraint (2) is the balance equation that sets the total production flow into a plant equal

to the total flow out of it for each product, with the implicit assumptions that to meet one unit

of demand for product n: (a) one unit of capacity is needed at each stage, and (b) all products

consume the same processing capacity at the plant. Constraint (3) guarantees that the total quantity

of production of a plant does not exceed its capacity. Constraint (4) avoids producing more than

needed. Constraint (5) ensures non-negativity of production flow.

Solving P2(A,B,d,q) is premised on first making capacity investments. To do this, the firm

considers a random demand vector D = (D1, D2, . . . , DI) and selects a production capacity matrix

µ =(µk

i

)which decides the distribution of corresponding random matrix Q(µ) =

(Qk

i (µki )

).

Profit is therefore a random variable, $(A,B,D,Q(µ)), that depends on the demand and capacity

distributions. For a given demand d and capacity q, profit is π(A,B,d,q), which is found by

solving P2(A,B,d,q). Hence, we can express the capacity investment decision faced by the firm

as solving the following problem, which we label P1(A,B,D):

V ∗(A,B,D) = maxµ

{ED,Q(µ)

[$(A,B,D,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µ

ki

}, (6)

where ED,Q(µ)

[$(A,B,D,Q(µ))

]is the expected profit. The expectation is over random demand

D and random capacity Q(µ), and V ∗(·) is the maximum value of expected profit minus capacity

investment cost. The matrix µ∗ that achieves V ∗(·) is called the optimal capacity investment strategy.

As shown in Jordan and Graves (1995) and Graves and Tomlin (2003), there are many ways a

single stage of a supply chain can be made flexible. Since the focus of this note is on the position,

rather than the type, of flexibility, we will focus on full flexibility and will assume a single stage of

flexibility. Full logistics flexibility is achieved at a stage kf if what is produced in each plant at stage

kf can be shipped to all plants at stage kf−1. We use A1full(kf ) to represent full logistics flexibility

configuration, where A1full(kf ) = {A1, . . . , Akf , . . . , AK}, with Ak = {(i, i) ∀ i = 1, 2, . . . , I} for

k 6= kf , and Akf = {(i, j) ∀ i = 1, 2, . . . , I, j = 1, 2, . . . , I}. It is worth emphasizing that in order to

make use of full logistics flexibility of stage kf , stage kf and all stages upstream to kf must have

process flexibility. As illustrated in Figure 1(c), in order to make use of full logistics flexibility of

stage 3, plants 1, 2 and 3 at stage 3 and all upstream stages must be able to process products A, B

and C.

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Full process flexibility is achieved at a single stage kf if all product types can be processed

in each plant of stage kf . We use B1full(kf ) to represent full logistics flexibility configuration,

where B1full(kf ) = {B1, . . . , Bkf , . . . , BK}, with Bk = {(i, i) ∀ i = 1, 2, . . . , I} for k 6= kf , and

Bkf = {(i, n) ∀ i = 1, 2, . . . , I, n = 1, 2, . . . , I}. In order to make use of process flexibility at stage

kf , stages kf and kf + 1 must have logistics flexibility so that plants at stage kf are supplied with

subassemblies of all products and are able to ship all products to the subsequent stage. As illustrated

in Figure 1(d), in order to make use of full process flexibility of stage 3, plants 1, 2 and 3 at stage

3 all need to ship products to all plants at stage 2. Also all plants must have supply from plants 1,

2 and 3 at stage 4. Therefore, full logistics flexibility is required at stages 3 and 4.

In the remainder of this note, we focus on the location of a single stage of full logistics (process)

flexibility. Also, when we say a stage is flexible, we mean it has full logistics (process) flexibility

and the corresponding process (logistics) flexibility to make it possible.

We assume that implementing full logistics flexibility at stage k incurs a fixed cost Λk ≥ 0,

(k = 1, 2, . . . ,K), to establish the shipping channels to all plants of stage k − 1. Also, full process

flexibility at stage k incurs a fixed cost Ψk ≥ 0, (k = 1, 2, . . . , K), to equip the plant with the

necessary tooling to process all types of products. We use ΛK+1 to denote the fixed cost incurred

for plants at stage K to establish inbound logistics flexibility (supply channels) to obtain all types of

raw material. Hence, to evaluate a flexibility configuration (A,B), we need to compute V ∗(A,B,D)

and subtract from it the fixed cost associated with the flexibility structure.

To develop our results on the optimal position of full logistics flexibility in a supply chain, we

first need to characterize the solution to P2(A1full(kf ),B,d,q). We define

mL,in(kf ) = rn −( kf−1∑

k=1

pknn +

K∑

k=kf

pkin

)−

[( kf−1∑

k=1

tknnn

)+ t

kf

inn +K∑

k=kf+1

tkiin

]

as the unit contribution margin for production flow from plant i of stage K to demand node n,

where rn is how much the company earns for selling one unit of the product,∑kf−1

k=1 pknn +

∑Kk=kf

pkin

is the production cost associated with the production flow, and∑kf−1

k=1 tknnn + tkf

inn +∑K

k=kf+1 tkiin

is the transportation cost. We can show (see Lemma 1 in On-line Appendix I) that, if a supply

chain has logistics flexibility only at a single stage, the production flow allocation problem in the

entire supply chain can be simplified to a single stage production flow allocation problem, where

production flow from plant i of stage K to demand node n is given by Ykf

in and is associated with

a unit profit margin, mL,in(kf ).

With respect to process flexibility, we define

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mP,in(kf ) = rn −[( K∑

k=1k 6=kf

pknn

)+ p

kf

in

]−

[( K∑

k=1k 6=kf ,kf+1

tknnn

)+ t

kf

inn + tkf+1nin

]

as the unit contribution margin for production flow of product n that is produced in plant i of

stage kf and in plant n of all other stages. The company earns rn for selling the product. At the

same time, this production flow involves production cost∑K

k=1k 6=kf

pknn + p

kf

in and transportation cost∑K

k=1k 6=kf ,kf+1

tknnn + tkf

jnn + tkf+1nin . We can show (see Lemma 2 in the On-line Appendix I) that, if a

supply chain has process flexibility at only a single stage, the production flow allocation problem

in the entire supply chain can be simplified to a single stage production flow allocation problem,

where flow of product n that is produced in plant i of stage kf and in plant n of all other stages is

given by Zkf

in and is associated with a unit profit margin, mP,in(kf ).

Note that margin mL,in(kf ) or mP,in(kf ) depends on the location (stage) of the logistics or

process flexibility structure, respectively, and is therefore a function of kf .

To generate our result, we will make use of the following assumptions throughout the paper.

Assumption 1. mL,in(k′) = mL,in(k) (mP,in(k′) = mP,in(k)) for all i and n and all k′ 6= k.

This assumption ensures that the unit contribution margin is the same regardless of where

logistics (process) flexibility is located, so that we can isolate the role of variability from the role of

cost.

Assumption 2. Unit variable production cost pkin and unit transportation cost tkijn are independent

of whether the capacity for product n at plant i of stage k is flexible or dedicated.

The main components of the unit variable production cost are material and labor, which usually

do not depend on whether the capacity is flexible or not. Consider, for instance, a flexible auto

assembly plant that produces models A and B, and a dedicated assembly plant that produces only

model A. In both plants, the material to produce model A is the same, and the labor skill required

(to install the doors, for example) is the same. Furthermore, the cost of shipping items from one

plant to another is clearly independent of whether those plants are flexible or not. Thus, Assumption

2 represents many systems in practice.

Assumption 3. mL,ii(kf ) > mL,in(kf ), mL,ii(kf ) > mL,ni(kf ), and mP,ii(kf ) > mP,in(kf ),

mP,ii(kf ) > mP,ni(kf ) ∀ n 6= i, i, n = 1, 2, . . . , I, 1 ≤ kf ≤ K.

This assumption reflects the cost of flexibility, logistics or process flexibility, in the form of more

sophisticated equipment, more highly trained staff, longer routes, etc., which reduces the margins

of products produced in flexible plants or shipped along non-standard distribution channels.

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Assumption 4. cki = ck

j , i = 1, 2, . . . , I, j = 1, 2, . . . , I, i 6= j, k = 1, 2, . . . ,K.

This assumption states that unit capacity investment costs are the same for all plants at the

same stage. This is a reasonable assumption in supply chains where plants at one stage of the supply

chain perform similar processing functions and therefore use similar equipment and facilities. For

example, in an electronics supply chain, semiconductor facilities feed board assembly plants, which

feed final assembly plants. Because the complexity of the technologies at each level is similar, the

capacity costs of plants within each level should also be similar. Of course, other supply chains may

involve very different processes, with very different capacity costs, at the same level. In such cases,

the differing capacity costs will obviously influence flexibility choices. But, since our focus is on the

influence of variability on the desired location for flexibility, we consider only supply chains without

a significant differences in capacity costs within stages.

The net effect of the above assumptions is that variability will drive flexibility location decisions,

not cost. Under these, we can show that for the single-stage logistics (process) flexibility configura-

tion, an optimal capacity configuration will have the same capacity at all stages without variability

or (logistics or process) flexibility (See proofs of lemma’s 3 and 4 in the On-line Appendix I).

3 Optimal Location for Logistics Flexibility

In systems where production facilities are already flexible, increasing system flexibility can be

achieved by introducing logistics flexibility. This is often the case in a pure distribution system

consisting of warehouses, depots and retail outlets, where all facilities can process all products. But

adding routes between facilities may entail fixed and variable costs. Within the Walmart distribu-

tion network, for example, personnel at any node (e.g., warehouse, depot, retail outlets) are able

to process all types of products, and so process flexibility can be realized without significant costs,

but logistics flexibility (e.g., supplying multiple stores from depots) is not costless.

It can be shown that necessary conditions for the optimal capacity investment strategies are

that if the single source of variability is downstream (upstream) of the flexible stage, then plants

downstream (upstream) of the flexible stage should have larger total capacity than plants upstream

(downstream) in order to hedge against the variability (see Lemma 3 in the On-line Appendix I).

This leads directly to our main results for the case of a single stage with logistics flexibility, which

we present in Theorem 1.

Theorem 1. For a multi-echelon supply chain with full logistics flexibility at only a single stage kf ,

process flexibility at all upstream stages of kf , and only a single source of variability at stage kv,

then:

8

(1) if kv + 1 ≤ kf ≤ K − 1 (i.e., the source of variability is downstream from the stage with

logistics flexibility), then logistics flexibility at stage kf achieves higher expected profit than

does logistics flexibility at stage kf + 1;

(2) if 2 ≤ kf ≤ kv (i.e., the source of variability is upstream from the stage with logistics flex-

ibility), then logistics flexibility at stage kf achieves higher expected profit than does logistics

flexibility at stage kf − 1.

Theorem 1 implies that if a supply chain has single source of variability, and all stages have (full)

process flexibility, it is most beneficial to place logistics flexibility closest to the source of variability.

The intuition behind this is as follows. Suppose we place the logistics flexible stage adjacent to the

source of variability on the downstream side (i.e., kv = kf ). This allows us to use any available

capacity of the plants at the stage with variability kv (and its upstream plants, since they feed

the plants in the stage with variability kv) to hedge against variability. Therefore, in an optimal

configuration, the capacity of plants at the stage with variability kv (and upstream of stage kv) will

be higher than that of the dedicated plants downstream of the stage with variability kv. If we move

the flexible stage one level further from the source of variability (i.e., kf = kv − 1), then we can

still use the plants at and upstream of the stage with variability kv to hedge against variability, but

now we must increase the capacity of the plants at an additional stage (i.e., stage kf , since plants

at stage kf should have the capacity to process items that they receive from plants at stage kv).

Hence, it costs more to get the same amount of protection as the flexible stage is moved away from

the source of variability.

From a management perspective, these results suggest that demand variability (i.e., kv = 0) pro-

vides motivation to make downstream stages of the supply chain flexible, while supply variability

(i.e., kv = K) provides motivation to make the upstream stages flexible. In systems where process

flexibility is inexpensive, this is a crisp insight. However, when process flexibility is costly, down-

stream logistics flexibility becomes more expensive (since it requires all upstream stages to have

process flexibility). So, when supply is variable, upstream logistics flexibility is clearly preferable

(since it is closer to source of variability and requires fewer upstream stages to have process flexi-

bility). But when demand is variable, we must balance the cost of the additional process flexibility

with the benefits of positioning the logistics flexibility further downstream. In On-Line Appendix II

we further investigate this tradeoff and we show that the flexibility location decision has a threshold

structure.

9

4 Optimal Location for Process Flexibility

In addition to facilitating logistics flexibility, process flexibility is effective in its own right. Indeed,

in systems where full logistics flexibility is inexpensive (e.g., material is shipped between facilities

via a third party logistics firm and so additional routes can be added without fixed cost), enhancing

flexibility is solely a matter of deciding where to add process flexibility. To gain insight into this

decision, we consider the problem of locating a single stage of full process flexibility.

It can be shown that the process flexible stage cannot have more (and may have less) capacity

than the other stages as a result of the ability to produce different products (See Lemma 4 in On-line

Appendix I for details).

We can now state our main results for the optimal location of process flexibility in Theorem

2. Note that Part (1) of the theorem holds when Assumptions 3 and 4 are relaxed, and with an

additional assumption:

Assumption 5. cki = ck′

i , i = 1, 2, . . . , I, k, k′ = 1, 2, . . . , K, k 6= k′.

This assumption states that unit capacity investment costs are the same for plants across stages.

If cki 6= ck′

i , then the optimal location of flexibility could be affected by both the location of variability

and the capacity investment cost structure. So we rule this out to focus exclusively on variability

location.

Theorem 2. For a multi-echelon supply chain with logistics flexibility at all stages,

(1) if only stage kv = 0 (demand) has variability, then the expected profit for a system with a

single stage of process flexibility at stage k is equal for any k = 1, 2, . . . , K.

(2) if only stage kv > 0 (plants) has variability, then expected profit is maximized if the single stage

with process flexibility is located at kv; for all other positions of the flexible stage, expected profit

is equal.

Theorem 2 characterizes the optimal location for process flexibility in a system where the fixed

cost of logistics flexibility is zero, and therefore it is costless to have logistics flexibility at all stages,

including stages adjacent to the stage with process flexibility.

The intuition behind the result of Theorem 2 is as follows. If the only source of variability is

demand, then the amount of demand for a product, say product n, that can be satisfied is restricted

by (1) the capacity of plants that produce product n at all dedicated (i.e., non-flexible) stages, and

(2) total capacity of plants at the flexible stage. No matter which stage has process flexibility, these

two restrictions are the same. Therefore, investing in process flexibility at any stage is equivalent.

10

In contrast, if the only source of variability is stage kv ≥ 1 (i.e., the capacity of stage kv is random),

then making the plants at stage kv flexible allows excess capacity of one plant at stage kv to make up

for a capacity shortage at another plant at that stage. If, instead, any stage other than kv has process

flexibility, then such pooling is not possible because production of each product is constrained by

the capacities at stage kv. Hence, investment in process flexibility at stage kv is optimal when it is

the only source of variability.

