4 COORDINATED REPLENISHMENTS cyclic schedules powers-of-two policies

4 COORDINATED REPLENISHMENTS

cyclic schedules powers-of-two policies

Q

Q

Q

Q

2

1

C

C *

**

T

T

T

T

2

1

C

C *

**

4.1 Powers-of-two policies

Cycle times are restricted to be powers of two times a certain basic period denoted as q . From (3.5) in Chapter 3, we have:

, (4.1)Let T = Q/d, T* = Q*/d

. (4.2)

q2T m

T2

T

T

T2

2

1

T

T

T

T

2

1

C

C *

*

*

**

Lot sizing problem

(4.3)Worst scenario:Because the cost is convex, the worst possible error must occur when two consecutive values of m, say m = k and m = k + 1 give the same error. Let T < T* correspond to m = k, and 2T > T* to m = k + 1.

. (4.4)

.2gives*

Setting xxT

T

2T

T2

T

T*

*

06.122

1

2

1

C

C*

, (4.5)

. (4.6)

Proposition 4.1 For a given basic period q, the maximum relative cost increase of a powers-of-two policy is 6 percent.

C/C*

T

2T2T* 2T 2T1T2 T1T

2*2 TT 2

*TT

2

1

2

1,]22[

2

1)(

2

1

2

1,2

2*

2

*

*

2/12/1

ixx

ii

i

ixi

i

xxeC

C

xT

TT

T

ii

i

Note that T1 gives better cost than

Note that 2T2 gives better cost than

Suppose possible to change q

For a single item, the optimal solution is to choose q equal to a power of two times T*. For N items, we can, in general, not fit q perfectly to all cycle times Ti*. The relative cost increase can be expressed as:

)(

)/(

1**

*

1

*

1

**

1

*

1*

i

N

ii

i

i

i

i

N

ii

N

iiii

N

ii

N

ii

xewC

C

C

C

C

CCC

C

C

C

C

*ii C/C

,2/1x2/1),22(2

1)x(e

C

Ci

xxi*

i

i ii

)2/1x2/1(2T/T ix*

iii

N

1i*i

*i C/C

. (4.7)

We know from (4.2) and (4.5) that for a given q, each

can be expressed as :

(4.8)

The weights wi for the different values of xi , can be seen as a probability distribution

F(x) on [-1/2, 1/2], i.e., F(-1/2) = 0 and F(1/2) = 1.

2/1

2/1*

)x(dF)x(eC

C

.1y0 . (4.9)Change q by multiplying by 2y , This means that a certain x is replaced by x + y for x + y ≤ ½, , and by x + y - 1 for x + y > ½.

**

'

*)(*

*

22

2222)(2

1)0(,10,2)(

22

2

22Each

iyx

mi

xymymm

i

y

imxi

x

mi

xi

TT

qyqT

qyqyqLet

TT

q

qTT

i

i

i

iii

ii

im

i

ii

.)1yu(dF)u(e)yu(dF)u(e

)x(dF)1yx(e)x(dF)yx(e)y(C

C

2/1y

2/1

2/1

2/1y

2/1

y2/1

y2/1

2/1*

(4.10)For a given distribution F(x) the minimum cost increase is obtained by minimizing (4.10) with respect to 0 ≤ y ≤ 1.

2

12

2

12*

2

1*2

2

1*2

2

1

'

1'

)(

)(

yxif

yxifi

i

yxifTT

yxifTTi

i

yix

i

yix

i

yix

i

i

yixi

T

yT

yT

du)1yu(dF)yu(dF)u(e

dy)1yu(dF)u(e)yu(dF)u(e)y(C

Cmin

2/1

2/1

1

2/1u

2/1u

0

1

0

2/1y

2/1

2/1

2/1y*1y0

02.12ln2

1du)u(edu)2/1(F)u(F)u(F)2/1(F)u(e

2/1

2/1

2/1

2/1

Proposition 4.2 If we can change the basic period q, the maximum relative cost increase of a powers-of-two policy is 2 percent.

Proof The average cost increase for 0 ≤ y ≤ 1 must be at least as large as the minimum

The worst case will occur when the distribution F(x) is uniform on , see (4.9).

A change of q will then not make any difference.

