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OPRE 6366. SCM : 4. Inventory Management 1. Inventory Management without Uncertainty You can study an operations management or a supply chain management book to learn about the Economic Order Quantity (EOQ) model. Here is a short summary. Suppose that the demand rate is constant and known and it is R per unit time. To meet the demand, a firm orders products in batches of Q. Whenever a batch is ordered, the firm pays an order processing/set up cost of S. The purchase cost of each unit in the batch is constant p. When the firm is holding inventory, it pays an inventory holding cost at the rate of h per dollar of inventory held and per time. The total cost rate is TC(Q) : = h 2 ( pQ)+ R Q S + R Q ( pQ)= h 2 ( pQ)+ R Q S + Rp. Since Rp is a fixed (sunk) cost that does not depend on Q, sometimes it is dropped from cost consideration and we write TC(Q)= pQh/2 + RS/Q. In the EOQ problem, we want to solve min Q0 TC(Q). For low values of Q, TC(Q) has a large ordering cost due to high R/Q. For high values of Q, TC(Q) has a large holding cost due to high pQ/2. So the cost TC must first be decreasing in Q and then increasing in Q. As soon as the cost stops decreasing and starts increasing, we have the optimal Q. That is, the derivative of TC vanishes at the optimal Q. Setting the derivative equal to zero, we obtain d dQ TC(Q)= h 2 ( p) R Q 2 S = 0 = Q = 2RS hp , where Q denotes the optimal order quantity (EOQ). The cost incurred by implementing the EOQ is TC(Q ) and it has a simple expression when the purchase cost Rp is not considered: TC(Q )= h 2 p 2RS hp + R 2RS hp S = RShp 2 + RShp 2 = 2RShp. 2. Long Term Quantity Discounts We present all-unit quantity discount and marginal-unit quantity discount. They are both quantity discount schemes, so the purchase price of a single unit drops or stays the same as more units are purchased. In other words, the purchase price of a single unit cannot increase as more is purchased. 2.1 All-unit Quantity Discount The price of each unit in an order of Q is the same. That price depends on the magnitude of Q as follows. Price per unit = p 1 if q 0 Q q 1 p 2 if q 1 < Q q 2 ... ... p N if q N1 < Q < q N 1

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OPRE 6366. SCM : 4. Inventory Management

1. Inventory Management without Uncertainty

You can study an operations management or a supply chain management book to learn about the EconomicOrder Quantity (EOQ) model. Here is a short summary.

Suppose that the demand rate is constant and known and it is R per unit time. To meet the demand, a firmorders products in batches of Q. Whenever a batch is ordered, the firm pays an order processing/set up costof S. The purchase cost of each unit in the batch is constant p. When the firm is holding inventory, it pays aninventory holding cost at the rate of h per dollar of inventory held and per time. The total cost rate is

TC(Q) :=h2(pQ) +

RQ

S +RQ(pQ) =

h2(pQ) +

RQ

S + Rp.

Since Rp is a fixed (sunk) cost that does not depend on Q, sometimes it is dropped from cost consideration andwe write TC(Q) = pQh/2 + RS/Q. In the EOQ problem, we want to solve

minQ≥0

TC(Q).

For low values of Q, TC(Q) has a large ordering cost due to high R/Q. For high values of Q, TC(Q) has alarge holding cost due to high pQ/2. So the cost TC must first be decreasing in Q and then increasing in Q. Assoon as the cost stops decreasing and starts increasing, we have the optimal Q. That is, the derivative of TCvanishes at the optimal Q. Setting the derivative equal to zero, we obtain

ddQ

TC(Q) =h2(p)− R

Q2 S = 0 =⇒ Q∗ =

√2RShp

,

where Q∗ denotes the optimal order quantity (EOQ).The cost incurred by implementing the EOQ is TC(Q∗) and it has a simple expression when the purchase

cost Rp is not considered:

TC(Q∗) =h2

p

√2RShp

+R√2RShp

S =

√RShp2

+

√RShp2

=√

2RShp.