From a management perspective, the above results suggest that, when demand is the major

source of variability (i.e., kv = 0), the impact of variability is not the key consideration in decisions

about locating process flexibility. Since process flexibility is equally effective at almost all stages,

it makes sense to install such flexibility wherever it is least expensive. In contrast, when supply is

variable (i.e., kv = K), then there is incentive to make the suppliers themselves flexible. In supply

chain terms, this suggests that multi-sourcing from flexible suppliers may be a helpful strategy for

mitigating problems of yield loss. But it may not be a particularly attractive option for dealing with

uncertain demand, since it may be cheaper to install (equally effective) flexibility at a downstream

stage.

5 Conclusions

In this note, we have focused specifically on the impact of variability on the optimal placement of

logistics and process flexibility in a multi-product, multi-echelon supply chain. Although we have

only discussed full flexibility, our insights about flexibility position generally carry over to other

configurations (e.g., the “chaining” structure suggested by Jordan and Graves (1995)), provided

that comparisons are made between the same configuration at different stages. To isolate the effect

of variability, we have considered systems in which the capacity investment cost is the same within

and across levels of the supply chain. For such systems, we have shown analytically that if there

is only a single source of variability (in supply, demand, or any intermediate stage of the supply

chain), then positioning logistics flexibility as close as possible to the source of variability or process

flexibility at the source of variability is optimal when the two types of flexibility are considered

separately (i.e., either process or logistics flexibility is costless and the problem is only to locate a

single stage of the other type of flexibility). When both types of flexibility are costly, the optimal

configuration is more complicated, but still exhibits a threshold structure that is informed by the

behavior of the cases where process and logistics flexibility are considered separately (see On-line

Appendix II for details).

In practical terms, our results imply that systems with a high degree of supply variability should

11

make use of upstream logistics flexibility provided process flexibility is inexpensive. For example,

supply chains that rely on recycled materials may be subject to uncontrollable fluctuations in their

inputs and hence would benefit from enhancing flexibility in this first stage of the network (e.g., by

using multiple recycling plants to supply each downstream production plant). In contrast, supply

chains subject to volatile customer demand may be better served by downstream logistics flexibility.

For example, the automotive supply chains that motivated the original Jordan and Graves (1995)

work must cope with fluctuations in individual model sales that occur after plant capacity decisions

have been made. By making the final assembly plants capable of supplying demands of different

models, their capacity can be used more efficiently to satisfy demand.

Of course, variability is only one factor affecting optimal flexibility configurations. Another

obvious factor is cost. For systems where flexibility is very expensive at upstream stages (e.g.,

electronics supply chains in which the first stage is a very costly and inflexible wafer fab), it may

make sense to use flexibility predominantly in downstream stages, regardless of the source of vari-

ability. In contrast, in systems where flexibility is very expensive at downstream stages (e.g., some

pharmaceutical supply chains, in which cost, specialization and regulations may restrict the extent

to which multiple products can be produced in the same finishing plant), it may make sense to use

more flexibility in upstream stages (e.g., commodity chemical plants). Further research is needed

to incorporate cost, variability, and the various process constraints of specific environments.

References

Aprile, D., A. C. Garavelli and I. Giannoccaro (2005). Operations planning and flexibility in a supply chain.Production Planning and Control 16 (1) 21-31.

Fine, C. H. and R. M. Freund (1990). Optimal investment in product-flexible manufacturing capacity.Management Science 36 (4) 449-466.

Graves, S. C. and B. T. Tomlin (2003). Process flexibility in supply chains. Management Science 49 (7)907-919.

Gupta, D., Y. Gerchak and J. A. Buzacott (1992). The optimal mix of flexiblity and dedicated manufacturingcapacities: hedging against demand variability. International Journal of Productin Economics 28 309-319.

Hopp, W.J. and M.L. Spearman (2008) Factory Physics: Foundations of Manufacturing Management, ThirdEdition. McGraw-Hill, New York.

Iravani, S.M.R., Sims, K. and M.P. Van Oyen (2005). Structural Flexibility: A new perspective on thedesign of manufacturing and service operations. Management Science 51, 151-166.

Jordan, W. C. and S. C. Graves (1995). Principles on the benefits of manufacturing process flexibility.Management Science 41 (4) 577-594.

Van Mieghem, J. A. (1998). Investment strategies for flexible resources. Management Science 44 (8)1071-1078.

12

ON-LINE APPENDIX IProofs of Analytical Results

Lemma 1. For a multi-echelon supply chain with full logistics flexibility only at stage kf , let Ykf

in be production

flow from plant i of stage K to demand node n of stage 0, and let mL(kf ) =(mL,in(kf )

)be the matrix where

element mL,in(kf ) denotes the element of the ith row and nth column, problem P2(A1full(kf ),B,d,q) canbe simplified to the following linear program, P2(mL(kf ),d,q):

π(mL(kf ),d,q) = maxY

kfin

{ I∑

i=1

I∑n=1

mL,in(kf )Y kf

in

}(7)

subject to :I∑

n=1

Ykf

in ≤ minkf≤k≤K

{qki

}i = 1, 2, . . . , I, (8)

I∑

i=1

Ykf

in ≤ min1≤k≤kf−1

{qkn, dn

}n = 1, 2, . . . , I. (9)

PROOF OF LEMMA 1:

Since a flow Xkijn can be non-zero only if (i, j) ∈ Ak, (i, n) ∈ Bk, and (j, n) ∈ Bk−1, for A1full(kf ), we have

Xkijn 6= 0 only if i = j = n, i, j, n = 1, 2, . . . , I, k = 1, 2, . . . , kf − 1, (10)

Xkijn 6= 0 only if j = n, i, j, n = 1, 2, . . . , I, k = kf , (11)

Xkijn 6= 0 only if i = j, i, j, n = 1, 2, . . . , I, k = kf + 1, kf + 2, . . . , K. (12)

Therefore constraint (2) of P2(A,B,d,q) becomes

X1nnn = · · · = X

kf−1nnn =

I∑

i=1

Xkf

inn n = 1, 2, . . . , I, (13)

Xkf

inn = Xkf+1iin = · · · = XK

iin i, n = 1, 2, . . . , I. (14)

And constraint (3) of P2(A,B,d,q) becomes

Xknnn ≤ qk

n n = 1, 2, . . . , I, ∀ 1 ≤ k < kf , (15)I∑

n=1

Xkf

inn ≤ qkf

i i = 1, 2, . . . , I, (16)

I∑n=1

Xkiin ≤ qk

i i = 1, 2, . . . , I, ∀ kf < k ≤ K. (17)

From (13) we have Xknnn =

∑Ii=1 X

kf

inn, n = 1, 2, . . . , I, ∀ 1 ≤ k < kf . substituting it into (15) we get

I∑

i=1

Xkf

inn ≤ qkn n = 1, 2, . . . , I, 1 ≤ k < kf . (18)

From (14) we have Xkiin = X

kf

inn, i, j = 1, 2, . . . , I, ∀ kf < k ≤ K. substituting it into (17) we get

I∑n=1

Xkf

inn ≤ qki i = 1, 2, . . . , I, kf < k ≤ K. (19)

13

Combining (19) with (16) we have

I∑n=1

Xkf

inn ≤ qki i = 1, 2, . . . , I, kf ≤ k ≤ K. (20)

If kf 6= 1, constraint (4) of P2(·) becomes

X1nnn ≤ dn n = 1, 2, . . . , I,

From (13) we have Xknnn =

∑Ii=1 X

kf

inn, n = 1, 2, . . . , I, ∀ 1 ≤ k < kf , which by substituting into the aboveinequality we get

I∑

i=1

Xkf

inn ≤ dn n = 1, 2, . . . , I. (21)

And if kf = 1, constraint (4) is exactly same as (21). On the other hand, (20) is equivalent to

I∑n=1

Xkf

inn ≤ minkf≤k≤K

{qki

}i = 1, 2, . . . , I. (22)

And combining (18) and (21) we have

I∑

i=1

Xkf

inn ≤ min1≤k≤kf−1

{qkn, dn

}n = 1, 2, . . . , I. (23)

We define Ykf

in as production flow from plant i at stage K to demand node n at stage 0. Since we havelogistics flexibility only at stage kf , the only possible path for flow Y

kf

in is from plant i to plant i at stagesK,K − 1, . . . , kf , then from plant i at stage kf to plant n at stage kf − 1, and then from plant n to plant n

at stages kf − 1, kf − 2, . . . , 1. Therefore, flow Ykf

in is exactly the same as the flow of product n that flexibleplant i at stage kf sends to plant n at stage kf − 1, i.e., Y

kf

in = Xkf

inn. Substituting Xkf

inn with Ykf

in in (22)and (23) we get (8) and (9) for P2(·).Now we derive the objective function of P2(·). Taking into account (10), (11) and (12), if kf 6= 1, objectivefunction (1) can be written as

I∑

i=1

rnX1nnn −

kf−1∑

k=1

I∑

i=n

(pk

nn + tknnn

)Xk

nnn −I∑

i=1

I∑n=1

(p

kf

in + tkf

inn

)X

kf

inn −K∑

k=kf+1

I∑

i=1

I∑n=1

(pk

in + tkiin

)Xk

iin.

In the above, by substituting Xkf

inn = Ykf

in , Xknnn =

∑Ii=1 X

kf

inn =∑I

i=1 Ykf

in , where n = 1, 2, . . . , I, and1 ≤ k < kf , and also Xk

iin = Xkf

inn = Ykf

in , where i, n = 1, 2, . . . , I, and kf < k ≤ K, we will have:

I∑n=1

[rn

I∑

i=1

Ykf

in

]−

kf−1∑

k=1

I∑n=1

[(pk

nn + tknnn

) I∑

i=1

Ykf

in

]

−I∑

i=1

I∑n=1

(p

kf

in + tkf

inn

)Y

kf

in −K∑

k=kf+1

I∑

i=1

I∑n=1

[(pk

in + tkiin

)Y

kf

in

](24)

=I∑

i=1

I∑n=1

rnYkf

in −I∑

i=1

I∑n=1

[( kf−1∑

k=1

(pk

nn + tknnn

))Y

kf

in

]

−I∑

i=1

I∑n=1

(p

kf

in + tkf

inn

)Y

kf

in −I∑

i=1

I∑n=1

[( K∑

k=kf+1

(pk

in + tkiin))

Ykf

in

]

=I∑

i=1

I∑n=1

rn −

( kf−1∑

k=1

pknn +

K∑

k=kf

pkin

)−

[( kf−1∑

k=1

tknnn

)+ t

kf

inn +K∑

k=kf +1

tkiin

]Y

kf

in .

14

Let

mL,in(kf ) = rn −( kf−1∑

k=1

pknn +

K∑

k=kf

pkin

)−

[( kf−1∑

k=1

tknnn

)+ t

kf

inn +K∑

k=kf+1

tkiin

],

then the objective function (1) can be written as∑I

i=1

∑In=1 mL,in(kf )Y kf

in , which is what we have in (7) forP2(·).If kf = 1, objective function (1) can be written as

I∑

i=1

I∑n=1

rnXkf

inn −I∑

i=1

I∑n=1

(p

kf

in + tkf

inn

)X

kf

inn −K∑

k=kf+1

I∑

i=1

I∑n=1

(pk

in + tkiin

)Xk

iin

substituting Xkf

inn = Ykf

in and Xkiin = X

kf

inn = Ykf

in , where i, n = 1, 2, . . . , I and kf < k ≤ K, in the above, wewill have:

I∑n=1

[rn

I∑

i=1

Ykf

in

]−

I∑

i=1

I∑n=1

(p

kf

in + tkf

inn

)Y

kf

in −K∑

k=kf+1

I∑

i=1

I∑n=1

[(pk

in + tkiin

)Y

kf

in

]

which is equivalent to (24) with the second term equal to 0. All the argument for kf 6= 1 case then follows.This completes the proof of Lemma 1.

Using Lemma 1, P1(A1full(kf ),B,D) can be simplified to P1(mL(kf ),D), where

V ∗(mL(kf ),D) = maxµ

{ED,Q(µ)

[$(mL(kf ),D,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

}.

We now introduce Lemma 2.

Lemma 2. For a multi-echelon supply chain with full process flexibility only at stage kf , let Zkf

in be theproduction flow from plant n at stage K to demand node n at stage 0 that is processed in plant i at stage kf

and in plant n at all the other stages, and let mP(kf ) =(mP,in(kf )

)be the matrix where element mP,in(kf )

denotes the element at the nth row and ith column, problemP2(A,B1full(kf ),d,q) can be simplified to the following linear program, P2(mP(kf ),d,q):

π(mP(kf ),d,q) = maxZ

kfin

{ n∑n=1

n∑

i=1

mP,in(kf )Zkf

in

}(25)

subject to :I∑

i=1

Zkf

in ≤ min1≤k≤Kk 6=kf

{qkn, dn

}n = 1, 2, . . . , I, (26)

I∑n=1

Zkf

in ≤ qkf

i i = 1, 2, . . . , I. (27)

Using Lemma 2, P1(A,B1full(kf ),D) can be simplified to P1(mP(kf ),D), where

V ∗(mP(kf ),D) = maxµ

{ED,Q(µ)

[$(mP(kf ),D,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

}.

15

PROOF OF LEMMA 2:

Since a flow Xkijn can be non-zero only if (i, j) ∈ Ak, (i, n) ∈ Bk, and (j, n) ∈ Bk−1, for B1full(kf ), we have

Xkijn 6= 0 only if i = n = j, i, j, n = 1, 2, . . . , I, ∀ k 6= kf , k 6= kf + 1, 1 ≤ k ≤ K, (28)

Xkijn 6= 0 only if j = n, i, j, n = 1, 2, . . . , I, k = kf , (29)

Xkijn 6= 0 only if i = n, i, j, n = 1, 2, . . . , I, k = kf + 1. (30)

Therefore constraint (2) of P2(A,B,d,q) becomes

X1nnn = · · · = X

kf−1nnn =

I∑

i=1

Xkf

inn n = 1, 2, . . . , I, (31)

Xkf+1nin = X

kf

inn n, i = 1, 2, . . . , I, (32)I∑

i=1

Xkf +1nin = X

kf +2nnn = · · · = XK

nnn n = 1, 2, . . . , I. (33)

And constraint (3) of P2(A,B,d,q) becomes

Xknnn ≤ qk

n n = 1, 2, . . . , I, ∀ k 6= kf , k 6= kf + 1, 1 ≤ k ≤ K, (34)I∑

n=1

Xkf

inn ≤ qkf

i i = 1, 2, . . . , I, (35)

I∑

i=1

Xkf+1nin ≤ q

kf+1n n = 1, 2, . . . , I. (36)

From (31) we have Xknnn =

∑Ii=1 X

kf

inn, for n = 1, 2, . . . , I, and 1 ≤ k < kf . Substituting it into (34) we get

I∑

i=1

Xkf

inn ≤ qkn n = 1, 2, . . . , I, 1 ≤ k < kf . (37)

From (32) and (33) we have Xknnn =

∑Ii=1 X

kf+1nin =

∑Ii=1 X

kf

inn, for n = 1, 2, . . . , I, and kf + 2 ≤ k ≤ K.Substituting it into (34) we get

I∑

i=1

Xkf

inn ≤ qkn n = 1, 2, . . . , I, kf + 2 ≤ k ≤ K. (38)

And substituting (32) in (36) we have

I∑

i=1

Xkf

inn ≤ qkf+1n n = 1, 2, . . . , I. (39)

If kf 6= 1, constraint (4) of P2(·) becomes

X1nnn ≤ dn n = 1, 2, . . . , I,

From (31) we have X1nnn =

∑Ii=1 X

kf

inn, for n = 1, 2, . . . , I, which by substituting into the above inequalitywe get

I∑

i=1

Xkf

inn ≤ dn n = 1, 2, . . . , I. (40)

16

And if kf = 1, constraint (4) is exactly same as (40). Combining (37), (38), (39) and (40) we have:

I∑

i=1

Xkf

inn ≤ min1≤k≤K,k 6=kf

{qkn, dn

}n = 1, 2, . . . , I. (41)

We define Zkf

in as production flow of product n that is produced in plant i at stage kf . This flow is alsoequal to the flow in plant n at all the other stages, since we have process flexibility only at stage kf . Thatis, Z

kf

in = Xkf

inn = Xkf+1nin . Substituting X

kf

inn and Xkf+1nin with Z

kf

in in (41) and (35) we get constraints (26)and (27) in Lemma 2.