2/1x2/1

1/2

-1/2

0

u

y

2

12

1

+=

-=

uy

yu

1

1

2/1

2/1)()(

ydyyudFue

2/1

0)()(

uduyudFue

Given y

Given u

duuFue

FNoteduFuFue

duyuFue

duyudFue

duyudFue

dyyudFue

u

u

u

y

)()(

0)2

1()]

2

1()([)(

)]([)(

)()(

)()(

)()(

2/1

2/1

2/1

2/1

2

1

0

2/1

2/1

2/1

0

2/1

2/1

2/1

2/1

2/1

0

1

0

2/1

2/1

u

1/2

-1/2

0 y2

12

1

uy

yu

1

1

2

1 )1()(u

duyudFue

Given y

2/1

2/1)1()(

ydyyudFue

Given u

duuFue

FNoteduFuFue

duyudFue

duyudFue

dyyudFue

y

u

y

u

y

)](1[)(

1)2

1()]

2

1()([)(

)1()(

)1()(

)1()(

2/1

2/1

2/1

2/1

1

2/1

2/1

2/1

2/1

2/1

1

2/1

1

0

2/1

2/1

4.2 Production smoothing

Mixed integer program (MIP) Manne (1958) Billington et al. (1983) Eppen and Martin (1987) Shapiro (1993)

Objective: Low inventory cost and smooth capacity utilization.Simplification: ignore stochastic variations ignore time-varying aspectMIP not used in practice

When the demands for different items are relatively stable, use cyclic schedules.

In a general case with many items and several production facilities, it can be extremely difficult to find suitable cyclic schedules.

It is also common in practice to smooth production outside the inventory control system.

Each item ordered periodically. Ordering period chosen to smoothen the load.

Example: 4 items, same demand

Item 1: 1,5,9 Item 2: 2,6,10

Item 3: 3,7,11 Item 4: 4,8,12

Periodic review: order up to S policy

Continuous review: large variation in production load resulting in long and uncertain lead time.

4.2.1 The Economic Lot Scheduling Problem (ELSP)

Cyclic schedules for a number of items with constant demands.

Backorders not allowed.

Finite production rate

Single production facility

Notation:N = number of items,hi = holding cost per unit and time unit for item i,Ai = ordering or setup cost for item i,di = demand per time unit,pi = production rate (pi > di),si = setup time in the production facility for item i, independent of the sequence of the items,Ti = cycle time for item i (the batch quantity Qi = Tidi).

Define:

i = di/pi ,

i = i Ti = production time per batch for item i excluding setup time,i = si + i = total production time per batch for item i.

The problem is to minimize subject to the constraint that all items should be produced in the common production facility.

Table 4.1 N = 10

Bomberger (1966) Table 4.1 Bomberger’s problem (time unit = one day).

2

T)1(dh

T

AC i

iiii

ii . (4.11)

N

1i iC

The optimal cycle time when disregardingthe capacity constraint is

)1(dh

A2T

iii

ii

)1(dhA2C iiiii

, (4.12)

. (4.13)Note that (4.11) and (4.12) are equivalent to (3.6) and (3.7) if we replace Ti by Qi/di , and i by di/pi .

Table 4.2 Independent solution of Bomberger’s problem.

Item 1 2 3 4 5 6 7 8 9 10

Ti167.5 37.7 39.3 19.5 49.7 106.6 204.3 20.5 61.5 39.3

Ci0.179 1.060 1.528 1.024 4.428 0.938 3.034 12.668 6.506 0.255

I2.36 2.01 3.56 4.29 2.49 1.67 3.04 5.87 11.20 1.17

10

1i i 62.31CC , is a lower bound for the total costs.

Solution not feasible: Consider items

4, 8 and 9.

Item 4 & 8 have cycle times 19.5 and 20.5. Must

be able to produce one batch of #4 and #8 in

[t,t+20.5], or [t+11.20,t+20.5]. But the length of

available time is 9.30, while 4+8 =10.16 >9.30.

Not feasible

t+61.5t+20.5t

Item 9

11.20

9.30Item 9

1N

1ii

N

1iiii

i

2

T)1(dh

T

AC

Does a feasible solution exist? If at least one setup time is positive an obvious necessary condition for a feasible solution to exist is

. (4.14)Condition (4.14) is also sufficient for feasibility.

Given the assumption of a common cycle time, the problem now is to minimize

, (4.15)

T)Ts(N

1iii

N

1ii

min

1

1

1T

sT N

ii

N

ii

with respect to the constraint that the common cycle must be able to accommodate production lots of all items

. (4.16)

, (4.17)a lower bound for the cycle time.

Need large enough T to squeeze setups in the slack

1-

N

ii

1

N

iiii

N

ii

dh

AT

1

1

)1(

2ˆ

Disregard (4.17), from (4.15)

. (4.18)Since (4.15) is convex in T the optimal solution,

, (4.19)For Bomberger’s problem, Tmin = 31.86, and consequently, . cost=41.17.

For problems where the indi vidual cycle times are reasonablysimilar, the common cycle approach gives a very goodapproximation.

,75.42ˆ TTopt 75.42T̂

)T,T̂(maxT minopt

17.4162.31 ** CCi

Two approaches for deriving better solutions.