2. Long Term Quantity Discounts

We present all-unit quantity discount and marginal-unit quantity discount. They are both quantity discountschemes, so the purchase price of a single unit drops or stays the same as more units are purchased. In otherwords, the purchase price of a single unit cannot increase as more is purchased.

2.1 All-unit Quantity Discount

The price of each unit in an order of Q is the same. That price depends on the magnitude of Q as follows.

Price per unit =

p1 if q0 ≤ Q ≤ q1p2 if q1 < Q ≤ q2. . . . . .pN if qN−1 < Q < qN

1

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where we set q0 = 0 and qN = ∞. Prices are monotonically ordered by pn−1 ≥ pn for 2 ≤ n ≤ N.All-Unit Quantity Discount Example: A shoe store sells each shoe at $60 but drops the price to $40 when

3 or more shoes are purchased by a single customer. Then

Price per shoe =

{60 if 0 ≤ Q ≤ 240 if 2 < Q < ∞

}.

Here there are N = 2 prices, and p1 = 60, p2 = 40, q0 = 0, q1 = 2 and q2 = ∞. If you buy 3 shoes as opposedto 2 shoes, not only the price of the third shoe but also the price of the first two shoes drop. That is why this isan all-unit discount scheme. If you look at the prices closely, you see that the cost of buying two shoes is thesame as the cost of buying three shoes. For Q = 3, this is commonly known as “buy-2-get-1-free” deal. ⋄

All-Unit Quantity Mark-up Example: IRS income tax rates depend on the income. If you are filing as a“single” individual for your 2013 taxable income, the tax rates are as follows depending on your 2013 incomeQ:

Tax per dollar of taxable income =

See tax tables if 0 ≤ Q ≤ 100, 000

28.0% if 100, 000 < Q ≤ 183, 25033.0% if 183, 250 < Q ≤ 398, 35035.0% if 398, 350 < Q ≤ 400, 00039.6% if 400, 000 < Q < ∞

.

Here N = 5, and q0 = 0, q1 = 100, 000, q2 = 183, 250, q3 = 398, 350, q4 = 400, 000 and q5 = ∞. Note thatthe tax on each dollar depends on the total taxable income. In this example, prices (tax rates) are increasingwith the taxable income, so we do not have a discount in tax rates but a mark-up. Accountants call this sortof increasing tax rates as a progressive tax regime. For example, if your taxable income is $183,251 instead of$183,250, your tax rate for each of 183,251 dollars jump up to 33% from 28%. This increases your tax from$51,310 to $60,472.83. The significant gap of $9,162.83=60,472.83-51,310 is due to the fact that the tax rate ismarked up for all-units (taxable income). In view of this gap, you may question the fairness of an all-unitquantity mark-up tax regime. ⋄

When we buy Q units, the purchasing cost is as follows.

Purchasing cost of Q units =

p1Q if 0 ≤ Q ≤ q1p2Q if q1 < Q ≤ q2. . . . . .pNQ if qN−1 < Q < qN

We coin the term region n for order sizes in (qn−1, qn]. If the order size is in region n, i.e., qn−1 ≤ Q < qn,

then define the total costs in region n as

TCn(Q) :=h2(pnQ) +

RQ

S +RQ(pnQ). (1)

Then the total cost over all regions is obtained by patching all TCn together, i.e.,

TC(Q) =

TC1(Q) if 0 ≤ Q ≤ q1TC2(Q) if q1 < Q ≤ q2. . . . . .TCN(Q) if qN−1 < Q < qN

In the all-units quantity discount problem, we want to solve

minQ≥0

TC(Q).

We propose the ordering algorithm in Table 1 to find the optimal order quantity. Note that this algorithmdoes not generate any candidate solution after step A2 and step C because the flag CandidatesComplete be-comes true and stops the while loop. We validate this algorithm in the exercises.