Now consider the objective function (1). Taking into account (28), (29) and (30), if kf 6= 1, objective function(1) can be written as

I∑n=1

rnX1nnn −

K∑

k=1k 6=kf ,kf+1

I∑n=1

(pk

nn + tknnn

)Xk

nnn −I∑

n=1

J∑

i=1

(p

kf

in + tkf

inn

)X

kf

inn −I∑

n=1

J∑

i=1

(p

kf+1nn + t

kf+1nin

)X

kf+1nin .

In the above function, by substituting Xkf

inn = Xkf+1nin = Z

kf

in , and Xknnn =

∑Ii=1 X

kf

inn =∑I

i=1 Zkf

in , ∀ 1 ≤k ≤ K, k 6= kf , kf + 1, we will have:

I∑n=1

[rn

I∑

i=1

Zkf

in

]−

K∑

k=1k 6=kf ,kf+1

I∑n=1

[(pk

nn + tknnn

) I∑

i=1

Zkf

in

]

−I∑

n=1

I∑

i=1

[(p

kf

in + tkf

inn

)Z

kf

in

]−

I∑n=1

I∑

i=1

[(p

kf+1nn + t

kf+1nin

)Z

kf

in

](42)

=I∑

n=1

I∑

i=1

rnZkf

in −I∑

n=1

I∑

i=1

[( K∑

k=1k 6=kf ,kf+1

(pk

nn + tknnn

))Z

kf

in

]

−I∑

n=1

I∑

i=1

[(p

kf

in + tkf

inn

)Z

kf

in

]−

I∑n=1

I∑

i=1

[((p

kf+1nn + t

kf+1nin

))Z

kf

in

]

=I∑

n=1

I∑

i=1

rn −[( K∑

k=1k 6=kf

pknn

)+ p

kf

in

]−

[( K∑

k=1k 6=kf ,kf+1

tknnn

)+ t

kf

inn + tkf+1nin

]

Zkf

in .

Let

mP,in(kf ) = rn −[( K∑

k=1k 6=kf

pknn

)+ p

kf

in

]−

[( K∑

k=1k 6=kf ,kf+1

tknnn

)+ t

kf

inn + tkf+1nin

],

then the objective function can be written as∑I

n=1

∑Ii=1 mP,in(kf )Zkf

in , which is the objective function inLemma 2.

If kf = 1, objective function (1) can be written as

I∑

i=1

I∑n=1

riXkf

inn −K∑

k=1k 6=kf ,kf+1

I∑n=1

(pk

nn + tknnn

)Xk

nnn −I∑

n=1

J∑

i=1

(p

kf

in + tkf

inn

)X

kf

inn −I∑

n=1

J∑

i=1

(p

kf+1nn + t

kf+1nin

)X

kf+1nin .

Substituting Xkf

inn = Xkf+1nin = Z

kf

in , and Xknnn =

∑Ii=1 X

kf

inn =∑I

i=1 Zkf

in , ∀ 1 ≤ k ≤ K, k 6= kf , kf + 1 inthe above function, we get (42). All the argument for kf 6= 1 case then follows. This completes the proof forLemma 2.

17

Using Lemma 2, P1(A,B1full(kf ),D) can be simplified to P1(mP(kf ),D), where

V ∗(mP(kf ),D) = maxµ

{ED,Q(µ)

[$(mP(kf ),D,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

}.

Before we present the proof for Theorem 1, we first need to prove the following lemma. Note that this lemmaholds when Assumptions 1 is relaxed.

Lemma 3. CASE 1 (downstream variability): Suppose a supply chain has variability only at stage kv (0 ≤kv ≤ K − 2) and logistics flexibility only at stage kf (kv + 2 ≤ kf ≤ K). Let µ =

(µk

i

)be the optimal

capacity matrix for the supply chain. Then the optimal capacity matrix µ satisfies the following:

(a) the optimal capacity configuration is the same for all deterministic stages downstream from the flexiblestage (i.e., there exist values qdwn

i such that µ1i = µ2

i = · · · = µkv−1i = µkv+1

i = · · · = µkf−1i = qdwn

i , i =1, 2, . . . , I);

(b) the optimal capacity configuration is the same for all stages upstream of the flexible stage (i.e., thereexist values qup

i such that µkf

i = µkf+1i = · · · = µK

i = qupi , i = 1, 2, . . . , I);

(c) qdwni ≤ di for 1 ≤ kv ≤ K − 2, i = 1, 2, . . . , I;

(d)∑I

i=1 qupi ≤ ∑I

i=1 qdwni .

CASE 2 (upstream variability): Suppose a supply chain has variability only at stage kv (2 ≤ kv ≤ K) andlogistics flexibility only at stage kf , which is located downstream of stage kv (i.e., 1 ≤ kf ≤ kv − 1). Let

µ =(µk

i

)be the optimal capacity matrix for the supply chain. Then for 2 ≤ kf ≤ kv − 1, the optimal

capacity matrix µ satisfies

(a) the optimal capacity configuration is the same for all stages downstream of the flexible stage ( i.e., thereexist values qdwn

i such that µ1i = µ2

i = · · · = µkf−1i = qdwn

i , i = 1, 2, . . . , I);

(b) the optimal capacity configuration is the same for all deterministic stages upstream of the flexible stage(i.e., there exist values qup

i such that µkf

i = µkf+1i = · · · = µkv−1

i = µkv+1i = · · · = µK

i = qupi , i =

1, 2, . . . , I);

(c) qdwni ≤ di, for i = 1, 2, . . . , I;

(d)∑I

i=1 qupi ≥ ∑I

i=1 qdwni .

For kf = 1, the optimal capacity matrix µ satisfies:

(a) the optimal capacity configuration is the same for all stages upstream of the flexible stage, except forstage kv, (i.e., there exist values qup

i such that µ1i = µ2

i = · · · = µkv−1i = µkv+1

i = · · · = µKi = qup

i , i =1, 2, . . . , I);

(b)∑I

i=1 qupi ≥ ∑I

i=1 di.

PROOF OF LEMMA 3 – Flexibility Upstream of Variability:

We prove Lemma 3 by the following two cases, namely, 1 ≤ kv ≤ K − 2 and kv = 0.

CASE 1: 1 ≤ kv ≤ K − 2

In order to prove the optimal capacity investment matrix µ satisfies conditions (a) to (d), we show that anyµ that does not satisfy any of the four conditions cannot be optimal, i.e., there exists a matrix µ̃, whichsatisfies the condition, and achieves higher total expected profit than µ.

18

Since the only variability is at stage kv (1 ≤ kv ≤ K − 2), for all k’s such that k 6= kv, we set Qki (µk

i ) = µki ,

as described in the paragraph following equation (6). Therefore, we have

Q(µ) =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

µ11 · · · µkv−1

1 Qkv1 µkv+1

1 · · · µK1

.... . .

......

.... . .

...µ1

I · · · µkv−1I Qkv

I µkv+1I · · · µK

I

,

where we consider q as the matrix of the realization of capacities as follows:

q =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

µ11 · · · µkv−1

1 qkv1 µkv+1

1 · · · µK1

.... . .

......

.... . .

...µ1

I · · · µkv−1I qkv

I µkv+1I · · · µK

I

.

And since demand is deterministic,

D = (D1, D2, . . . , DI) = (d1, d2, . . . , dI) = d.

For full logistics flexibility configuration at stage kf , where kv + 2 ≤ kf ≤ K, by Lemma 1, the optimalcapacity investment µ is the solution to P1(mL(kf ),d):

V ∗(mL(kf ),d) = maxµ

{V (mL(kf ),d,Q(µ))

}

= maxµ

{EQ(µ)

[$(mL(kf ),d,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

},

where $(mL(kf ),d,Q(µ)) is calculated by solving P2(mL(kf ),d,q) and taking expectation over Q(µ):

π(mL(kf ),d,q) = maxY

kfij

{ I∑

i=1

I∑

j=1

[mL,ij(kf ) · Y kf

ij

]}(43)

subject to:I∑

j=1

Ykf

ij ≤ minkf≤k≤K

{µk

i

}i = 1, 2, . . . , I, (44)

I∑

j=1

Ykf

ji ≤ min1≤k≤kf−1,k 6=kv

{µk

i , di, qkvi

}i = 1, 2, . . . , I, (45)

where

mL,ij(kf ) = rj −( kf−1∑

k=1

pkjj +

K∑

k=kf

pkij

)−

[( kf−1∑

k=1

tkjjj

)+ t

kf

ijj +K∑

k=kf+1

tkiij

].

CASE 1 – Part (a) and Part (b):

Assume that µ does not satisfy condition (a) or (b). Let, for i = 1, 2, . . . , I,

qdwnmin,i = min

1≤k≤kf−1,k 6=kv

{µki } and qup

min,i = minkf≤k≤K

{µki }

Now consider the capacity matrix µ̃ as:

µ̃ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

qdwnmin,1 · · · qdwn

min,1 µkv1 qdwn

min,1 · · · qdwnmin,1 qup

min,1 · · · qupmin,1

.... . .

......

.... . .

......

. . ....

qdwnmin,I · · · qdwn

min,I µkv

I qdwnmin,I · · · qdwn

min,I qupmin,I · · · qup

min,I

.

19

If µ does not satisfy condition (a), then

µki > qdwn

min,i for some i (1 ≤ i ≤ I) and some k (1 ≤ k ≤ kf − 1, k 6= kv).

If µ does not satisfy condition (b), then

µki > qup

min,i for some i (1 ≤ i ≤ I) and some k (kf ≤ k ≤ K).

In either case we have

K∑

k=1

I∑

i=1

cki µ̃k

i <

K∑

k=1

I∑

i=1

cki µk

i .

which implies that capacity matrix µ̃ has a lower capacity investment than that of µ. On the other hand,since

min1≤k≤kf−1,k 6=kv

{µ̃k

i

}= qdwn

min,i = min1≤k≤kf−1,k 6=kv

{µki } i = 1, 2, . . . , I,

and

minkf≤k≤K

{µ̃k

i

}= qup

min,i = minkf≤k≤K

{µki } i = 1, 2, . . . , I,

then, the feasible region for problem P2 (i.e., constraints (44) and (45)) are the same for capacity matrices µand µ̃. On the other hand, since the objective function of problem P2(mL(kf ),d,q), i.e., (43), is independentof capacity matrices; hence, we have

$(mL(kf ),d,Q(µ̃)) = $(mL(kf ),d,Q(µ)) for all realization of Q.

Therefore, since µ̃ has a lower capacity investment, it achieves higher expected net profit than µ, and henceµ cannot be optimal solution to P1(mL(kf ),d). Thus, the optimal capacity matrix µ must follow bothcondition (a) and (b).

CASE 1 – Part (c):

To prove part (c), suppose

µ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

qdwn1 · · · qdwn

1 µkv1 qdwn

1 · · · qdwn1 qup

1 · · · qup1

.... . .

......

.... . .

......

. . ....

qdwnI · · · qdwn

I µkv

I qdwnI · · · qdwn

I qupI · · · qup

I

satisfies condition (a) and (b), but does not satisfy (c), i.e., qdwni′ > di′ for some i′ (1 ≤ i′ ≤ I). Let

µ̃ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

qdwn1 · · · qdwn

1 µkv1 qdwn

1 · · · qdwn1 qup

1 · · · qup1

.... . .

......

.... . .

......

. . ....

di′ · · · di′ µkv

i′ di′ · · · di′ qupi′ · · · qup

i′...

. . ....

......

. . ....

.... . .

...qdwn

I · · · qdwnI µkv

I qdwnI · · · qdwn

I qupI · · · qup

I

,

that is, substituting all qdwni′ in µ with di′ . Since qdwn

i′ > di′ for i′ (1 ≤ i′ ≤ I), we have

K∑

k=1

I∑

i=1

cki µ̃k

i <

K∑

k=1

I∑

i=1

cki µk

i .

20

On the other hand, the only difference between capacity matrices µ and µ̃ is in plant i′ at stages downstream ofkf , except for stage kv, hence the righthand-side of constraint (45) for capacity matrices µ and µ̃, respectively,for i′ is

min1≤k≤kf−1,k 6=kv

{µk

i′ , di′ , qkv

i′}

= min{qdwn

i′ , di′ , qkvi

}= min

{di′ , q

kvi

},

min1≤k≤kf−1,k 6=kv

{µ̃k

i′ , di′ , qkv

i′}

= min{di′ , di′ , q

kv

i′}

= min{di′ , q

kv

i′}.

Therefore, the feasible region, i.e., constraints (44) and (45), are the same for capacity matrices µ and µ̃. So,similar to Cases (a) and (b), we have

$(mL(kf ),d,Q(µ̃)) = $(mL(kf ),d,Q(µ)) for all realization of Q.

Therefore, µ̃ achieves higher expected net profit than µ, and hence µ cannot be optimal solution to P1(mL(kf ),d).Thus, the optimal capacity matrix µ must follow condition (c).

CASE 1 – Part (d):

Let

µ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

qdwn1 · · · qdwn

1 µkv1 qdwn

1 · · · qdwn1 qup

1 · · · qup1

.... . .

......

.... . .

......

. . ....

qdwnI · · · qdwn

I µkv

I qdwnI · · · qdwn

I qupI · · · qup

I

µ̃ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

qdwn1 · · · qdwn

1 µkv1 qdwn

1 · · · qdwn1 qdwn

1 · · · qdwn1

.... . .

......

.... . .

......