I. Dynamic programming model

Bomberger (1966) AssumingTi = niW

W should be able to accommodate production of all items.

)w(F)Wn(Cmin)w(F iiiin

1ii

Fi(w) = minimum cost of producing items

i +1,i+2,..., N when the vailable capacity in the basic period is w, i.e., W - w has been used for items 1, 2,...,i. , (4.20)where Ci(niW) are the costs (4.11) for item i with Ti = niW,

i = si + iniW, and the integer ni is subject to the constraint

. (4.21)orNote that the upper bound in (4.21) is equivalent to i w.

FN(w) = 0 for all w 0. F0(W) gives the minimum costs when

the basic period is equal to W.

W/)sw(n1 iii Wnsw iii

Minimize over W. Bomberger’s solution

C=36.65, W=40, ni=1 for i 7, n7 =3.

Serious Assumption: W should be able to accommodate production of all items.

II. Heuristic

Doll and Whybark, 1973).

The procedure is to successively im prove the multipliers ni and the basic period W according to the following iterative procedure:

1. Determine the independent solution and use the shortest cycle time as the initial basic period W.

2. Given W, choose powers-of-two multipliers, (ni = 2mi, mi 0), to minimize the item costs (4.11).

N

1iiiii

ii

2

Wn)1(dh

W

n/AC

N

1iiiii

N

1iii

n)1(dh

n/A2W

3. Given the multipliers ni , minimize the total costs

with respect to W.

4. Go back to Step 2 unless the procedure has converged. In that case, check whether the obtained solution is feasible. If the solution is infeasible, try to adjust the multipliers and then go back to Step 3.

Compare with independent solution.

Apply the heuristic to Bomberger’s problem.

Table 4.3 Solution of Bomberger’s problem with W = 23.42.

Item 1 2 3 4 5 6 7 8 9 10

ni8 2 2 1 2 4 8 1 2 2

i2.62 2.47 4.19 5.12 2.37 1.50 2.87 6.63 8.71 1.37

Table 4.4 Feasible production plan.

Basic periodItems Production

time1 4, 8, 2, 9 22.93  

2 4, 8, 3, 5, 10, 1 22.30  

3 4, 8, 2, 9 22.93  

4 4, 8, 3, 5, 10, 6 21.18  

5 4, 8, 2, 9 22.93  

6 4, 8, 3, 5, 10, 7 22.55  

7 4, 8, 2, 9 22.93  

8 4, 8, 3, 5, 10, 6 21.18  

In case of stochastic demand, one possible approach is to first solve a deterministic problem based on averages, and then try to adapt the solution to the stochastic case by adding suitable safety stocks.

4.2.2 Production smoothing and batch quantities

Adjust the batch quantities to obtain a reasonably smooth load.

Karmarkar (1987, 1993). Axsäter (1980, 1986),

Bertrand (1985), and Zipkin (1986).

Consider a machine in a large multi-center shop.

D = average output of material (demand), units per time unit,P = average processing rate, units per time unit,Q = batch quantity,t = setup time,T = average time in the system for a batch,h = holding cost per unit and time unit after processing.

Assume:The batches arrive at the machine as a Poisson process with rate = D/Q. Thus Av demand=Q=D.The processing time is exponentially distributed. average processing time for a batch is 1/ = t + Q/P. Service rate = . = l / = Dt/Q + D/P. The average time spent in the M/M/1 system is

P/DQ/Dt1

P/Qt

1

/1T

. (4.22)

The average cycle stock is approximated as Q/2.

P/D1

Dt2Q*

Assume that the average holding cost per unit and time unit for work-in-process is exactly half of the holding cost h after the process. Av cycle stock = Q/2. Total holding cost after the process=hQ/2.Work-in-process TD=Av time in the system * Av demand

.(4.23)

. (4.24)

For low values of D, Q* is essentially linear in D. For larger values, Q* grows very rapidly.

QP/DQ/Dt1

P/DQDt

2

hmin)QTD(

2

hmin

0Q0Q

4.3 Joint replenishments

4.3.1 A deterministic model

Setup costs: Individual setup costs for each item, and a joint setup cost for the whole group of items. Reason: joint setup costs, quantity discounts, coordinated transports.

constant continuous demand. No backorders. batch quantities are constant. production time is disregarded. No lead time or the lead time is same for all items.

Notation:

N = number of items,hi = holding cost per unit and time unit for item i,A = setup cost for the group,ai = setup cost for item i,di = demand per time unit for item i,Ti = cycle time for item i.i = hidi

Assume all demands equal to one. Items areordered so that a1/1 a2/ 2 ... aN/N .Note that increasing setup costs and decreasingholding costs mean increasing lot sizes and cycletimes.

i

ii

a2T

2

)n(T

T

n/aaAC

N

2iii11

1

N

2iii1

Approach 1. An iterative technique

If there were no joint setup cost

, (4.25)i.e., T1 would be the smallest cycle time. Assume other cycle

times of items 2, 3... N are integer multiples ni of the cycle time

for item 1,

, i = 2, 3,..., N. (4.26)Our objective is to minimize w.r.t T1, n2, n3, ... nN the cost.