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Initialize n := N and CandidatesComplete = False.While n ≥ 1 and CandidatesComplete = False do

Find EOQn.If qn−1 < EOQn ≤ qn

A1: Add EOQn to the set of candidate solutions;A2: CandidatesComplete = True;

else if EOQn ≤ qn−1B: Add qn−1 to the set of candidate solutions;

else ifC: CandidatesComplete = True;

If CandidatesComplete = Falsen:=n-1;

Evaluate the total cost at each candidate solution.Pick the order quantity yielding the minimum cost as the optimal.

Table 1: All-unit discount ordering algorithm

2.2 Marginal-unit Quantity Discount

Marginal-unit quantity discounts apply only to the additional units bought on top of Q units:

Price for an additional unitgiven that Q is already purchased

=

p1 if 0 ≤ Q ≤ q1p2 if q1 < Q ≤ q2. . . . . .pN if qN−1 < Q < qN

Let Vn be the cost of buying exactly qn units. Then V0 = 0, V1 = p1(q1− q0) and V2 = p1(q1− q0) + p2(q2−

q1). In general,Vn = p1(q1 − q0) + p2(q2 − q1) + · · ·+ pn(qn − qn−1).

Also note that the recursive equation Vn = Vn−1 + pn(qn − qn−1) holds. Since qN = ∞, so is VN . But computa-tions do not require VN , we only need V1, . . . , VN−1.

Marginal-Unit Quantity Discount Example: A shoe store sells each shoe at $60 when 1 or 2 shoes arebought but drops the price of the third or more shoes to $40. Then

Price per additional shoe =

{60 if 0 ≤ Q ≤ 240 if 2 < Q < ∞

}.

Here, there are N = 2 prices, and p1 = 60, p2 = 40, q0 = 0, q1 = 2 and q2 = ∞. Also V0 = 0, V1 =V0 + p1(q1 − q0) = 0 + 60(2) = 120, V2 = ∞, If you buy 3 shoes as opposed to 2 shoes, only the price of thethird shoe drops. That is why this is a marginal-unit discount scheme. ⋄

Marginal-Unit Quantity Mark-up Example: Consider an alternative tax regime where

Tax per additional dollar of taxable income =

28.0% if 0 < Q ≤ 183, 25033.0% if 183, 250 < Q ≤ 398, 35035.0% if 398, 350 < Q ≤ 400, 00039.6% if 400, 000 < Q < ∞

.

Here N = 4, and q0 = 0, q1 = 183, 250, q2 = 398, 350, q3 = 400, 000 and q4 = ∞. If your taxable income is$183,251 instead of $183,250, your tax rate for your first $183,250 remains 28% and the last dollar has a rate of

3

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33%. If your taxable income increases from $183,250 to $183,251, your tax increases by $0.33. Compare thisincrease of $0.33 to the increase of $9,162.83 under the currently used all-unit progressive tax regime of IRS,which is fairer? ⋄

Now,

Purchasing cost of Q units =

p1Q if 0 ≤ Q < q1V1 + p2(Q− q1) if q1 ≤ Q < q2. . . . . .VN−1 + pN(Q− qN−1) if qN−1 ≤ Q < qN

If qn−1 ≤ Q < qn, define the total costs in region n as

TCn(Q) :=h2(Vn−1 + pn(Q− qn−1)) +

RQ

S +RQ(Vn−1 + pn(Q− qn−1). (2)

It is instructive to compare, (2) to (1). The total cost over all regions is obtained by patching all TCn together,i.e.,

TC(Q) =

TC1(Q) if 0 ≤ Q < q1TC2(Q) if q1 ≤ Q < q2. . . . . .TCN(Q) if qN−1 ≤ Q < qN

.

In the marginal-units quantity discount problem, we want to solve

minQ≥0

TC(Q).

Note that the optimization problem is always minimization of TC(Q), what changes from no discount to all-unit discount and then to marginal-unit discount is TC(Q).