. . ....

qdwnI · · · qdwn

I µkv

I qdwnI · · · qdwn

I qdwnI · · · qdwn

I

both of which satisfy condition (a), (b) and (c). Suppose∑I

i=1 qupi >

∑Ii=1 qdwn

i , i.e. µ does not satisfycondition (d). It is clear that

K∑

k=1

I∑

i=1

cki µk

i −K∑

k=1

I∑

i=1

cki µ̃k

i =K∑

k=kf

I∑

i=1

cki (qup

i − qdwni ) > 0, (46)

since cki = ck

j , i, j = 1, 2, . . . , I, i 6= j, and k = 1, 2, . . . , K. Thus, capacity matrix µ̃ has a lower capacityinvestment cost than that of µ.

For capacity investment vector µ̃, $(mL(kf ),d,Q(µ̃)) is calculated by solving P2(mL(kf ),d,q) and takingexpectation over Q(µ̃):

π(mL(kf ),d,q) = maxY

kfij

{ I∑

i=1

I∑

j=1

[mL,ij(kf ) · Y kf

ij

]}

subject to:I∑

j=1

Ykf

ij ≤ qdwni i = 1, 2, . . . , I, (47)

I∑

j=1

Ykf

ji ≤ min{qdwn

i , di, qkvi

}= min

{qdwn

i , qkvi

}(since (c) holds) i = 1, 2, . . . , I. (48)

21

On the other hand, since according to Assumption 3,

mL,ii(kf ) > mL,ij(kf ) and mL,ii(kf ) > mL,ji(kf ) ∀ j 6= i, i = 1, 2, . . . , I,

the most profitable way of allocating flows (Y kf

ij ’s) is to use dedicated ones. Since by comparing right-hand-side of (47) and (48) we have qdwn

i ≥ min{qdwn

i , qkvi

}, we get the maximum profit for µ̃ by setting

Yii = min{qdwn

i , qkvi

}and Y

kf

ij = 0 ∀ i 6= j:

π(mL(kf ),d,q) =I∑

i=1

[mL,ii(kf ) ·min

{qdwn

i , qkvi

}].

On the other hand, for capacity investment vector µ, $(mL(kf ),d,Q(µ)) is calculated by solving P2(mL(kf ),d,q)and taking expectation over Q(µ):

π(mL(kf ),d,q) = maxY

kfij

{ I∑

i=1

I∑

j=1

[mL,ij(kf ) · Y kf

ij

]}

subject to:I∑

j=1

Ykf

ij ≤ qupi i = 1, 2, . . . , I

I∑

j=1

Ykf

ji ≤ min{

qdwni , qkv

i

}i = 1, 2, . . . , I

Since∑I

i=1 qupi >

∑Ii=1 qdwn

i , we have∑I

i=1 qupi >

∑Ii=1 min

{qdwn

i , qkvi

}for any q, hence

I∑

i=1

I∑

j=1

mL,ij(kf ) · Y kf

ij =I∑

i=1

I∑

j=1

mL,ji(kf ) · Y kf

ji

≤I∑

i=1

[mL,ii(kf )

I∑

j=1

Ykf

ji

]

≤I∑

i=1

[mL,ii(kf ) ·min(qdwn

i , qkvi )

].

Therefore

$(mL(kf ),d,Q(µ)) ≤ $(mL(kf ),d,Q(µ̃)) for all realization of Q (49)

By (46) and (49),

V (mL(kf ),d,Q(µ)) < V (mL(kf ),d,Q(µ̃)),

and hence µ cannot be optimal solution to P1(mL(kf ),d). Thus, the optimal capacity matrix µ must followcondition (d).

CASE 2: kv = 0

The proof for the case with kv = 0 is similar to that for kv 6= 0, except for the following minor modifications,given that now demand is the only source of variability and capacity is deterministic:

(1) Q(µ) = q = µ;

(2) Since D is stochastic and Q(µ) is deterministic, to calculate V ∗, the expectation is taken over Dinstead of Q(µ);

22

(3) Stage kv with uncertain capacity is missing in capacity matrices Q(µ), q, µ and µ̃, and in constraintssuch as (45);

(4) In part (d) constraint (48), instead of qkvi , we have di.

PROOF OF LEMMA 3 – Flexibility Downstream of Variability:

CASE 1: 2 ≤ kf ≤ kv − 1

In order to prove the optimal capacity vector µ satisfies conditions (a) to (d), similar to the proof of theprevious case, we show that for any capacity matrix µ which does not satisfy any of the four conditions,there exists a capacity matrix µ̃, which satisfies the condition and achieves higher total expected profit thanµ; therefore, µ cannot be optimal.

Since the only variability is at stage kv (2 ≤ kv ≤ K), we have

Q(µ) =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

µ11 · · · µkv−1

1 Qkv1 µkv+1

1 · · · µK1

.... . .

......

.... . .

...µ1

I · · · µkv−1I Qkv

I µkv+1I · · · µK

I

,

and the matrix of the realization of the above capacity matrix is:

q =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

µ11 · · · µkv−1

1 qkv1 µkv+1

1 · · · µK1

.... . .

......

.... . .

...µ1

I · · · µkv−1I qkv

I µkv+1I · · · µK

I

.

Since D is deterministic, we have

D = (D1, D2, . . . , DI) = (d1, d2, . . . , dI) = d.

For supply chain with variability at stage kv and full logistics flexibility at stage kf , by Lemma 1, the optimalcapacity investment µ is solution to P1(mL(kf ),d):

V ∗(mL(kf ),d) = maxµ

{V (mL(kf ),d,Q(µ))

}

= maxµ

{EQ(µ)

[$(mL(kf ),d,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

},

where $(mL(kf ),d,Q(µ)) is calculated by solving P2(mL(kf ),d,q) and taking expectation over Q(µ):

π(mL(kf ),d,q) = maxY

kfij

{ I∑

i=1

I∑

j=1

[mL,ij(kf ) · Y kf

ij

]}

subject to:I∑

j=1

Ykf

ij ≤ minkf≤k≤K,k 6=kv

{µk

i , qkvi

}i = 1, 2, . . . , I, (50)

I∑

j=1

Ykf

ji ≤ min1≤k≤kf−1

{µk

i , di

}i = 1, 2, . . . , I. (51)

23

CASE 1 – Part (a) and Part (b):

Assume that µ does not satisfy condition (a) or (b). Let, for i = 1, 2, . . . , I,

qdwnmin,i = min

1≤k≤kf−1{µk

i } and qupmin,i = min

kf≤k≤K,k 6=kv

{µki } i = 1, 2, . . . , I,

and

µ̃ =

stage 1 · · · stage (kf−1) stage kf· · · stage (kv−1) stage kv stage (kv+1) · · · stage K

qdwnmin,1 · · · qdwn

min,1 qupmin,1 · · · qup

min,1 µkv1 qup

min,1 · · · qupmin,1

.... . .

......

. . ....

......

. . ....

qdwnmin,I · · · qdwn

min,I qupmin,I · · · qup

min,I µkv

I qupmin,I · · · qup

min,I

.

If µ does not satisfy condition (a), then

µki > qdwn

min,i for some i (1 ≤ i ≤ I) and some k (1 ≤ k ≤ kf − 1).

If µ does not satisfy condition (b), then

µki > qup

min,i for some i (1 ≤ i ≤ I) and some k (kf ≤ k ≤ K, k 6= kv).

In either case we haveK∑

k=1

I∑

i=1

cki µ̃k

i <

K∑

k=1

I∑

i=1

cki µk

i .

which implies that µ̃ has a lower capacity investment cost than µ. On the other hand, since

min1≤k≤kf−1

{µ̃k

i

}= qdwn

min,i = min1≤k≤kf−1

{µki } i = 1, 2, . . . , I,

and

minkf≤k≤K,k 6=kv

{µ̃k

i

}= qup

min,i = minkf≤k≤K,k 6=kv

{µki } i = 1, 2, . . . , I,

constraints (50) and (51) result in the same feasible region for both capacity matrices µ and µ̃. So we have

$(mL(kf ),d,Q(µ̃)) = $(mL(kf ),d,Q(µ)) for all realization of Q.

Therefore,

V (mL(kf ),d,Q(µ̃)) > V (mL(kf ),d,Q(µ)),

and µ̃ achieves higher expected net profit than µ, and hence µ cannot be optimal solution to P1(mL(kf ),d).Thus, the optimal capacity matrix µ must follow both condition (a) and (b).

CASE 1 – Part (c):

To prove (c), suppose

µ =

stage 1 · · · stage (kf−1) stage kf· · · stage (kv−1) stage kv stage (kv+1) · · · stage K

qdwn1 · · · qdwn

1 qup1 · · · qup

1 µkv1 qup

1 · · · qup1

.... . .

......

. . ....

......

. . ....

qdwnI · · · qdwn

I qupI · · · qup

I µkv

I qupI · · · qup

I

,

which satisfies conditions (a) and (b), but not (c), i.e., qdwni′ > di′ for some i′ (1 ≤ i′ ≤ I). Now consider

capacity matrix µ̃ as follows:

µ̃ =

stage 1 · · · stage (kf−1) stage kf· · · stage (kv−1) stage kv stage (kv+1) · · · stage K

qdwn1 · · · qdwn

1 qup1 · · · qup

1 µkv1 qup

1 · · · qup1

.... . .

......

. . ....

......

. . ....

di′ · · · di′ qupi′ · · · qup

i′ µkv

i′ qupi′ · · · qup

i′...

. . ....

.... . .

......

.... . .

...qdwn

I · · · qdwnI qup

I · · · qupI µkv

I qupI · · · qup

I

,

24

that is, substituting all qdwni′ in µ with di′ and get µ̃. Since qdwn

i′ > di′ for i′ (1 ≤ i′ ≤ I), we have

K∑

k=1

I∑

i=1

cki µ̃k

i <

K∑

k=1

I∑

i=1

cki µk

i .

On the other hand, the only difference between capacity matrices µ and µ̃ is in plant i′ at stages downstreamof kf ; hence, the right-hand-side of constraint (51) for capacity matrix µ and µ̃, respectively, for i′ is

min1≤k≤kf−1

{µk

i′ , di′}

= min{qdwn

i′ , di′}

= di′ ,

min1≤k≤kf−1

{µ̃k

i′ , di′}

= min{di′ , di′

}= di′ .

Hence constraints (50) and (51) are the same for capacity matrices µ and µ̃. So we have

$(mL(kf ),d,Q(µ̃)) = $(mL(kf ),d,Q(µ)) for all realization of Q.

Therefore,

V (mL(kf ),d,Q(µ̃)) > V (mL(kf ),d,Q(µ)),

which implies that µ cannot be optimal solution to P1(mL(kf ),d). Thus, the optimal capacity matrix µmust follow condition (c).

CASE 1 – Part (d):

For this part, we establish a new unit margin matrix mmin based on the smallest margin value of the originalunit margin matrix m. We first prove that a capacity matrix which satisfies condition (d) achieves higherexpected profit than a capacity matrix that does not under this new margin matrix. And then we show thatthis is also true under original unit margin matrix m.

For proof in this part, we model the uncertain capacity of plant i at stage kv as Qkvi (µkv

i ) = µkvi + εkv

i ,where µkv

i is capacity investment level decided by the firm and εkvi is an additive random term to account

for variability in the capacity of plant i at stage kv. Let ε be the corresponding random vector. Let ε beobservation of ε.

Assume that capacity matrix µ is:

µ =

stage 1 · · · stage (kf−1) stage kf· · · stage (kv−1) stage kv stage (kv+1) · · · stage K

qdwn1 · · · qdwn

1 qup1 · · · qup

1 µkv1 qup

1 · · · qup1

.... . .

......

. . ....

......

. . ....

qdwnI · · · qdwn

I qupI · · · qup

I µkv

I qupI · · · qup

I

,

which satisfies conditions (a) to (c), but not (d). That is,∑I

i=1 qupi <

∑Ii=1 qdwn

i . Now consider anothercapacity matrix

µ̃ =

stage 1 · · · stage (kf−1) stage kf· · · stage (kv−1) stage kv stage (kv+1) · · · stage K

qdwn1 · · · qdwn

1 µ̃1 · · · µ̃1 µkv1 µ̃1 · · · µ̃1

.... . .

......

. . ....

......

. . ....

qdwnI · · · qdwn

I µ̃I · · · µ̃I µkv

I µ̃I · · · µ̃I

that satisfies conditions (a) to (d), with∑I

i=1 µ̃i =∑I

i=1 qdwni . In the following paragraphs, we refer to

the change from µ to µ̃, i.e., increasing the sum of capacities at stages upstream of flexibility, as increasingcapacity from µ to µ̃.

Let mL,min = min1≤i,j≤I{mL,ij(kf )} and

mLmin =

mL,min · · · mL,min

.... . .

...mL,min · · · mL,min

.

25

Under the new unit margin matrix mLmin, for capacity investment µ,

π(mLmin,d,q) = maxY

kfij

{ I∑

i=1

I∑

j=1

(mL,min · Y kf

ij )}

= maxY

kfij

{mL,min ·

I∑

i=1

I∑

j=1

Ykf

ij

}.

Since µ follows condition (b), minkf≤k≤K,k 6=kv

{µk

i , qkvi

}= min

{qup

i , qkvi

}, from (50) we have

I∑

i=1

I∑

j=1

Ykf

ij ≤I∑

i=1

min{qup

i , qkvi

}. (52)

Similarly, since µ follows condition (a) and (c), min1≤k≤kf−1

{µk

i , di

}= min

{qdwn

i , di

}= qdwn

i , from (51)we have

I∑

i=1

I∑

j=1

Ykf

ji ≤I∑

i=1

qdwni . (53)

Since∑I

i=1 qupi <

∑Ii=1 qdwn

i , then∑I

i=1 min{qup

i , qkvi

}<

∑Ii=1 qdwn

i , combining with (52) and (53), themaximum value that

∑Ii=1

∑Ij=1 Y

kf

ij can achieve is∑I

i=1 min{qup

i , qkvi

}. Thus, the optimal solution to

problem P2(mmin,d,q) is

π(mLmin,d,q) = maxY

kfij

{mL,min ·

I∑

i=1

I∑

j=1

Ykf

ij

}= mL,min ·

I∑

i=1

min{qup

i , qkvi

}.

Hence

V (mLmin,d,Q(µ)) = EQ(µ)

[$(mLmin,d,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

= mL,min · EQ(µ)

[ I∑

i=1

min{qup

i , Qkvi (µkv

i )}]−

K∑

k=1

I∑

i=1

cki µk

i

= mL,min ·I∑

i=1

EQ(µ)

[min

{qup

i , Qkvi (µkv

i )}]−

K∑

k=1

I∑

i=1

cki µk

i

= mL,min

I∑

i=1

(∫ qupi

0

qkvi f(qkv

i )dqkvi +

∫ ∞

qupi

qupi f(qkv

i )dqkvi

)

−kf−1∑

k=1

I∑

i=1

cki qdwn

i −I∑

i=1

ckvi µkv

i −K∑

k=kf ,k 6=kv

I∑

i=1

cki qup

i ,

where f(qkvi ) is probability density function (pdf) of Qkv

i (µkvi ).