,(4.27)

1ii TnT

Fix cost ith item holding cost=Tinii/2

N

2iii1

N

2iii1

N32*1

n

)n/aaA(2)n,...,n,n(T

N

2i

N

2iii1ii1N32

* )n)(n/aaA(2)n,...,n,n(C

)aA(

an

1

1

i

ii

N

2iii11 a2)aA(2C

Given n2, n3... nN,

, (4.28)

. (4.29)Note that T1 is not chosen according to (4.25). If we disregard

n2, n3... nN to be integers, then from (4.29),

. (4.30)From (4.30) and (4.29), the lower bound for the costs:

.(4.31)

HEURISTIC 1. Determine start values of n2, n3... nN by rounding

(4.30) to the closest positive integers.

2. Determine the corresponding T1 from (4.28).

3. Given T1, minimize (4.27) with respect to n2, n3... nN.

This means that we are choosing ni as the positive

integer satisfying

.;2

0:

)1(2

)1(

21

2

21

convexisCT

angives

dn

dCNote

nnT

ann

i

ii

i

iii

iii

. (4.32)Return to Step 2, if any multiplier ni has changed since the last

iteration.

3n,1n,1350

105000

101000

50n 432

1155.031001070010100010500010

)3/505050350(2T1

Example 4.1 N = 4 , A = 300, a1 = a2 = a3 = a4 = 50, h1 = h2 = h3

= h4 = 10, d1 = 5000, d2 = 1000, d3 = 700, and

d4 = 100. As re quested, ai/i is nondecreasing with

i. When applying the heuristic we obtain

again n2 = 1, n3 = 1, n4 = 3, i.e., the algorithm has already

converged.

N

0i ii

ii

T,...,T,T T

1a

2

Tmin

N10

0i TT

Approach 2. Roundy´s 98 percent approximationthe joint setups have cycle time T0 0,

, i = 1, 2, ..., N, (4.33)where ki is a nonnegative integer. Let a0 = A, and 0 = 0,

, (4.34)subject to the constraints (4.33). replacing (4.33) by

, i = 1, 2, ..., N. (4.35)

02 TT iki

N

1ii0i

N

0i ii

ii

T,...,T,T,...,,)TT(

T

1a

2

Tminmax

N10N21

N

1ii00 2

1

N

The resulting solution will give a lower bound for the costs. I lagrangian relaxation

, (4.36)where i are nonnegative.

define

= 1 - 21,

. . (4.37) .

=N - 2N.

N

0i ii

ii

T,...,T,T T

1a

2

Tmin

N10

q2T imi

the optimal solution must have all The optimal solution of the relaxed problem can be obtained by solving N + 1 independent classical lot sizing problems.

(4.38)without any constraints on the cycle times.rounding the cycle times of the relaxed problem,

(4.39)

.0i

for some number q > 0. From Proposition 4.1, if q is given, the maximum cost increase is at most 6 percent. If we adjust q to get a better approximation the cost increase is at most 2 percent according to Propo sition 4.2. We have now obtained Roundy’s solution of the problem.

I. A simpler technique instead of Lagrangian Relaxation

From (4.34) and (4.35) the optimal cycle times in the relaxed prob lem are nondecreasing with i,

1ii TT , i = 1, 2, ..., N. (4.40)Since 0 = 0 we will always aggregate items 0 and 1. After

aggregation we have an item with cost parameters A + a1 and

1. Next we check whether a2/2 < (A + a1)/1. If this is the

case the aggregate item should include also item 2, etc. When no more aggregations are possible we can optimize the resulting aggregate items individually.

Example 4.2

N = 4, A = 300, a1 = a2 = a3 = a4 = 50, h1 = 50000, h2 = 10000, h3 = 7000, and h4 = 1000.

Both of the approaches considered assume constant demand, but can also be used in case of stochastic demand.

4.3.2 A stochastic model

independent, stationary, integer demand, complete backordering.

N = number of items,

hi = holding cost per unit and time unit for item i,

b1, i = shortage cost per unit and time unit for item i,A = setup cost for the group,

ai = setup cost for item i,

Li = constant lead-time for item i.

Viswanathan (1997)

First step: disregard the joint setup cost and consider the items individu ally for a suitable grid of review periods T. For each review period, determine the optimal individual (s, S) policies for all items and the corresponding average costs.