3. Preliminaries from Probability

3.1 Numerical events and random variables

Consider tossing a coin once. We do not know what the outcome will exactly be but we know that it is eitherHead (H) or Tail (T). The set of possible outcomes is called a sample space and is denoted by S. For a singletoss, S = {H, T}. For two tosses in a row, S = {HH, HT, TH, TT}. A real valued function X from the samplespace S to real variables is a random variable. For a single coin toss, X1 given as X1(H) = 0 and X1(T) = 1 israndom variable. X2(H) = 1 and X2(T) = 0 is another random variable. X1 denotes the number of tails andX2 denotes the number of heads in a single toss.

Example: Consider tossing a coin twice. The sample space is S = {HT, TH, HH, TT}. Let X1 be the num-ber of tails. Then X1(HT) = 1, X1(TH) = 1, X1(HH) = 0, X1(TT) = 2. Let X2 be the number of heads. ThenX2(HT) = 1, X2(TH) = 1, X2(HH) = 2, X1(TT) = 0. Let X3 = X1 − X2, then X3(HT) = 0, X3(TH) = 0,X3(HH) = −2, X3(TT) = 2. How would you define X3 in English. Suppose that X4 denotes the earnings in abetting situation where we earn $5 if H comes up and earn $10 if T comes up, i.e. X4(HT) = 15, X4(TH) = 15,X4(HH) = 10, X4(TT) = 20. If you let X1

4 and X24 be the earnings in the first and the second toss then

X14(HT) = 5, X1

4(TH) = 10, X14(HH) = 5, X1

4(TT) = 10 and X24(HT) = 10, X2

4(TH) = 5, X24(HH) = 5,

X24(TT) = 10. Note that X4 = X1

4 + X24 . Clearly, one can define various random variables using the same

sample space.

A discrete random variable can assume only a finite or countably infinite number of distinct values. All therandom variables presented above are discrete because they take a finite number of values. Here is an example

4

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of a discrete random variable with countably infinite values: Suppose that we are tossing a coin until H comesup, then S = {H, TH, TTH, . . . , T..TH, . . . }. Let X be the number of tosses until H appears so X(H) = 1,X(TH) = 2, X(TTH) = 3, etc. Clearly X assumes all positive integers. There infinite number of integers butthey are countable. Consequently, X is discrete. Some other examples of discrete random variables:

1. Number of oil reserves to be found in Texas in this year

2. Number of machines breaking down in a plant today

3. Number of BS students graduating this year

We represent probabilities associated with a random variable in a handy way by P(X = a) where a is aconstant. P(X = a) is the probability that the random variable X would assume value a. P(X = a) is calledthe probability mass function (pmf for short).

Example: Consider tossing a coin twice. Let X be the number of Tails coming up. The sample space, therandom variable and associated probabilities are:

S X ProbabilityHT 1 1/4TH 1 1/4TT 2 1/4

HH 0 1/4

P(X = a) =

1/4 if a = 01/2 if a = 11/4 if a = 20 if a = 3

Exercise: Suppose that 4 people including you and your friend line up at random. Find the pmf for X,

the number of people standing between you and your friend. Solution: P(X = 0) = 3/6, P(X = 1) = 2/6,P(X = 2) = 1/6. There are 4! = 24 ways of ordering 4 people. If you and your friend are back to back, we canstick two of you together and treat you as a super human. Then we have 3! = 6 ways of ordering the superhuman and the remaining two people. These 6 ways do not consider whether you are the first or your friendis. Taking who is first between you two into account, we obtain 12=2*6 ways. So in 12 ways out of 24, you willbe next to each other. That gives a probaility of P(X = 0) = 12/24.

Repeat this question with a total of 5 people. Solution: P(X = 0) = 4/10, P(X = 1) = 3/10, P(X = 2) =2/10 and P(X = 3) = 1/10.

Averages are generally used to summarize data. We often summarize a random variable with the expectedvalue and variance:

E(X) = ∑a

a · P(X = a) Var(X) = E(X− E(X))2

Expected value has a physical meaning: If you put weights of P(X = a) at each point a, then E(X) is thecenter of the gravity. No such intuitive reasoning can be provided for variance, except for saying that it is theexpected value of the square of the variation from the mean. Surely other measures than variance can be builtto measure variation.