Since Qkvi (µkv

i ) = µkvi + εkv

i.= r(εkv

i ), we have εkvi = Qkv

i (µkvi )− µkv

i.= s(Qkv

i (µkvi )). Let g(εkv

i ) be pdf of εkvi ,

then

g(εkvi ) = f(r(εkv

i ))

∣∣∣∣∣dr(εkv

i )dεkv

i

∣∣∣∣∣ = f(µkvi + εkv

i ),

and

f(qkvi ) = g(s(qkv

i ))

∣∣∣∣∣ds(qkv

i )dqkv

i

∣∣∣∣∣ = g(qkvi − µkv

i ).

So 1

1Note that for the sake of calculation, we assume the integration is from −∞ to ∞. The approach can be easilyrevised for cases where the random capacity is bounded from below and from above. For those cases one can set theprobability of the random capacity being less than the lower bound and larger than the upper bound to zero in ourcalculation.

26

V (·) = mL,min

I∑

i=1

(∫ qupi −µkv

i

−∞(µkv

i + εkvi )f(µkv

i + εkvi )dεkv

i +∫ ∞

qupi −µkv

i

qupi f(µkv

i + εkvi )dεkv

i

)

−kf−1∑

k=1

I∑

i=1

cki qdwn

i −I∑

i=1

ckvi µkv

i −K∑

k=kf ,k 6=kv

I∑

i=1

cki qup

i

= mL,min

I∑

i=1

(∫ qupi −µkv

i

−∞(µkv

i + εkvi )g(εkv

i )dεkvi +

∫ ∞

qupi −µkv

i

qupi g(εkv

i )dεkvi

)

−kf−1∑

k=1

I∑

i=1

cki qdwn

i −I∑

i=1

ckvi µkv

i −K∑

k=kf ,k 6=kv

I∑

i=1

cki qup

i .

The optimal µkvi

∗satisfies

∂V

∂µkvi

∣∣∣∣µkv

i =µkvi

∗=

[mL,min

d(qupi − µkv

i )dµkv

i

(µkv

i + qupi − µkv

i

)g(εkv

i ) + mL,min

∫ qupi −µkv

i

−∞

d(µkvi + εkv

i )dµkv

i

g(εkvi )dεkv

i

−mL,mind(qup

i − µkvi )

dµkvi

qupi g(εkv

i )− ckvi

]∣∣∣∣µkv

i =µkvi

∗

= mL,min

∫ qupi −µkv

i

∗

−∞g(εkv

i )dεkvi − ckv

i

= 0. (54)

On the other hand,

∂V

∂qupi

= mL,mind(qup

i − µkvi )

dqupi

(µkv

i + qupi − µkv

i

)g(εkv

i )

−mL,mind(qup

i − µkvi )

dqupi

qupi g(εkv

i ) + mL,min

∫ ∞

qupi −µkv

i

g(εkvi )dεkv

i −K∑

k=kf ,k 6=kv

cki

= mL,min

∫ ∞

qupi −µkv

i

g(εkvi )dεkv

i −K∑

k=kf ,k 6=kv

cki .

Therefore, by envelope theory,

∂V ∗

∂qupi

=∂V

∂qupi

∣∣∣∣µkv

i =µkvi

∗

= mL,min

∫ ∞

qupi −µkv

i

∗g(εkv

i )dεkvi −

K∑

k=kf ,k 6=kv

cki ,

but by (54) we know that

mL,min

∫ ∞

qupi −µkv

i

∗g(εkv

i )dεkvi = mL,min −mL,min

∫ qupi −µkv

i

∗

−∞g(εkv

i )dεkvi = mL,min − ckv

i ,

27

hence,

∂V ∗

∂qupi

= mL,min − ckvi −

K∑

k=kf ,k 6=kv

cki

= mL,min −K∑

k=kf

cki

> mL,min −K∑

k=1

cki

> 0,

since we assume the supply chain makes profit even through the least profitable arc.

Therefore, for a capacity matrix µ with∑I

i=1 qupi <

∑Ii=1 qdwn

i , profit can be improved by increasing qupi for

some i until the sum of capacities for the upstream and downstream stages equate. Hence

V (mLmin,d,Q(µ̃)) > V (mLmin,d,Q(µ)).

That is, under the profit margin matrix mLmin, increasing capacity from µ to µ̃ increases profit V (·).Now consider the original profit margin matrix mL. We know that the profit V (mL,d,Q(µ)) consists oftwo parts, revenue EQ(µ)

[$(mL,d,Q(µ))

], and cost

∑Kk=1

∑Ii=1 ck

i µki . Compared with the case under

margin matrix mLmin, increasing capacity from µ to µ̃ under profit margin matrix mL increases E[π(·)] evenmore because of higher profit margin. On the other hand, the extra cost of increasing capacity from µ toµ̃,

∑Kk=1

∑Ii=1 ck

i µ̃ki −

∑Kk=1

∑Ii=1 ck

i µki , is independent of the margin. Therefore, under the profit margin

matrix mL, increasing capacity from µ to µ̃ increases profit V (·) more than the case under profit marginmatrix mLmin , i.e.,

V (mL,d,Q(µ̃))− V (mL,d,Q(µ)) ≥ V (mLmin,d,Q(µ̃))− V (mLmin,d,Q(µ))> 0.

and so µ is not optimal solution to P1(mL(kf ),d).

CASE 2: kf = 1

The proof for this case is similar to the kf > 1 case, with the following minor modifications:

(1) Since kf = 1, there is no plant stages downstream of kf , i.e., stages 1, 2, . . . , kf − 1 are missing incapacity matrices Q(µ), q, µ and µ̃, and in constraints such as (51);

(2) Proof of Part (b) is similar to proof of Part (d) in Case 1, except that all qdwni are replaced by di.

This completes the proof of Lemma 3.

With respect to the statements in Lemma 3, we would like to add the following:

• Limits of kf for fixed kv:

– Lemma 3 CASE 1 is for flexibility upstream of variability (kv+1 ≤ kf ≤ K). As stated in Theorem1 (1), when flexibility is upstream of variability, the optimal location of flexibility is at the stageclosest to variability, i.e., kf = kv + 1, and all the other locations for flexibility are suboptimal.Lemma 3 describes property of structures that are suboptimal, i.e., kf : kv + 2 ≤ kf ≤ K.

– Lemma 3 CASE 2 is for flexibility downstream of variability (1 ≤ kf ≤ kv). As stated in Theorem1 (2), when flexibility is downstream of variability, the optimal location of flexibility is at thestage closest to variability, i.e., kf = kv, and all the other locations for flexibility are suboptimal.Lemma 3 describes property of structures that are suboptimal, i.e., kf : 1 ≤ kf ≤ kv − 1.

28

• The two optimal structures, namely kf = kv + 1 for flexibility upstream of variability, and kf = kv forflexibility downstream of variability, are not included in the Lemma, because of the following:

– The proof of Theorem 1 does not need property of optimal structures, namely kf = kv +1 for thecase with flexibility upstream of variability, and kf = kv for the case with flexibility downstreamof variability.

– The optimal structures, namely kf = kv + 1 for the case with flexibility upstream of variability,and kf = kv for the case with flexibility downstream of variability, have the similar propertystated in Lemma 3. But including these structures in the lemma, while not needed, will lead tomore subcases and thus complicate the statement of the lemma.

• Limits of kv:

– Limits of kv is such that the corresponding kf does not exceed limits of the system, i.e., 1 ≤ kf ≤K. For CASE 1, kv + 2 ≤ kf ≤ K requires kv ≤ K − 2. And kv ≥ 0 is to make sure kv does notexceed the supply chain. For CASE 2, 1 ≤ kf ≤ kv − 1 requires kv ≥ 2. And kv ≤ K is to makesure kv does not exceed the supply chain.

PROOF OF THEOREM 1, PART (1):

In order to prove

V ∗(mL(kf ),D) ≥ V ∗(mL(kf + 1),D) ∀ kf : kv + 1 ≤ kf ≤ K − 1,

by sample path method, it suffices to show that for each possible optimal µ for system with flexibility atstage kf + 1, there exists at least one corresponding µ̃ for system with flexibility at stage kf , such that

V (mL(kf ),D,Q(µ̃)) ≥ V (mL(kf + 1),D,Q(µ)),

which in turn requires

π(mL(kf ),d,q)−K∑

k=1

I∑

i=1

cki µ̃k

i ≥ π(mL(kf + 1),d,q)−K∑

k=1

I∑

i=1

cki µk

i for all d,q. (55)

By “possible optimal” capacity we refer to capacity matrices that satisfy conditions in Lemma 3.

From Assumption 1, we have

mL,ij(k) = mL,ij(k + 1), i, j = 1, 2, . . . , I, ∀ k : 1 ≤ k < K.

Therefore, to simplify notation we use mL,ij = mL,ij(k) = mL,ij(k + 1), and hence we have

mL(k) = mL ∀ k : 1 ≤ k ≤ K.

By Assumption 3, we write

cki = ck, i = 1, 2, . . . , I, ∀ k : 1 ≤ k ≤ K.

CASE 1: 1 ≤ kv ≤ K − 2

In this case, the capacity of plants at stage kv are sources of variability. By Lemma 3, in order for µ to beoptimal when flexibility is located at stage kf + 1, µ must have the form

µ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage kf stage (kf+1) · · · stage K

qdwn1 · · · qdwn

1 µkv1 qdwn

1 · · · qdwn1 qup

1 · · · qup1

.... . .

......

.... . .

......

. . ....

qdwnI · · · qdwn

I µkv

I qdwnI · · · qdwn

I qupI · · · qup

I

(56)

29

with

qdwni ≤ di for i = 1, 2, . . . , I, and

I∑

i=1

qupi ≤

I∑

i=1

qdwni .

We claim that for all such µ (i.e., certain qdwni , qup

i and µkvi for all i) in the structure with flexibility at stage

kf + 1, there exists a capacity matrix µ̃ (with element µ̃ki ) in the structure with flexibility at stage kf , where

µ̃ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

qdwn1 · · · qdwn

1 µkv1 qdwn

1 · · · qdwn1 qup

1 · · · qup1

.... . .

......

.... . .

......

. . ....

qdwnI · · · qdwn

I µkv

I qdwnI · · · qdwn

I qupI · · · qup

I

, (57)

and µ̃ results in a higher profit for any realization of capacity Qkvi for all i.

When flexibility is at stage kf , for any realization of capacity matrix Q(µ̃), we have from Lemma 1,

π(mL,d,q) = maxY

kfij

{ I∑

i=1

I∑

j=1

[mL,ij · Y kf

ij

]}(58)

subject to:I∑

j=1

Ykf

ij ≤ minkf≤k≤K

{µ̃k

i

}i = 1, 2, . . . , I, (59)

I∑

j=1

Ykf

ji ≤ min1≤k≤kf−1,k 6=kv

{µ̃k

i , di, q̃kvi

}i = 1, 2, . . . , I, (60)

Using (57) for i = 1, 2, . . . , I, we have:

µ̃ki =

{qup

i : kf ≤ k ≤ K,qdown

i : 1 ≤ k ≤ kf − 1, k 6= kv,

then, constraints (59) and (60) becomes

I∑

j=1

Ykf

ij ≤ qupi i = 1, 2, . . . , I, (61)

I∑

j=1

Ykf

ji ≤

min{di, q

kvi

}: if kv = 1, kf = 2,

min{qdwn

i , di, qkvi

}: otherwise,

i = 1, 2, . . . , I. (62)

When flexibility is at stage kf + 1, for any realization of capacity matrix Q(µ), we have

π(mL,d,q) = maxY

kfij

{ I∑

i=1

I∑

j=1

[mL,ij · Y kf

ij

]}(63)

subject to:I∑

j=1

Ykf

ij ≤ minkf+1≤k≤K

{µk

i

}i = 1, 2, . . . , I, (64)

I∑

j=1

Ykf

ji ≤ min1≤k≤kf ,k 6=kv

{µk

i , di, qkvi

}i = 1, 2, . . . , I. (65)

30

Using (56), for i = 1, 2, . . . , I, we have:

µki =

{qup

i : kf + 1 ≤ k ≤ K,qdown

i : 1 ≤ k ≤ kf , k 6= kv,

then, constraints (64) and (65) becomes

I∑

j=1

Ykf

ij ≤ qupi i = 1, 2, . . . , I, (66)

I∑

j=1

Ykf

ji ≤ min{di, q

kvi , qdwn

i

}i = 1, 2, . . . , I. (67)

Compare constraints (61) and (62) with (66) and (67), the feasible region defined by constraints (61) and(62) is either larger than (when kv = 1, kf = 2) or equivalent to (when kv 6= 1, kf 6= 2) the feasible regiondefined by (66) and (67). Since the objective functions (58) and (63) have the same structure, therefore

$(mL,d,Q(µ̃)) ≥ $(mL,d,Q(µ)) for all realization of Q. (68)

With respect to the capacity investment cost,

K∑

k=1

I∑

i=1

cki µ̃k

i =kf−1∑

k=1,k 6=kv

I∑

i=1

cki qdwn

i +K∑

k=kf

I∑

i=1

cki qup

i +I∑

i=1

ckvi µkv

i ,

K∑

k=1

I∑

i=1

cki µk

i =kf∑

k=1,k 6=kv

I∑

i=1

cki qdwn

i +K∑

k=kf+1

I∑

i=1

cki qup

i +I∑

i=1

ckvi µkv

i .

And therefore

K∑

k=1

I∑

i=1

cki µ̃k

i −K∑

k=1

I∑

i=1

cki µk

i =I∑

i=1

ckf

i qupi −

I∑

i=1

ckf

i qdwni

= ckf ·( I∑

i=1

qupi −

I∑

i=1

qdwni

)≤ 0. (69)

Considering (68) and (69), it becomes clear that capacity matrix µ̃ has a lower capacity investment andresults in a higher profit than µ for any realization of Qkv

i . Therefore, we have proven (55) and hence

V ∗(mL(kf ),D) ≥ V ∗(mL(kf + 1),D) ∀ kf : kv + 1 ≤ kf ≤ K − 1. (70)

CASE 2: kv = 0

The proof for the kv = 0 case is similar to the kv 6= 0 case, except for the following minor modifications,given that now demand is the only source of variability and capacity is deterministic:

(1) Q(µ) = q = µ;

(2) Since D is stochastic and Q(µ) is deterministic, to calculate V ∗, the expectation is taken over Dinstead of Q(µ);

(3) Stage kv with uncertain capacity is missing in capacity matrices µ and µ̃. And qkvi is omitted in all

the constraints regarding capacities, since demand is the only source of variability.

PROOF OF THEOREM 1, PART (2):

In order to prove

V ∗(mL(kf ),D) ≥ V ∗(mL(kf − 1),D) ∀ kf : 2 ≤ kf ≤ kv,

31

by sample path method, it suffices to show that for each possible optimal µ for system with flexibility atstage kf − 1, there exists at least one corresponding µ̃, such that

V (mL(kf ),D,Q(µ̃)) ≥ V (mL(kf − 1),D,Q(µ)),

which in turn requires

π(mL(kf ),d,q)−K∑

k=1

I∑

i=1

cki µ̃k

i ≥ π(mL(kf − 1),d,q)−K∑

k=1

I∑

i=1

cki µk

i ) for all d,q. (71)

By “possible optimal” capacity we refer to capacity matrices that satisfy conditions in Lemma 3.From Assumption 1, we have

mL,ij(k) = mL,ij(k − 1), i, j = 1, 2, . . . , I, ∀ k : 1 < k ≤ K.