Ci(T) = average costs per time unit for item i when using the optimal individual (s, S) policy with a review interval of T time units.

Second step: determine the review period T by minimizing,

(4.41)

Note that the actual costs are lower than the costs according to (4.41), since the major setup cost A is not incurred at reviews where none of the items are ordered.can-order policies. (Si, Q) policy.

N

1ii )T(CT/A)T(C

A B

Distribution: Warehouse Store

Production: Subassembly Final product

Figure 5.1 An inventory system with two coupled inventories.

Reduces the length and uncertainty of lead times.

5 MULTI-ECHELON SYSTEMS 5.1 Inventory systems in distribution and production

Central warehouse

Retailers

5.1.1 Distribution inventory systems

Distribution system or arborescent systemSerial system

Figure 5.2 Distribution inventory system.

5.1.2 Production inventory systems

convergent flow

Figure 5.3 An assembly system.

It is, in general, considerably easier to deal with serial systems than with other types of multi-echelon systems. Raw material are low volume items. Higher setup cost at earlier stages, large bathes.

1

3

2

6

5

4

8

7

Figure 5.4 A general multi-echelon inventory system.

Figure 5.5 Bill of material corresponding to the inventory system in Figure 5.4.

5.2 Different ordering systems

5.2.1 Installation stock reorder point policies

(R, Q) policies; special case R=S-1, Q=1(S policy with discrete units.)

Installation stock (R, Q) policy (on hand + outstanding orders - backorders)

(s, S) policy is not the optimal policy for a multi-echelon system

KANBAN policySmoothing aspect and local decentralized controls.

Example 5.1

Table 5.1 Installation and echelon stock inventory positions in Figure 5.6.

Item Installation stock inventory position

Echelon stock inventory position

1 5 5

2 2 7

3 3 3

4 5 28

echelon stock reorder point policy will generally use larger reorder points than an installation stock policy to achieve similar control.5 + 2*5 + 2*2 + 3*3=285 + 2*7 +3*3=28

13 2N .......

5.2.3 Comparison of installation and echelon stock policies

Figure 5.7 Serial inventory system with N installations.

Raw material Final product

Increasing batch sizes

1nnn QjQ

inIP

enIP i

1i

1nin IP...IPIP

inR

enR

Qn = batch quantity at installation n.

, (5.1)where jn is a positive integer.

Notation:

= installation inventory position at installation n,

=

= echelon stock inventory position at installation n,

= installation stock reorder point at installation n,

=echelon stock reorder point at installation n.

0inIP 0e

nIP

nen

0en

enn

in

0in

in QRIPR,QRIPR

0inIP i

nR

initial inventory positions and satisfying

(5.2)An installation stock policy is always nested. Installation n may order only if (n-1) has just ordered. Echelon IP at n is only changed by final demand at 1 and replenishment order at n.Assuming: is an integer multiple of Qn-1.

All demands at installation n are for multiples of Qn-1

all replenishments are also multiples of Qn-1

)QR(IP k

n

1k

ik

en

)QR(RR k

1n

1k

ik

in

en

Proposition 5.1 An installation stock reorder point policy can always be replaced by an equivalent echelon stock reorder point policy.Proof When n orders, it means 1,2,…,n-1 has ordered. Then

. (5.3)

. (5.4)

If is not a multiple of Qn-1, change by an amount

less than Qn-1 .So that is a multiple of Qn-1.

Unit demand. Orders would be triggered exactly at the same time. When (n-1) orders Qn-1, its inventory reaches the level Rn-

1+Qn-1. At this point, level n inventory goes from Rn+Qn-1 to Rn. And then n orders.

inRi

n0i

n RIP in

0in RIP

e1

i1 RR

111 nenn

en

en

en

in QRQRIPIPIP

1, 1111 nQRRQIPRRR nen

enn

in

in

ei

Proposition 5.2 An echelon stock reorder point policy which is nested can always be replaced by an equivalent installation stock reorder point policy.Proof For installation 1,

For installation n > 1, immediately after ordering

(5.5)

(5.6)

Therefore,

Example 5.2 Consider a serial system with N = 3 installations and batch quantities Q1 = 5, Q2 = 10, and Q3 = 20. Assume that the initial inventory positions are , , and . Assume furthermore that the demand for item 1 is one unit per time unit. Consider first an installation stock policy with reorder points , , and Note that our assumptions that Qn as well as are multiples of Qn-1 are satisfied. It is easy to check that installation 1 will order at times 5, 10, 15,..... The demand at installation 2 is consequently 5 units at each of these times. This means that installation 2 will order at times 10, 20, 30,.... Demands for 10 units at installation 3 at these times will trigger orders at times 20, 40, 60,.....