Example: Consider tossing a coin twice. Let X be the number of Tails coming up.

E(X) = 0 · 1/4 + 1 · 1/2 + 2 · 1/4 = 1

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Var(X) = (0− 1)2 · 1/4 + (1− 1)2 · 1/2 + (2− 1)2 · 1/4 = 1/2

Expected value of functions of random variables can be computed using:

E(g(X)) = ∑a

g(a) · P(X = a)

where g is any function of the random variable X. The next example illustrates this concept.

Example: Reconsider the 4 people lining up at random, including you and your friend. Suppose you wantto pass a full glass of water to your friend and 1/4 of the water in the glass spills over at every pass. What isthe expected percentage of water your friend receives. Let us construct g first, let g(X) be the percentage ofthe water left in the glass after X passes. g(X = 1) = 3/4, g(X = 2) = (3/4)2 and g(X = 3) = (3/4)3. Then,

E(g(X)) = (3/4)(3/6) + (3/4)2(2/6) + (3/4)3(1/6) = 0.63

63% of the water remains and 37% is spilt. It is not hard to imagine applications of these ideas to communica-tion networks where data is distorted as it moves through the network.

Most commonly used discrete distributions are: Binomial and Poisson. The former is used to model num-ber of successes in a certain number of trials where success probability of each trial is constant and is indepen-dent of other trials. The latter is used to model the number of occurrences, say number of arrivals to a bankteller in a day. There are various other useful discrete densities such as geometric and multinomial that areused in other fields like quality control, reliability, etc.

A continuous random variable takes any value in an interval. Inches of rainfall in Dallas this month and theelapsed time until you have a flat tire are examples of continuous random variables. Let X denote any (discreteor continuous) random variable, the distribution function is F(x) and is defined by

F(x) = P(X ≤ x) for −∞ < x < ∞.

Discrete random variables have distribution function that look like a step function whereas continuous vari-ables have continuous distribution functions. In the case of a continuous random variable, when the derivativeof F can be obtained, it is called the probability density function or pdf for short and denoted by f (x):

f (x) =dF(x)

dx.

For a continuous random variable X, P(X = a) = 0 for any value of a. This is equivalent to the area of ofthe height of pdf at a, certainly the area of a line is zero. Indeed if P(X = a) > 0 then X is not continuous at a.For a continuous variable X:

P(a ≤ X ≤ b) = P(a < X ≤ b) = P(a ≤ X < b) = P(a < X < b) =∫ b

af (x)dx = F(b)− F(a)

Expected value and variances of continuous random variables are computed analogously to discrete ran-dom variables. If X is a continuous random variable,

E(X) =∫ ∞

−∞x · f (x)dx Var(X) = E(X− E(X))2 E(g(X)) =

∫ ∞

−∞g(x) · f (x)dx

where g is again a function of random variable X.

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Example: Let us find the expected value of a random variable X given by its pdf f (x):

f (x) ={

3x2/2 + x if 0 ≤ x ≤ 10 otherwise

}You may first want to check that this density is valid, i.e. it sums up to one:∫ ∞

−∞f (x)dx =

∫ 1

03x2/2 + xdx = (1/2)x3 + (1/2)x2|x=1

x=0 = 1.

Now the expected value:

E(X) =∫ ∞

−∞x · f (x)dx =

∫ 1

0x(3x2/2 + x)dx = (3/8)x4 + (1/3)x3|x=1

x=0 = 17/24.

Two common continuous distributions are Uniform and Normal. We say that X has a uniform distributionover the interval (a, b) and write X ∼ U(a, b) if the probability of X taking any value in the interval is constant,i.e. pdf is constant. To make sure that probabilities over (a, b) sum up to 1, we choose pdf as:

f (x) ={

1/(b− a) if a ≤ x ≤ b0 otherwise

}Also note that

F(x) =

0 if x < ax/(b− a) if a ≤ x ≤ b1 if x > b

Uniform distribution arises in practice if the knowledge about the random variable is very limited to the levelthat we only know its range (a, b) and nothing else. In this case, we assume that each outcome over the inter-val is equally likely. Many companies use such an approach in demand forecasting; Future demand forecastsare modelled by a uniform distribution over (a, b) where a (b) stands for the low (high) demand scenario.Although uniform distribution is simple to work with, it lacks a very important property that limits its justi-fication. That is, sum of two uniform random variables is not a uniform random variable. That is why otherdistributions especially Normal finds an extensive use in the practice.