Therefore, to simplify notation we use mL,ij = mL,ij(k) = mL,ij(k − 1), and hence we have

mL(k) = mL ∀ k : 1 ≤ k ≤ K.

By Assumption 3, we write

cki = ck, i = 1, 2, · · · , I, ∀ k : 1 ≤ k ≤ K.

CASE 1: 2 < kf ≤ kv

In this case, capacity of plants at stage kv are sources of variability. By Lemma 3, in order for µ to be optimalwhen logistics flexibility is located at stage kf − 1, µ must have the form

µ =

stage 1 · · · stage (kf−2) stage (kf−1) · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

qdwn1 · · · qdwn

1 qup1 · · · qup

1 µkv1 qup

1 · · · qup1

.... . .

......

. . ....

......

. . ....

qdwnI · · · qdwn

I qupI · · · qup

I µkv

I qupI · · · qup

I

(72)

with

qdwni ≤ di i = 1, 2, . . . , I, and

I∑

i=1

qupi ≥

I∑

i=1

qdwni .

We claim that for all such µ in the structure with flexibility at stage kf − 1, there exists a capacity matrixµ̃ in the structure with flexibility at stage kf , where

µ̃ =

stage 1 · · · stage (kf−1) stage kf· · · stage (kv−1) stage kv stage (kv+1) · · · stage K

qdwn1 · · · qdwn

1 qup1 · · · qup

1 µkv1 qup

1 · · · qup1

.... . .

......

. . ....

......

. . ....

qdwnI · · · qdwn

I qupI · · · qup

I µkv

I qupI · · · qup

I

, (73)

and µ̃ results in a higher profit for any realization of capacity Qkvi for all i.

When flexibility is at stage kf , for any realization of capacity matrix Q(µ̃), we have

π(mL,d,q) = maxY

kfij

{ I∑

i=1

I∑

j=1

[mL,ij · Y kf

ij

]}

subject to:I∑

j=1

Ykf

ij ≤ minkf≤k≤K,k 6=kv

{µ̃k

i , qkvi

}i = 1, 2, . . . , I, (74)

I∑

j=1

Ykf

ji ≤ min1≤k≤kf−1

{µ̃k

i , di

}i = 1, 2, . . . , I. (75)

32

Based on (73), we have for i = 1, 2, . . . , I:

µ̃ki =

qupi : kf ≤ k ≤ K, k 6= kv,

qdowni : 1 ≤ k ≤ kf − 1,

µki : k = kv,

and constraints (74)and (75) become

I∑

j=1

Ykf

ij ≤

qkvi : if kv = kf = K

min{qup

i , qkvi

}: otherwise

i = 1, 2, . . . , I, (76)

I∑

j=1

Ykf

ji ≤ qdwni i = 1, 2, . . . , I. (77)

When flexibility is at stage kf − 1, for any realization of capacity matrix Q(µ), we have

π(mL,d,q) = maxy

{ I∑

i=1

I∑

j=1

[mL,ij · Y kf

ij

]}

subject to:I∑

j=1

Ykf

ij ≤ minkf−1≤k≤K,k 6=kv

{µk

i , qkvi

}i = 1, 2, . . . , I, (78)

I∑

j=1

Ykf

ji ≤ min1≤k≤kf−2

{µk

i , di

}i = 1, 2, . . . , I. (79)

Using (72) for i = 1, 2, . . . , I we have:

µki =

{qup

i : kf − 1 ≤ k ≤ K, k 6= kv,qdown

i : 1 ≤ k ≤ kf − 2,

therefore, constraints (78)and (79) become

I∑

j=1

Ykf

ij ≤ min{qup

i , qkvi

}i = 1, 2, . . . , I, (80)

I∑

j=1

Ykf

ji ≤ qdwni i = 1, 2, . . . , I. (81)

Comparing constraints (76) and (77) with (80) and (81), we see that the feasible region defined by constraints(76) and (77) is either larger than or equivalent to the feasible region defined by (80) and (81). Therefore

$(mL,d,Q(µ̃)) ≥ $(mL,d,Q(µ)) for all realization of Q. (82)

With respect to the capacity investment cost,

K∑

k=1

I∑

i=1

cki µ̃k

i =kf−1∑

k=1

I∑

i=1

cki qdwn

i +K∑

k=kf ,k 6=kv

I∑

i=1

cki qup

i +I∑

i=1

ckvi µkv

i ,

K∑

k=1

I∑

i=1

cki µk

i =kf−2∑

k=1

I∑

i=1

cki qdwn

i +K∑

k=kf−1,k 6=kv

I∑

i=1

cki qup

i +I∑

i=1

ckvi µkv

i .

33

And therefore

K∑

k=1

I∑

i=1

cki µ̃k

i −K∑

k=1

I∑

i=1

cki µk

i =I∑

i=1

ckf−1i qdwn

i −I∑

i=1

ckf−1i qup

i

= ckf−1 ·[ I∑

i=1

qdwni −

I∑

i=1

qupi

]

≤ 0. (83)

Considering (82) and (83), it becomes clear that capacity matrix µ̃ has a lower capacity investment andresults in a higher profit than µ for any realization of Qkv

i . Therefore, we have proven (71) and hence

V ∗(mL(kf ),D) ≥ V ∗(mL(kf − 1),D) ∀ kf : 2 < kf ≤ kv. (84)

CASE 2: kf = 2

Proof for this case is similar to the 2 < kf ≤ kv case. But now in order for µ to be optimal when flexibilityis at stage 1, by Lemma 3 Case(2): kf = 1, it must have the form

µ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

qup1 · · · qup

1 µkv1 qup

1 · · · qup1

.... . .

......

.... . .

...qup

I · · · qupI µkv

I qupI · · · qup

I

(85)

with

I∑

i=1

qupi ≥

I∑

i=1

di.

We claim that for all such µ, there exists a capacity matrix µ̃ in the structure with flexibility at stage 2,where

µ̃ =

stage 1 stage 2 · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

d1 qup1 · · · qup

1 µkv1 qup

1 · · · qup1

......

. . ....

......

. . ....

dI qupI · · · qup

I µkv

I qupI · · · qup

I

, (86)

and µ̃ results in a higher profit for any realization of capacity Qkvi for all i.

All arguments for the 2 < kf ≤ kv case follows by replacing qdwni with di.

Before we present the proof for Theorem 2, we first need to present Lemma 4. Note that Lemma 4 holdswhen Assumptions 1 is relaxed.

Lemma 4. Suppose a supply chain has variability only at stage kv (0 ≤ kv ≤ K) and process flexibility onlyat stage kf (1 ≤ kf ≤ K). Let µ =

(µk

i

)be the optimal capacity investment matrix for the supply chain.

Then the optimal capacity matrix µ satisfies the following:

(a) the optimal capacity configuration is the same for all stages except for the flexible and variable stage(i.e., there exist values qi such that qi = µk

i , ∀k 6= kf , kv, k = 1, 2, . . . , K, i = 1, 2, . . . , I);

(b) qi ≤ di, for 1 ≤ kv ≤ K, i = 1, 2, . . . , I;

(c) µkf

i ≤ qi, for i = 1, 2, . . . , I.

34

PROOF OF LEMMA 4:

We prove the lemma by proving the following two cases, namely, 1 ≤ kv ≤ K and kv = 0.

CASE 1: 1 ≤ kv ≤ K

In order to prove the optimal capacity matrix µ satisfies conditions (a) to (c), we show that any µ that doesnot satisfy any of the three conditions cannot be optimal, i.e., there exists a vector µ̃, which satisfies theseconditions, and achieves higher total expected profit than µ.

Since the only variability is at stage kv (1 ≤ kv ≤ K), for all k’s such that k 6= kv, we set Qki (µkv

i ) = µki .

Therefore, we have

Q(µ) =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

µ11 · · · µkv−1

1 Qkv1 µkv+1

1 · · · µK1

.... . .

......

.... . .

...µ1

I · · · µkv−1I Qkv

I µkv+1I · · · µK

I

,

q =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage K

µ11 · · · µkv−1

1 qkv1 µkv+1

1 · · · µK1

.... . .

......

.... . .

...µ1

I · · · µkv−1I qkv

I µkv+1I · · · µK

I

.

And since demand is deterministic,

D = (D1, D2, . . . , DI) = (d1, d2, . . . , dI) = d.

For full process flexibility configuration at stage kf , by Lemma 2, the optimal capacity matrix µ is thesolution to P1(mP (kf ),d):

V ∗(mP (kf ),d) = maxµ

{V (mP (kf ),d,Q(µ))

}

= maxµ

{EQ(µ)

[$(mP (kf ),d,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

},

where $(mP (kf ),d,Q(µ)) is calculated by solving P2(mP (kf ),d,q) and taking expectation over Q(µ):

π(mP (kf ),d,q) = maxZ

kfij

{ I∑

i=1

I∑

j=1

[mP,ij(kf ) · Zkf

ij

]}(87)

subject to:I∑

j=1

Zkf

ij ≤ min1≤k≤Kk 6=kf ,kv

{µk

i , qkvi , di

}i = 1, 2, . . . , I, (88)

I∑

j=1

Zkf

ji ≤ µkf

i i = 1, 2, . . . , I, (89)

where

mP,ij(kf ) = ri −[( K∑

k=1k 6=kf

pkii

)+ p

kf

ji

]−

[( K∑

k=1k 6=kf ,kf+1

tkiii)

+ tkf

jii + tkf +1iji

].

35

CASE 1 – Part (a):

Assume that µ does not satisfy condition (a). Let, for i = 1, 2, . . . , I,

qmin,i = min1≤k≤Kk 6=kf ,kv

{µki }.

Now consider the capacity matrix µ̃ as:

µ̃ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

qmin,1 · · · qmin,1 µkv1 qmin,1 · · · qmin,1 µ

kf

1 · · · qmin,1

.... . .

......

.... . .

......

. . ....

qmin,I · · · qmin,I µkv

I qmin,I · · · qmin,I µkf

I · · · qmin,I

.

If µ does not satisfy condition (a), then

µki > qmin,i for some i (1 ≤ i ≤ I) and some k (1 ≤ k ≤ K, k 6= kf , kv).

So we haveK∑

k=1

I∑

i=1

cki µ̃k

i <

K∑

k=1

I∑

i=1

cki µk

i .

which implies that capacity matrix µ̃ has a lower capacity investment cost than that of µ. On the otherhand, since

min1≤k≤Kk 6=,kf ,kv

{µ̃k

i

}= qmin,i = min

1≤k≤Kk 6=,kf ,kv

{µki } i = 1, 2, . . . , I,

Then, the feasible region for problem P2 (i.e., constraints (88) and (89)) are the same for capacity matricesµ and µ̃. On the other hand, since the objective function of problem P2, i.e., (87), is independent of capacitymatrices; hence, we have

$(mP (kf ),d,Q(µ̃)) = $(mP (kf ),d,Q(µ)) for all realization of Q.

Therefore, since µ̃ has a lower capacity investment, it achieves higher expected net profit than µ, and henceµ cannot be optimal solution to P1(mP (kf ),d). Thus, the optimal capacity matrix µ must follow condition(a).

CASE 1 – Part (b):

To prove part (b), consider

µ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

q1 · · · q1 µkv1 q1 · · · q1 µ

kf

1 · · · q1

.... . .

......

.... . .

......

. . ....

qi′ · · · qi′ µkv

i′ qi′ · · · qi′ µkf

i′ · · · qi′

.... . .

......

.... . .

......

. . ....

qI · · · qI µkv

I qI · · · qI µkf

I · · · qI

.

which satisfies condition (a), but does not satisfy (b), i.e., qi′ > di′ for some i′ (1 ≤ i′ ≤ I). Let

µ̃ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

q1 · · · q1 µkv1 q1 · · · q1 µ

kf

1 · · · q1

.... . .

......

.... . .

......

. . ....

di′ · · · di′ µkv

i′ di′ · · · di′ µkf

i′ · · · di′

.... . .

......

.... . .

......

. . ....

qI · · · qI µkv

I qI · · · qI µkf

I · · · qI

,

36

that is, substituting all qi′ in µ with di′ . Since qi′ > di′ for i′ (1 ≤ i′ ≤ I), we have

K∑

k=1

I∑

i=1

cki µ̃k

i <

K∑

k=1

I∑

i=1

cki µk

i .

On the other hand, the only difference between capacity matrices µ and µ̃ is in plant i′ at all stages exceptfor stages kf and kv, and for these capacities, we have

min1≤k≤Kk 6=kf ,kv

{µk

i′ , di′}

= di′ = min1≤k≤Kk 6=kf ,kv

{µ̃k

i′ , di′}.

Hence, the feasible region, i.e., constraints (88) and (89), are the same for capacity matrices µ and µ̃. So,similar to Part (a), we have

$(mP (kf ),d,Q(µ̃)) = $(mP (kf ),d,Q(µ)) for all realization of Q.

Therefore, µ̃ achieves higher expected net profit than µ, and hence µ cannot be optimal solution to P1(mP (kf ),d).Thus, the capacity matrix that does not satisfy condition (b) cannot be optimal.

CASE 1 – Part (c):

First we need to show that∑I

i=1 µkf

i ≤ ∑Ii=1 qi. Suppose

µ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

q1 · · · q1 µkv1 q1 · · · q1 µ

kf

1 · · · q1

.... . .

......

.... . .

......

. . ....

qI · · · qI µkv

I qI · · · qI µkf

I · · · qI

satisfies (a) and (b), but with∑I

i=1 µkf

i >∑I

i=1 qi. Then

µ̃ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

q1 · · · q1 µkv1 q1 · · · q1 q1 · · · q1

.... . .

......

.... . .

......

. . ....

qI · · · qI µkv

I qI · · · qI qI · · · qI

has a lower capacity investment cost, as

K∑

k=1

I∑

i=1

cki µk

i −K∑

k=1

I∑

i=1

cki µ̃k

i =K∑

k=kf

I∑

i=1

cki (µkf

i − qi) > 0, (90)

since cki = ck

j , for i, j = 1, 2, . . . , I, i 6= j, and k = 1, 2, . . . , K.