Using (5.4) we obtain the equivalent echelon stock policy as , , and Recall that when considering the echelon stock the inventory positions at all installations are reduced by the final demand, i.e., by one unit each time unit, and not by the internal system orders. The initial inventory positions are , , and . It is easy to verify that the orders will be triggered at the same times as with the installation stock policy.

Assume then that we change the echelon stock reorder point at installation 3 to This will not change the orders at installations 1 and 2, but the orders at installation 3 will be triggered 2 time units earlier, i.e., at times 18, 38, 58,.... Recall that the echelon stock is reduced by one unit at a time. The resulting echelon stock policy is not nested and it is impossible to get the same control by an installation stock policy.

Propositions 5.1 and 5.2 show that in any serial inventory system an installation stock policy is simply a special case of an echelon stock policy. This is also true for assembly systems.

Installation policy: Only local information needed. The higher generality of an echelon stock policycan be an advantage in certain situations.

Stock on hand at installation 2

0 1 2 3 4 Time

50

Stock on hand at installation 1

0

1 2 3 4 Time

50

5

Example 5.3 a serial system (Figure 5.7), N = 2, Q1 = 50, Q2 =100.

final demand at installation 1 =50. lead-time at installation 1 is one, at installation 2 is 0.5. No shortages allowed, holding costs at installation 1 are higher than at installation 2.

Figure 5.8 Inventory development in the optimal solution.

LT=0.5, R2e =25+50=75,

IP20(0.5)=0.

LT=1, R1e=50, IP1

i(0.5)=75.

Echelon Policy is more general. At t=0.5- , IP2i=0,IP1

i=25.

The dominance of echelon stock policies for serial and assembly systems does not carry over to distribution systems.

5.2.4 Material Requirements Planning

periodic review, rolling horizon

Master Production Schedule (MPS).

External demands of other items

A bill of material for each item specifying all of its immediate components and their numbers per unit of the parent.

Inventory status for all items

Constant lead-times for all items.

Rules for safety stocks and batch quantities.

MPS. A production or program of final products. Must cover total system lead time.

1

2(1)

3(1)

Figure 5.9 Considered product structure.Table 5.2 Net requirements of item 1.

Item 1 Period 1 2 3 4 5 6 7 8

Lead-time = 1 Gross requirements

10 25 10 20 5 10

Order quantity = 25 Scheduled receipts 25

Safety stock = 5

Projected inventory

22 12 37 12 2 -18 -23 -23 -33

Net requirements 3 20 5 10

Table 5.4 MRP record for item 1 with planned orders in periods 3 and 5.

Item 1 Period 1 2 3 4 5 6 7 8

Lead-time = 1 Gross requirements 10 25 10 20 5 10

Order quantity = 25 Scheduled receipts 25

Safety stock = 5

Projected inventory

22 12 37 12 27 7 27 27 17

Planned orders 25 25

reorder point = the lead-time demand plus the safety stock minus one

Figure 5.10 Material requirements planning for items 1, 2, and 3 in Figure 5.9.

safety time Table 5.5 MRP record for item 1 with a safety time.

Item 1 Period 1 2 3 4 5 6 7 8

Lead-time = 1 Gross requirements 10 25 10 20 5 10

Order quantity = 25 Scheduled receipts 25

Safety stock = 5

Projected inventory

22 12 37 37 27 32 27 27 17

Safety time = 1

Planned orders 25 25

Figure 5.11 Product structures.

Figure 5.12 Gross requirements from different sources.

MRP is often referred to as a push system since orders are in a sense triggered in anticipation of future needs. Reorder point systems and KANBAN systems are similarly said to be pull systems because orders are triggered when downstream installations need them.

The MRP logic is simple. Yet the computational effort can be very large if there are thousands of items and complex multi-level product structures.

“nervousness”

DRP, Distribution Requirements Planning.Manufacturing Resource Planning- MRP II

Rough Cut Capacity Planning (RCCP)Capacity Requirements Planning (CRP)Enterprise Resource Planning (ERP)

Figure 5.13 Manufacturing Resource Planning.

5.2.5 Ordering system dynamics

bullwhip effect, Forrester (1961)decentralized installation stock policies in multi-echelon systems can yield very large demand variations early in the material flow, even though the final demand is very stable.

Note: both the information delays and the problems of large demand variations at upstream facilities are largely due to long lead-times and large batch quantities.

5.3 Order quantities

Assumptions:demand is known and deterministic. In case of stochastic demand, replace the stochastic demand by its mean and use a deterministic model when determining batch quantities. all lead-times are zero.

1 2

5.3.1 A simple serial system with constant demand

Figure 5.14 A simple serial two-echelon inventory system. Example 5.4

item 1 is produced from one unit of the component 2.

d = 8, A1 = 20, A2 = 80, h1 = 5, h2 = 4.