Before we study Normal distribution let us see why it is so popular:

• Most observations are expected to be around the mean. Extreme observations are very rare. Because ofthis, normal is called a light-tail distribution.

• Sum of independent normal random variables is another normal random variable.

• Technically speaking averages of all independent random variables (as the number of variables in the av-erage grows) converge to normal random variable (this is known as central limit theorem). This propertyis extensively used in statistics.

Let us elucidate why we are interested in the sums of random variables. Suppose we are modelling the weeklydemand for a supplier. One approach is to figure out the weekly demands of each customer of the supplierand sum them up. This is demand aggregation over customers. Another approach is to find daily demands forthe supplier and sum them up over a week. This is temporal aggregation over days. Clearly one can aggregatethe demand over both customers and days. Important point is that we often deal with aggregated numberscoming from different sources. Then assuming a normal distribution for sum/average of numbers is very

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justifiable as such sum/average converges to a normal distribution.

If X is a normal random variable with mean µ and variance σ2, its pdf is

f (x) =1√

2π σexp(− (x− µ)2

2σ2 ) for −∞ < x < ∞

where exp is the exponential function. We write X ∼ N(µ, σ2).

Using this pdf form we can argue that linear transformations of normal random variables are normal butwith an appropriate scaling of parameters. Namely if α, β are constants and X ∼ N(µ, σ2) then αX + β ∼N(αµ+ β, α2σ2). This property, known as scaling of normal, becomes useful in computing normal probabilities:

P(a ≤ X ≤ b) = P(a− µ

σ≤ X− µ

σ≤ b− µ

σ) = P(

a− µ

σ≤ Z ≤ b− µ

σ) =

∫ b−µσ

a−µσ

ϕ(z)dz

where Z is a standard normal variable, i.e. Z ∼ N(0, 1) and ϕ(z) is the pdf for the standard normal variable.We use Φ for cdf of standard normal variable. Because of the above equality, we can use standard normaldensities to compute probabilities for any normal random variable. This is the reason why books tabulate onlystandard normal probabilities.

It also follows that we can use scaling to compute cumulative density function,

F(x) = P(X ≤ x) = P(X− µ

σ≤ x− µ

σ) = P(Z ≤ x− µ

σ) =

∫ x−µσ

−∞ϕ(z)dz = Φ

(x− µ

σ

)Because of this scaling normal probabilities for all normal variables can be counted via standard normal vari-able or its cdf Φ, which is often tabulated at the end of books.

Another set of equalities useful for us in inventory deals with computation of expected stockouts. Sup-pose that we currently hold ROP units on hand and facing random demand X. Expected stockout is E(X −ROP)+ := E max(X− ROP, 0). In general, expected shortages are found by numerically integrating∫ ∞

ROP(x− ROP) f (x)dx.

It may be conceptually useful to rewrite expected stockouts as∫ ∞

ROP(x− ROP) f (x)dx =

∫ ∞

ROP

∫ ROP

xdu f (x)dx =

∫ ∞

ROP

∫ ∞

uf (x)dxdu =

∫ ∞

ROP1− F(u)du

where we change the order of integration to obtain the second equality. However, these manipulations cannotsave us from numerical integration. In the case of Normal demand, we can avoid integrals, see the Formulason p.220 of Chopra.

As you may have already realized by now, manipulations with normal density is not easy. For example,F(x) cannot be put into a closed form expression. However, it is possible to obtain it with numerical integrationand all modern software -including Excel- has built-in functions to compute these probabilities.