For capacity matrix µ̃, $(mP(kf ),d,Q(µ̃)) is calculated by solving P2(mP(kf ),d,q) and taking expectationover Q(µ̃):

π(mP (kf ),d,q) = maxZ

kfij

{ I∑

i=1

I∑

j=1

[mP,ij(kf ) · Zkf

ij

]}

subject to:I∑

j=1

Zkf

ij ≤ min1≤k≤Kk 6=kf ,kv

{µk

i , qkvi , di

}= min

{qi, q

kvi

}(since (a) and (b) hold) i = 1, 2, . . . , I,(91)

I∑

j=1

Zkf

ji ≤ µkf

i = qi i = 1, 2, . . . , I. (92)

37

Since according to Assumption 3,

mP,ii(kf ) > mP,ij(kf ) and mP,ii(kf ) > mP,ji(kf ) ∀ j 6= i, i = 1, 2, . . . , I,

the most profitable way of allocating flows (Zkf

ij ’s) is to use dedicated plants. By comparing right-hand-sideof (91) and (92) we have qi ≥ min

{qi, qkv

i

}. Therefore, we get the maximum gross profit for µ̃ by setting

Zii = min{qi, qkv

i

}and Z

kf

ij = 0, ∀ i 6= j:

π(mP (kf ),d,q) =I∑

i=1

[mP,ii(kf ) ·min

{qi, qkv

i

}].

On the other hand, for capacity investment vector µ, $(mP(kf ),d,Q(µ)) is calculated by solving P2(mP(kf ),d,q)and taking expectation over Q(µ):

π(mP (kf ),d,q) = maxZ

kfij

{ I∑

i=1

I∑

j=1

[mP,ij(kf ) · Zkf

ij

]}

subject to:I∑

j=1

Zkf

ij ≤ min{qi, q

kvi

}i = 1, 2, . . . , I,

I∑

j=1

Zkf

ji ≤ µkf

i i = 1, 2, . . . , I.

Hence

I∑

i=1

I∑

j=1

mP,ij(kf ) · Zkf

ij =I∑

i=1

I∑

j=1

mP,ji(kf ) · Zkf

ji

≤I∑

i=1

[mP,ii(kf )

I∑

j=1

Zkf

ji

]

≤I∑

i=1

[mP,ii(kf ) ·min

{qi, q

kvi

}]

Therefore

$(mP (kf ),d,Q(µ)) ≤ $(mP (kf ),d,Q(µ̃)) for all realization of Q (93)

By (90) and (93),

V (mP (kf ),d,Q(µ)) < V (mP (kf ),d,Q(µ̃)),

and hence µ cannot be optimal solution to P1(mP (kf ),d). Thus, the optimal capacity matrix µ must follow∑Ii=1 µ

kf

i ≤ ∑Ii=1 qi.

Now we show that µkf

i ≤ qi for all i = 1, 2, . . . , I. Suppose µkf

i′ > qi′ for some i′, since∑I

i=1 µkf

i ≤ ∑Ii=1 qi,

there has to be some i′′, such that µkf

i′′ < qi′′ . Without loss of generality, we assume that plant i′′ is the onlyplant at stage kf with such property. Let us design µ̃ such that µ̃

kf

i′ = qi′ and µ̃kf

i′′ = µkf

i′′ + µkf

i′ − qi′ , that is,at stage kf , µ

kf

i′ − qi′ units of capacity is moved to plant i′′. From Assumption 4, we have cki = ck

j , ∀ i, j =

1, 2, . . . , I, i 6= j, ∀ k = 1, 2, . . . , K. So µ̃ and µ have same capacity investment cost. Only now µkf

i′ −qi′ unitsof capacity can be used through dedicated arc from plant i′′, rather than flexible arcs from plant i′. SinceAssumption 3 assures that dedicated arc is more profitable than flexible ones, µ̃ gives higher revenue than

38

µ. Hence µ cannot be optimal. Therefore the optimal capacity investment matrix has property µkf

i ≤ qi forall i = 1, 2, . . . , I.

CASE 2: kv = 0

The proof for the kv = 0 case is similar to that for the kv 6= 0 case, except for the following minor modifications.Given that now demand is the only source of variability and capacity is deterministic:

(1) Q(µ) = q = µ;

(2) Since D is stochastic and Q(µ) is deterministic, to calculate V ∗, the expectation is taken over Dinstead of Q(µ);

(3) Stage kv with uncertain capacity is missing in capacity matrices Q(µ), q, µ and µ̃, and in constraintssuch as (88);

(4) We have di instead of qkvi in constraint (91) of Part (c) and the arguments following the constraint.

PROOF OF THEOREM 2:

From Assumption 1, we have mP,ij(k) = mP,ij(k′), ∀ i, j, ∀ k 6= k′. To simplify notation we use mP,ij =mP,ij(k), and hence we have

mP (k) = mP ∀ k : 1 ≤ k ≤ K.

PART (1) - Variability in Demand:

To prove Part (1), we show that when demand is variable, the total expected profit of all configurations inwhich only one stage has process flexibility are the same. Consider two flexibility configurations (kf 6= k′f ):

(I) Full process flexibility at stage kf ,

(II) Full process flexibility at stage k′f ,

we show

V ∗(mP (kf ),D) = V ∗(mP (k′f ),D) ∀ 1 ≤ kf , k′f ≤ K.

Let µ be a candidate for the optimal capacity investment matrix for configuration (I). By Lemma 4,

µ =

stage 1 · · · stage (kf−1) stage kf stage (kf+1) · · · stage K

q1 · · · q1 µkf

1 q1 · · · q1

.... . .

......

.... . .

...qI · · · qI µ

kf

I qI · · · qI

,

and

V (mP (kf ),D,Q(µ)) = ED,Q(µ)

[$(mP (kf ),D,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

= ED

[$(mP ,D,q(µ))

]−

K∑

k=1

I∑

i=1

ciµki

= ED

[$(mP ,D,q(µ))

]−

I∑

i=1

ci

[(K − 1)qi + µ

kf

i

], (94)

39

where ED

[$(mP ,D,q(µ))

]is calculated by solving the following problem P2(mP ,d,q) and taking expec-

tation over D:

π(mP ,d,q) = maxZ

kfij

{ I∑

i=1

I∑

j=1

[mP,ij · Zkf

ij

]}(95)

subject to:I∑

j=1

Zkf

ij ≤ min1≤k≤Kk 6=kf

{µk

i , di

}= min

{qi, di

}i = 1, 2, . . . , I,

I∑

j=1

Zkf

ji ≤ µkf

i i = 1, 2, . . . , I.

Note that by Assumption 5, we write cki = ci, i = 1, 2, · · · , I, ∀ k : 1 ≤ k ≤ K.

For configuration (II), build a capacity matrix µ̃ using elements of µ as follows:

µ̃ =

stage 1 · · · stage (k′f−1) stage k′f stage (k′f+1) · · · stage K

q1 · · · q1 µkf

1 q1 · · · q1

.... . .

......

.... . .

...qI · · · qI µ

kf

I qI · · · qI

,

we have

V (mP (k′f ),D,Q(µ̃)) = ED,Q(µ̃)

[$(mP (k′f ),D,Q(µ̃))

]−

K∑

k=1

I∑

i=1

cki µ̃k

i

= ED

[$(mP ,D,q(µ̃))

]−

K∑

k=1

I∑

i=1

ciµ̃ki

= ED

[$(mP ,D,q(µ̃))

]−

I∑

i=1

ci

[(K − 1)qi + µ

kf

i

], (96)

where ED

[$(mP ,D,q(µ̃))

]is calculated by solving P2(mP ,d,q) and taking expectation over D:

π(mP ,d,q) = maxZ

kfij

{ I∑

i=1

I∑

j=1

[mP,ij · Zkf

ij

]}(97)

subject to:I∑

j=1

Zkf

ij ≤ min1≤k≤Kk 6=kf

{µ̃k

i , di

}= min

{qi, di

}i = 1, 2, . . . , I,

I∑

j=1

Zkf

ji ≤ µ̃kf

i = µkf

i i = 1, 2, . . . , I.

Comparing (94) and (95) with (96) and (97), we conclude that for all candidate optimal capacity investmentmatrix µ for configuration (I), there exists a corresponding capacity matrix µ̃ for configuration (II), such thatV (mP (kf ),D,Q(µ)) = V (mP (k′f ),D,Q(µ̃)) for all realization of D. And this is true for the other directionas well. Therefore we have

V ∗(mP (kf ),D) = V ∗(mP (k′f ),D) ∀ 1 ≤ kf , k′f ≤ K. (98)

40

In other words, configurations (I) and (II) result in the same total expected profit. This completes the proof.

PART (2) - Variability at Stage kv:

First we prove that investing in process flexibility at stage kf is optimal. We consider the following twoconfigurations:

(I) Full process flexibility at stage kv;

(II) Full process flexibility at stage kv 6= kf .

Let µ be a candidate for the optimal capacity investment matrix for configuration (II), assuming kv < kf

without loss of generality. By Lemma 4,

µ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

q1 · · · q1 µkv1 q1 · · · q1 µ

kf

1 · · · q1

.... . .

......

.... . .

......

. . ....

qI · · · qI µkv

I qI · · · qI µkf

I · · · qI

with qi ≤ di and µkf

i ≤ qi, ∀i, which gives

V (mP (kf ),D,Q(µ)) = ED,Q(µ)

[$(mP (kf ),D,Q(µ))

]−

K∑

k=1

I∑

i=1

cki µk

i

= EQ(µ)

[$(mP ,d,Q(µ))

]−

K∑

k=1

I∑

i=1

ciµki

where EQ(µ)

[$(mP ,d,Q(µ))

]is calculated by solving the following P2(mP ,d,q) and taking expectation

over Q(µ):

π(mP ,d,q) = maxZ

kfij

{ I∑

i=1

I∑

j=1

[mP,ij · Zkf

ij

]}(99)

subject to:I∑

j=1

Zkf

ij ≤ min1≤k≤Kk 6=kf ,kv

{µk

i , qkvi , di

}= min

{qi, q

kvi

}i = 1, 2, . . . , I, (100)

I∑

j=1

Zkf

ji ≤ µkf

i i = 1, 2, . . . , I. (101)

For configuration (I), consider capacity matrix µ̃ which is exactly the same as µ:

µ̃ = µ =

stage 1 · · · stage (kv−1) stage kv stage (kv+1) · · · stage (kf−1) stage kf· · · stage K

q1 · · · q1 µkv1 q1 · · · q1 µ

kf

1 · · · q1

.... . .

......

.... . .

......

. . ....

qI · · · qI µkv

I qI · · · qI µkf

I · · · qI

,

Since process flexibility is only at stage kv, we have

V (mP (kv),D,Q(µ̃)) = ED,Q(µ̃)

[$(mP (kv),D,Q(µ̃))

]−

K∑

k=1

I∑

i=1

cki µ̃k

i

= EQ(µ̃)

[$(mP ,d,Q(µ̃))

]−

K∑

k=1

I∑

i=1

ciµ̃ki

= EQ(µ)

[$(mP ,d,Q(µ))

]−

K∑

k=1

I∑

i=1

ciµki ,

41

where EQ(µ̃)

[$(mP ,d,Q(µ̃))

]is calculated by solving the following problem P2(mP ,d,q) and taking ex-

pectation over Q(µ̃):

π(mP ,d,q) = maxZ

kfij

{ I∑

i=1

I∑

j=1

[mP,ij · Zkf

ij

]}(102)

subject to:I∑

j=1

Zkf

ij ≤ min1≤k≤Kk 6=kv

{µ̃k

i , di

}= min

{qi, µ

kf

i , di

}= µ

kf

i (by Lemma 4) i = 1, 2, . . . , I, (103)

I∑

j=1

Zkf

ji ≤ qkvi (since flexibility and variability are both at stage kv) i = 1, 2, . . . , I. (104)

Comparing problems (99) and (102), it is clear that constraint (101) is the same as constrain (103), andconstraint (100) is more strict than constraint (104). Therefore, for each realization of Q, problem (102)achieves higher value than (99), i.e., π(mP ,d,q(µ̃)) ≥ π(mP ,d,q(µ)). Consequently, for all candidateoptimal capacity investment matrix µ for configuration (II), there exists a corresponding capacity matrix µ̃for configuration (I), such that V (mP (kv),d,Q(µ̃)) ≥ V (mP (kf ),d,Q(µ)) for all realization of Q. Therefore,we have

V ∗(mP (kv),d) ≥ V ∗(mP (kf ),d) ∀ kf 6= kv. (105)

In other words, configurations (I) outperforms configuration (II). Since kf and kv are arbitrarily chosen, thiscompletes the proof.

Now we prove that investing in process flexibility at all the stages other than stage kf gives the same profit.The proof is similar to PART (1), with the following minor modifications:

(1) Since Q is stochastic and D is deterministic, to calculate V ∗, the expectations are taken over Q insteadof D;

(2) In capacity matrices µ and µ̃, there is a stage kv with capacity investment µkvi ;

(3) In constraints with di, di is replaced by qkvi , which is one realization of random capacity Qkv

i .

42

ON-LINE APPENDIX IIMulti-Stage Flexibility Configurations: The Impact of

Cost and Variability

Many facilities (e.g., automotive assembly plants, chemical plants, etc.) require investment to make theirprocesses flexible. When the fixed cost of process flexibility is non-zero, it may not be feasible to have processor logistics flexibility at all stages. On the other hand, there are cases where flexibility can be located inmore than one stage in a supply chain. To study the impact of cost and location of variability on the optimallocation of process and logistics flexibility, in this Appendix, we focus on the optimal location of process andlogistics flexibility in a two-echelon supply chain with two products when flexibility can be located in morethan one stage, and the flexibility costs are non-zero.

Taking into account the interdependency of logistics and process flexibility, for a two-product two-echelonsystem, we depict the possible five flexibility configurations in Figure 2. Note that these are the only poten-tially optimal configurations for a two-product, two-plant system. (There are other configurations in which acertain stage of logistics (process) flexibility has been installed but is not in use because it is not accompaniedby the necessary process (logistics) flexibility. Those configurations are suboptimal, since the fixed cost oflogistics (process) flexibility is non-zero, and therefore not included in our analysis.) In these configurationswe also show the raw material inventory and we include its cost of logistics flexibility (if established) in ouranalysis. Note that in order to make use of full process flexibility at stage K = 2, all plants at that stage needto be supplied with all types of raw material and therefore logistics flexibility is required from raw materialto stage K, as depicted in the figure. Structure (1) has no flexibility, structure (5) has full flexibility, whilestructures (2), (3) and (4) have partial flexibility (i.e., at least one stage does not have flexibility). Comparingthe five structures we observe that if we vary costs of logistics and process flexibility and keep all the otherparameters fixed, the optimal flexibility configuration follows a threshold structure. Depending on differentparameter values, for demand side variability case, the optimal threshold structure results in a 2-reagin, ora 3-regain, or a 4-region policy. Figure 3(Left) presents the case where the threshold structure results in a4-region policy.

For the supply side variability case, the optimal threshold structure results in a 2- or 3-region policy. Figure3(Right) presents the case where the optimal threshold results in a 3-region policy. To save space, in Theorem3 we only present the results for the structures in Figure 3. The proof of this theorem along with the proofsfor other threshold structures are presented at the end of this Appendix.

Theorem 3. For a two-product two-stage supply chain, consider V ∗1 to V ∗

5 to be the optimal expected profitfor configurations 1 to 5, respectively, and let Λk = Λ 6= 0, and Ψk = Ψ 6= 0 for k = 1, 2.