11

111

Q

dA

2

QhC

8h

dA2Q

1

11

40dhA2C 111

I.Treat the two installations independently

. (5.7)

,

.

, (5.8) where k is a positive integer.

12 QkQ

Echelon stock, item 2

Installation stock, item 2

Time

Installation stock = echelon stock, item 1

Time

Figure 5.15 illustrates the behavior of the inventory levels for k = 3.

Figure 5.15 Inventory levels for k = 3.

12

122

Qk

dA

2

Q)1k(hC

2

2

1

*

h

dA2

Q

1k

, (5.9)

If k* < 1 it is optimal to choose k = 1. If k* > 1. Let k´ be the largest integer less or equal to k*, i.e., k´ k* < k´ + 1, it is optimal to choose k = k´ if k*/k´ (k´ + 1)/k*, otherwise k = k´+ 1.

k* 2.24, k = 2. C2 = 56, C = C1 + C2 = 40 + 56 = 96.

1

21

12121

Q

d)

k

AA(

2

Q)h)1k(h(CCC

11

11

e1 Q

dA

2

QeC

12

12

e2 kQ

dA

2

kQeC

1

21

121

e2

e1 Q

d)

k

AA(

2

Q)eke(CCC

II. Treat the two installations together

. (5.10)Alternatively, use the echelon holding costs e1 = h1 - h2,

and e2 = h2,

, (5.11)

. (5.12)

, (5.13)

(5.10) and (5.13) are equivalent.

21

21

1eke

d)k

AA(2

Q

)eke(d)k

AA(2)k(C 21

21

)k

eAkeAeAeA(d2)k(C 12

2122112

21

12*

eA

eAk

. (5.14)

. (5.15)

(5.16)

. (5.17)If k* < 1 it is optimal to choose k = 1. If k* > 1. Let k´ be the largest integer less or equal to k*, i.e., k´ k* < k´ + 1, it is optimal to choose k = k´ if k*/k´ (k´ + 1)/k*, otherwise k = k´+ 1.

1A

1h

Example 5.5 d = 8, A1 = 20, A2 = 80, h1 = 5, h2 = 4.

e1 = h1 - h2 = 1, e2 = h2 = 4. From (5.17), k* = 1.

Applying (5.14) and (5.15), ,

and C* 89.44, about 7 percent lower than the costs obtained in Example 5.4.

(5.14) and (5.15) are essentially equivalent to the corresponding expressions (3.3) and (3.4) for the classical economic order quantity model.

= A1 + A2/k , (5.18)

= e1 + ke2. (5.19)

89.17Q*1 89.17QkQQ *

1*1

*2

Stochastic Inventory Model

Proportional Cost Models:

x: initial inventory,

y: inventory position (on hand + on order-backorder),

: random demand, () , (),

(y- )+: ending inventory position, N.B.L,

(y- ) : ending inventory position, B.L,

=1/(1+r) : discount factor,

ordering cost : c(y-x),

holding cost : h (y- )+

penalty cost : p( -y)+

salvage cost : - s(y- )+

Minimum cost f(x) satisfies:

L(y) convex, L’(0) < -c (otherwise never order) L′ eventually becomes positive

)2()()(min

)()(

)()()()(min)(0

yLxyc

dy

dyshxycxf

xy

y

psp

y

xy

SxIfxSotherwisexq

Sxxy

cSL

,,0

*

*

)(

},max{)(

PolicyStockBase

)4(0)('

)5()()]1([

)1()(.

)5()()(

)(..

0))(1)(()()(

bcc

c

shccp

cpyLB

acc

c

shccp

cpyLBN

yyshc

ou

u

ou

u

psp

Example

c=$1, h=1¢ per month, =0.99, p=$2(NBL), p=$0.25(BL),

s=50 ¢, c+h- s=51.5 ¢,

NBL: p-c = 100 ¢, BL: p-c(1- )=24 ¢,

)32.0(,32.05.5124

24)()(

)66.0(,66.05.51100

100)()(

1

1

yyii

yyi

Set up cost K

L(x) if we order nothing

K+c(S-x)+L(S) if we order upto S If we order, L’(S)+c=0.