Computing probabilities for X ∼ N(µ, σ2)Functions Matlab Excelpd f , f (x) normpd f (x, µ, σ) normdist(x, µ, σ, 0)cd f , F(x) normcd f (x, µ, σ) normdist(x, µ, σ, 1)inverse cd f , F−1(p) norminv(p, µ, σ) norminv(p, µ, σ)

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For a given p (0 ≤ p ≤ 1), inverse cdf finds the outcome value x such that the probability of observing smalleroutcomes than x is exactly p. Note that if p = F(x) then x = F−1(p). Note that Excel uses a single functionto compute both pdf and cdf, use the last argument of normdist to specify whether you want pdf or cdf.Discussion here has been purposefully very sketchy. More information on probability can be found in [4], [5]and [3].

4. Inventories with Uncertainty: ℜϵαδ the textbook

4.1 Cycle Service Level vs. Fill Rate

Cycle service level is a probability while fill rate is a ratio. Unfortunately, the third edition of Chopra & Meindlsays: “Fill rate is equivalent to the probability that product demand is supplied from available inventory” onp.307. This sentence is not accurate as it gives the impression that the fill rate is a probability. Please ignorethis sentence.

5. Postponement

Postpone each product as much as possible, ideally until the random demand for that product is observed.Recall the discussion about how the variances of the forecast for a given period reduce as time passes, seechapter SC03. By postponing, it is possible to learn more information about demand which effectively reducesdemand forecast variability. As we know from the above discussion, safety stocks are used to counter thevariability in the demand. Thus, lower safety stocks are needed when the variability is low. Consequently,postponement is used to reduce safety stock holding costs.

If we can single out a single customization — generally the highest value adding — stage, we can splitsupply chain into two before the customization and after the customization. Inventories before the customiza-tion have two significant properties. First, before the customization, raw materials or semi-finished assemblieshave low manufacturing or material costs. Consequently, inventory holding costs are lower for these items,so even when a high level of inventory is held, the cost may be relatively insignificant. Second, before thecustomization, raw material or semi-finished assemblies have more or less the same characteristics and func-tionality so they can be pooled into broader categories of inventories. Inventory pooling reduces the variabilityof the aggregated demand and hence the safety stock levels.

A word of warning is in order here. Reduction in demand variability is due to both the resolution ofuncertainty and inventory pooling. Many works in the literature fail to identify the distinction between thesetwo factors. They go ahead with a partial analysis of studying only inventory pooling effects on the aggregateddemand. Such a partial analysis could be defended by saying that resolution of uncertainty is not measured inalmost all of the instances in practice. However, this defense points out the need for resolution of uncertaintydata rather than the correctness of the partial analysis.

We use the term product family to refer to a group of products which are differentiated at the customizationstage. Postponement saves significant amount of money under the following conditions.

1. Products are almost indistinguishable before customization. That is, the higher the component common-ality before the customization, the more the variance reduction due to inventory pooling.

2. Customization stage adds a significant value to the product. This condition will make inventory hold-ing cost rates significantly low before the customization and high after the customization. Therefore,postponement becomes more of a cost cutting strategy. For example LCD displays add a high value tolaptops. If the stage where displays are assembled is the customization stage, savings due to postpone-ment will be higher.

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3. Demand is highly random and is fluctuating so forecasts are very variable. Therefore, safety stocks aresignificant and so will the savings in safety stock holding costs be, when postponement is implemented.

4. SC is flexible to allow for postponements to various stages/times. Resolution of uncertainty in customerdemands determine where to customize products in a product family. Customization should happenafter much of the uncertainty in the demand is resolved. The timing of the resolution of uncertainty canbe different. In Figure 1, uncertainty resolution in product family A happened before that in productfamily C. Thus, product family A can be customized earlier while the customization of product familyC needs to postponed further. Product in family C should be kept in a generic form and be customizedright before (or perhaps after, which is make-to stock policy) the customer demand. If the supply chain isnot flexible enough to postpone the customization of family C until after the last production stage, thenwe cannot reap all the potential benefits of postponement.