(1) (Variability only in demand): If V ∗3 − V ∗

1 > V ∗5 − V ∗

3 , V ∗4 −V ∗12 > V ∗

5 − V ∗4 and V ∗

3 − V ∗1 +V ∗42 > V ∗

5 − V ∗4 ,

then optimal flexibility configuration follows a 4-region policy as depicted in Figure 3(Left), which canbe described as follows:

(a) if Ψ + Λ < V ∗5 − V ∗

3 , and Λ < V ∗5 − V ∗

4 , then configuration (5) is optimal;

(b) if Ψ + Λ <V ∗4 −V ∗1

2 , Λ > V ∗5 − V ∗

4 , and Ψ < V ∗4 − V ∗

3 , then configuration (4) is optimal;

(c) if Ψ + Λ > V ∗5 − V ∗

3 , 2Λ + Ψ < V ∗3 − V ∗

1 , and Ψ > V ∗4 − V ∗

3 , then configurations (3) and (2) areoptimal;

(d) if Ψ + Λ >V ∗4 −V ∗1

2 , and 2Λ + Ψ > V ∗3 − V ∗

1 , then configuration (1) is optimal;

(2) (Variability only in Supply): If V ∗3 − V ∗

1 > V ∗5 − V ∗

3 and V ∗3 −V ∗12 < V ∗

5 − V ∗3 , then optimal flexibility

configuration follows a 3-region policy as depicted in Figure 3(Right), which can be described as follows:

(a) if Ψ + Λ < V ∗5 − V ∗

3 , and 2Ψ + 3Λ < V ∗5 − V ∗

1 , then configuration (5) is optimal;

(b) if Ψ + Λ > V ∗5 − V ∗

3 and 2Λ + Ψ < V ∗3 − V ∗

1 , then configuration (3) is optimal;

(c) if 2Λ + Ψ > V ∗3 − V ∗

1 , and 2Ψ + 3Λ > V ∗5 − V ∗

1 , then configuration (1) is optimal;

43

Demand

Demand

Plant 1

Plant 2

Plant 1

Plant 2

Stage 0Stage 1Stage 2

FLEXIBILITY CONFIGURATION (3)

A

B

AB

AB

A

A

A

B

Raw Material

Raw Material

Demand

Demand

Plant 1

Plant 2

Plant 1

Plant 2

Stage 0Stage 1Stage 2

FLEXIBILITY CONFIGURATION (1)

A

B

A

B

A

B

A

B

Raw Material

Raw Material

Demand

Demand

Plant 1

Plant 2

Plant 1

Plant 2

Stage 0Stage 1Stage 2

FLEXIBILITY CONFIGURATION (2)

A

B

A

B

AB

AB

A

B

Raw Material

Raw Material

Demand

Demand

Plant 1

Plant 2

Plant 1

Plant 2

Stage 0Stage 1Stage 2

FLEXIBILITY CONFIGURATION (4)

A

B

AB

AB

AB

AB

A

B

Raw Material

Raw Material

Demand

Demand

Plant 1

Plant 2

Plant 1

Plant 2

Stage 0Stage 1Stage 2

FLEXIBILITY CONFIGURATION (5)

A

B

AB

AB

AB

AB

A

B

Raw Material

Raw Material

Figure 2: Flexibility Configurations for Two-Product Two-Plant System: (1) No flexibility; (2) Full processflexibility at stage 1 with minimally required logistics flexibility; (3) Full process flexibility at stage 2 withminimally required logistics flexibility; (4) Full process flexibility at stages 1 and 2 with minimally requiredlogistics flexibility; (5) Full process flexibility and full logistics flexibility at stages 1 and 2.

As both threshold structures in Figure 3 show, when the cost of flexibility is low, both types of flexibility areused at both stages. When the cost of flexibility is sufficiently high, no flexibility is used. For intermediatecosts of process and logistics flexibility, thresholds define boundaries between use and nonuse of flexibility.Notice, however, that supply side variability still favors configurations with upstream logistics and processflexibility (Configuration 3) over those with downstream process and/or logistics flexibility (Configurations 2and 4). In contrast, demand variability favors configurations with downstream logistics flexibility (Configu-ration 4) over configurations with upstream logistics flexibility (Configuration 3), unless the cost of processflexibility is high, which increases the cost of downstream logistics flexibility and offsets its natural advan-tage. Furthermore, for the case of demand variability, the location of process flexibility does not matter (i.e.,Configurations 2 and 3 are equally good). These observations are consistent with our analytical results inSections 3 and 4.

As the number of products and stages in the supply chain increases, the number of possible flexibilityconfigurations also increases. While we cannot depict the optimal policy with a two-dimensional figure, wecan still use the approach of Theorem 3 to demonstrate that the optimal policy consists of structured regionsdivided by thresholds.

PROOF OF THEOREM 3:

To evaluate a supply chain configuration, we introduce the total expected profit for a configuration i, TP (i),which is defined as expected profit minus fixed cost associated with the investment in process and/or logisticsflexibility. We list this total expected profit for all the five configurations as follows:

• TP (1) = V ∗1 ;

• TP (2) = V ∗2 − 2Λ−Ψ;

• TP (3) = V ∗3 − 2Λ−Ψ;

• TP (4) = V ∗4 − 2Λ− 2Ψ;

• TP (5) = V ∗5 − 3Λ− 2Ψ.

44

FLEX. CONFG.

FLEX. CONFG.

FLEX. CONFG.

FLEX. CONFG.(1)

V*- V* V*- V*

V*- V*

V*- V*

Cost of Process Flexibility13

(Ψ)35

5

4

4

1

2

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

(2) and (3)(5)

(4)

FLEX. CONFG.

FLEX. CONFG.(1)

V*- V* V*- V*

V*- V*

Cost of Process Flexibility13

(Ψ)35

1

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

FLEX. CONFG.(5)

(3)

5

3

Figure 3: Threshold Structure for Optimal Flexibility Decision: (Left) Demand Side Variability (Right)Supply Side Variability

PART (1):

For demand side variability, Theorem 2 shows that V ∗2 = V ∗

3 , since both configurations 2 and 3 incur fixedcost 2Λ + Ψ for flexibility, we have TP (2) = TP (3). Therefore, we only need to compare configurations 1, 3,4 and 5.

From Theorem 1, we know that V ∗4 ≥ V ∗

3 . Since configuration 5 has full logistics and process flexibilityat both stages, configuration 1 has no flexibility, configurations 3 has one stage of process flexibility, andconfiguration 4 has two stages of process flexibility, it is clear that V ∗

1 ≤ V ∗3 ≤ V ∗

4 ≤ V ∗5 . Therefore, we have

TP (1) > TP (3) if 2Λ + Ψ > V ∗3 − V ∗

1 , TP (3) > TP (1) otherwise;TP (1) > TP (4) if 2Λ + 2Ψ > V ∗

4 − V ∗1 , TP (4) > TP (1) otherwise;

TP (1) > TP (5) if 3Λ + 2Ψ > V ∗5 − V ∗

1 , TP (5) > TP (1) otherwise;TP (3) > TP (4) if Ψ > V ∗

4 − V ∗3 , TP (4) > TP (3) otherwise;

TP (3) > TP (5) if Λ + Ψ > V ∗5 − V ∗

3 , TP (5) > TP (3) otherwise;TP (4) > TP (5) if Λ > V ∗

5 − V ∗4 , TP (5) > TP (4) otherwise.

We use Figure 4 to show the above relationship in the space of Λ and Ψ. Note that Lines D, E and F crossat point (V ∗

4 −V ∗3 , V ∗

5 −V ∗4 ), Line A, E, and C cross at (2V ∗

5 +V ∗1 − 3V ∗

3 , 2V ∗3 −V ∗

1 −V ∗5 ), and Lines B, F, C

cross at ( 3V ∗4 −V ∗1 −2V ∗52 , V ∗

5 − V ∗4 ). Depending on the value of V ∗

1 to V ∗5 , we have the six threshold structures

shown in Figure 5.From the figures, we observe the following relationship between the values of V ∗

1 to V ∗5 , and the threshold

structures in Figure 5.

1. If V ∗3 − V ∗

1 ≤ V ∗5 − V ∗

3 and V ∗4 −V ∗12 ≤ V ∗

5 − V ∗4 , then threshold structure is shown in Figure 5 [6];

2. If V ∗3 − V ∗

1 ≤ V ∗5 − V ∗

3 and V ∗4 −V ∗12 > V ∗

5 − V ∗4 , then threshold structure is shown in Figure 5 [5];

3. If V ∗3 − V ∗

1 > V ∗5 − V ∗

3 and V ∗4 −V ∗12 ≤ V ∗

5 − V ∗4 , then threshold structure is shown in Figure 5 [4];

4. If V ∗3 − V ∗

1 > V ∗5 − V ∗

3 , V ∗4 −V ∗12 > V ∗

5 − V ∗4 and V ∗

3 − V ∗1 +V ∗42 < V ∗

5 − V ∗4 , then threshold structure is

shown in Figure 5 [3];

5. If V ∗3 − V ∗

1 > V ∗5 − V ∗

3 , V ∗4 −V ∗12 > V ∗

5 − V ∗4 and V ∗

3 − V ∗1 +V ∗42 = V ∗

5 − V ∗4 , then threshold structure is

shown in Figure 5 [2];

6. If V ∗3 − V ∗

1 > V ∗5 − V ∗

3 , V ∗4 −V ∗12 > V ∗

5 − V ∗4 and V ∗

3 − V ∗1 +V ∗42 > V ∗

5 − V ∗4 , then threshold structure is

shown in Figure 5 [1].

45

*

1

*

3 VV −

2

*

1

*

3 VV −

2

*

1

*

4 VV −

2

*

1

*

4 VV −

2

*

1

*

5 VV −

3

*

1

*

5 VV −

*

3

*

4 VV −

*

4

*

5 VV −

*

3

*

5 VV −

*

3

*

5 VV −

→Ψ →Ψ →Ψ

→Ψ →Ψ →Ψ

↑Λ ↑Λ ↑Λ

↑Λ ↑Λ ↑Λ

(a) (b) (c)

(d) (f)(e)

A BC

D

F

E

)4()1( TPTP >

)1()4( TPTP >

)4()5( TPTP >

)5()4( TPTP >

)3()5( TPTP >

)5()3( TPTP >)3()4( TPTP >

)4()3( TPTP >

)1()5( TPTP >

)5()1( TPTP >

)1()3( TPTP >

)3()1( TPTP >

Figure 4: Performance Comparison between Configuration Pairs

PART (2):

When variability is in supply, it can be shown using Theorem 1 that configuration 3 outperforms configuration4. Theorem 2, on the other hand, implies that configuration 3 outperforms configuration 2. Therefore, weonly compare configurations 1, 3 and 5 in the following paragraphs.

Since configuration 5 has full logistics and process flexibility at both stages, configuration 1 has no flexibility,and configurations 3 has partial flexibility, it is clear that V ∗

1 ≤ V ∗3 ≤ V ∗

5 . Therefore, we have

TP (1) > TP (3) if 2Λ + Ψ > V ∗3 − V ∗

1 , TP (3) > TP (1) otherwise;TP (1) > TP (5) if 3Λ + 2Ψ > V ∗

5 − V ∗1 , TP (5) > TP (1) otherwise;

TP (3) > TP (5) if Λ + Ψ > V ∗5 − V ∗

3 , TP (5) > TP (3) otherwise.

Diagrams (a), (c) and (e) in Figure 4 show the above relation. Depending on the values of V ∗1 , V ∗

3 and V ∗5 ,

it can be concluded from diagrams (a), (c) and (e) in Figure 4 that there exists four threshold structures foroptimal policy, which are shown in Figure 6. From the figures, we observe the following relationship betweenthe values of V ∗

1 to V ∗5 and the thresholds:

1. If V ∗3 − V ∗

1 ≤ V ∗5 − V ∗

3 , then threshold structure is shown in Figure 6 [4];

2. If V ∗3 − V ∗

1 > V ∗5 − V ∗

3 and V ∗3 −V ∗12 < V ∗

5 − V ∗3 , then threshold structure is shown in Figure 6 [3];

2. If V ∗3 −V ∗12 = V ∗

5 − V ∗3 , then threshold structure is shown in Figure 6 [2];

3. If V ∗3 −V ∗12 > V ∗

5 − V ∗3 , then threshold structure is shown in Figure 6 [1].

46

FLEX. CONFG.

FLEX. CONFG.

FLEX. CONFG.

FLEX. CONFG.(1)

V*- V* V*- V*

V*- V*

V*- V*

Cost of Process Flexibility13

(Ψ)35

5

4

4

1

2

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

(2) and (3)(5)

(4)

[1]

A

B

DF

E

FLEX. CONFG.(1)

V*- V* V*- V*

V*- V*

Cost of Process Flexibility13

(Ψ)35

4 1

2

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

(2) and (3)

[2]

FLEX. CONFG.(5) FLEX. CONFG.

CONFG.(4)

V*- V*5 4

FLEX.

A

B

F

E

FLEX. CONFG.(1)

V*- V* V*- V*

V*- V*

Cost of Process Flexibility13

(Ψ)35

4 1

2

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

(2) and (3)

FLEX. CONFG.(5) FLEX. CONFG.

CONFG.(4)

V*- V*5 4

FLEX.

A

B

F

E

C

[3]

FLEX. CONFG.(1)

V*- V* V*- V*

Cost of Process Flexibility13

(Ψ)35

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

FLEX. CONFG.(5)

A

E

C

(2) and (3)FLEX. CONFG.

1V*- V*

[4]

5

3

FLEX. CONFG.(1)

V*- V*

V*- V*

Cost of Process Flexibility1

(Ψ)

4 1

2

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

CONFG.(4)

V*- V*5 4

FLEX. B

F

[5]

C

(5)FLEX. CONFG.

52

FLEX. CONFG.(1)

V*- V*

Cost of Process Flexibility1

(Ψ)

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

52

C

(5)FLEX. CONFG.

[6]

31V*- V*5

Figure 5: Optimal Threshold Structures for Demand Side Variability.

FLEX. CONFG.(1)

V*- V*

Cost of Process Flexibility1

(Ψ)3

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

V*- V*3

5V*- V*

5

(5)CONFG.FLEX. FLEX. CONFG.

(3)

3

A

[1]

V*- V*

E

3 1

2 FLEX. CONFG.(1)

V*- V*

Cost of Process Flexibility1

(Ψ)3

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

V*- V*35

FLEX. CONFG.(3)

A

V*- V*3 1

2

[2]

FLEX. CONFG.

(5)

E

FLEX. CONFG.(1)

V*- V* V*- V*

Cost of Process Flexibility13

(Ψ)35

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

FLEX. CONFG.(5)

A

E

C

FLEX. CONFG.

1V*- V*5

3

[3]

(3)

FLEX. CONFG.(1)

V*- V*

Cost of Process Flexibility1

(Ψ)

Cos

t of

Log

istic

s F

lexi

bilit

y(Λ

)

52

C

(5)FLEX. CONFG.

[4]

31

V*- V*5

Figure 6: Optimal Threshold Structures for Supply Side Variability.

47