Use the cheaper of alternatives L(x) and K+ c(S-x)+L(S)

)()()(min)( yLxycxyKxfxy

S x

cost

L(x)

KK

s S x

L(x)+cx

c

s

K+c(S-x)+L(S)

Two-bin or (s,S) policy

order y-x if x s

order nothing if x > s

Multiperiod models

Infinite Horizon (f1000 & f1001 cannot be different)

dyfyLxycxfxy

)()()()(min)(0

12

)9()()()()(min)(0

dyfyLxycxf

xy

Taking derivative of {}

)10()()(')('00

dSfSLc

If f convex, find S the base stock level, then

)11()()()()()(0

dSfSLxScxf

It is possible to show thatf’(x)=-c for x ≤ S (12)

)18(0))(1()('

N.B.Lfor similarly

(13) B.L0)1()('

toreducewhich

ScSL

cSL

Proportional costs:

So that

dypdyhyLy

y)()()()()(

0

policyS)(s,stilld,complicatemorecostSetuptime,Lead

:Remark

)20()1()1(

)1()(:.

)20()1()(

)(:..

(18),and(13)into(19)Substitute

)19()()()('

bcc

c

chcp

cpSLB

acc

c

chcp

cpSLBN

pSphSL

ou

u

ou

u

Example 4:

1800

5

4

205

20

)(

5 ,20

S

cc

cS

cc

ou

u

ou

Example 5:

5.73.05.1882

)(25

1

2

3)(

25

1

10

3

)()()()()(

0,2

3p(z) 0,z ,

10

3)(,1,15,

25

1)(

25

252

25

0

0

25 2

ye

deydey

dypdyhyL

zzzzhcKe

y

y

y

y

y

5.805.1010

2525

25

25

q

:policy optimal The

5.80

:ionapproximatSuccesive

5.73.05.18825.101155.73.05.1882

)()(

5.101

025.753.01)(

3.025.75)(

xifxotherwise

SS

S

Sy

y

s

Seses

SLcSKsLCs

S

edy

ydLc

edy

ydL

Multiperiod models: No Setup Cost Begin with two periods

Demand D1, D2, i.i.d

Density: ()

L(y) = expected one period holding+ shortage penalty cost;

strictly convex with linear cost and () >0,

c purchase cost /unit

c1(x1) optimal cost with 1 period to go;

c+L’(y10)=0

while y10 is the optimal base stock level.

gotoperiods2withlevelbasestocky

convexiswhich)]([)()(min)(

)()]()([)()(

)()())((

)()(

)(

02

11222222

0122

01

02

02111

)(

)()(

22111

)(

)()(11

22

012

012

012222

0122

0122

01

0111

011

011

01

xcEyLxycxc

dyLyycdyL

dycxcE

Dycxc

xc

xy

yy

yy

yDyifDyL

yDyifyLDyyc

yxifxL

yxifyLxyc

Example: c=10, h=10, p=15 the demand density is

Solution:

100

10

1

0)(

if

otherwise

2

0

10

01

010

1

01

)4/5(1575

10

)(10

10

)(15)(

2,10

)(

5

1

1015

1015)(

zz

dz

dz

zL

yy

ySince

hp

cpy

z

z

3/3592/194/)(24/)(

)4/5(1575)(10min)(

3/3592/194/)(24/)(

10

1)](1070[

10

1]))(4/5()(1575[

10

1]2)4/5(2*1575)2(10[

10

1]))(4/5()(1575[)]([

22

23

2

222222

22

23

2

10

22

2

0

222

10

2

22

2

0

22211

222

2

2

2

2

yyy

yyxyxc

yyy

dy

dyy

dy

dyyxcE

xy

y

y

y

y

2201

5502

02

2202

02

02

202

02

2

2

11

22

:policy optimal The

5.ywithvalue

smalleratoleads)(xcinto6yand5yngSubstituti

42.5

0])(8

122/29[

{}

zerotoequalitsetting,ytorespectwithderivativeTake

xifxotherwise

xifxotherwise

q

q

y

yydy

d

Multi-Period Dynamic Inventory Model with no Setup Cost

Cn(xn): n periods to go,

: discount factor.

DP equations:

SS

xcxc

dycyLxycxc

S

SSSSSS

xc

DycEyLxycxc

nn

nn

Dxy

nn

nnnnnnxy

nnnn

lim)3

)(lim)(bysatisfied

)()()()(min)(2)

0;)-c(1)(L'

where,...................... 1)

:Properties

0)(

])([)()(min)(

0

1321

00

1

Multi-Period Dynamic Inventory Model with Setup Cost

nnnn

nn

nn

nn

nn

SxifxSSxifn

xyifKxyifnn

nnnnnnnxy

nn

q

xyK

dycyLxycxyKxc

0

nn

n

0

01

:)policyS,(soptimalThe

.Sfindthenconvex,isL(y)If

)(

)()()()()(min)(

Multi-Period Dynamic Inventory Model with Lead TimesLead time:

0)1()('

horizoninfinite

)()()(

:follows as timelead 0 toormcan transf

positioninventoryu

)()()()()()(min)(

0

n

01

0

cS

dyLy

dyfdyLuycuyKuf

D

DnnDnnnnnuy

nnnn