Family A

Family B

Family C

DueDate

Time

DemandVariability

Resolution ofUncertainty inthe Demand of

Family A

Resolution ofUncertainty inthe Demand of

Family B

Resolution ofUncertainty inthe Demand of

Family C

CustomizeFamily A

CustomizeFamily B

CustomizeFamily C

Make-to-orderFor Family C��������

Figure 1: Resolution of uncertainty (decrease in the variance) in product families A, B and C, and the appro-priate customization stage for them.

The flexibility we are referring to here is SC flexibility. Thus it includes flexibility of production lines,e.g. the extent to which we can modify the lines to postpone some operations. It also includes supplierflexibility, e.g. the extent to which supplier can quickly deliver orders. Since some operations are delayedpurposefully with the postponement strategy, the risk of missing customer deadlines is higher. It maybe necessary and also wise to incur some extra costs to obtain a better supplier service, such as frequentdeliveries from the supplier or high product availability at the supplier. Postponement puts similar“burdens” on procurement operations, sales departments, personnel, etc. For example H&P uses itswarehouses also as packing facilities to achieve delayed customization. This adds some extra responsi-bility to warehouse personnel and complicates the operations at the warehouses. Shortly, postponement

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stretches out SC, sometimes to the point of breaking. A functional IT infrastructure is also needed tosatisfy the needs of a stretched SC.

5. Products inside a product family have negative correlations. That is, they are substitute products. Neg-ative correlations further reduce the variance of the aggregated demand and lead to lower safety stocks.

Postponement, when implemented properly, has the potential to save inventory holding costs, obsoles-cence costs and to increase sales, forecast accuracy. On the other hand, modifying SC to implement post-ponement has its own costs. These costs range from redesigning products for postponement to inefficienciesintroduced into SC to achieve component commonality and to training cost of the personnel. Reader shouldnot get the impression that postponement always reduces total costs. Rather it is the proper analysis of thepostponement, trading off cost savings against extra costs due to postponement, which has the potential toreduce total costs.

6. Product Availability: ℜϵαδ the textbook

7. Optimal Wholesale Pricing PhD

Consider a supplier(S) and a retailer(R) subject to random demand D with pdf f , cdf F and F := 1− F. Let usdefine the following cost parameters:

• c: Supplier’s cost of supplying 1 unit. If supplier manufactures this unit, c is the marginal manufacturingcost. If the supplier simply buys the unit, c is the purchasing price for the supplier.

• w: Wholesale price that supplier charges to retailer.

• p: Market price that retailer charges to customers.

Naturally c ≤ w ≤ p. From supplier’s perspective, c and p are constant. However, the supplier can determinew. We will study finding optimal w to maximize the supplier’s profits. Note supplier’s profits depend on howmany units retailer purchases, call this y.

For every given value of w retailer reacts to set how many units it will buy from the supplier. For example, ifthe wholesale price is large, the retailer will buy less, i.e. y(w) decreases in w. Let us quantify this observation,first let ΠR(y, D) denote the profit of retailer if it buys y and the demand is D.

ΠR(y, D) = p(y ∧ D)− wy

ΠR(y, D) is a random function, taking its expected value:

ΠR(y) := E(ΠR(y, D)) = p∫ ∞

0(y ∧ D) f (D)dD− wy = p

∫ ∞

0

∫ y∧D

x=0dx f (D)dD− wy

= p∫ y

0

∫ ∞

xf (D)dDdx− wy = p

∫ y

0F(D)dD− wy.

It is easy to show that ΠR(y) is concave, so it is maximized at y∗ = F−1(w/p). In order to emphasize thedependence of the retailer’s order size, we use the notation y∗(w) = F−1(w/p).

We now study supplier’s profit ΠS(w),

ΠS(w) = (w− c)(

arg maxy

ΠR(y))= (w− c)y∗(w) = (w− c)F−1(w/p)

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The first order condition for a critical w is:

dΠS(w)

dw= F−1(w/p) +

w− cpF′(F−1(w/p))

Now remember that F′ = − f , y∗ = F−1(w/p) and w = pF(y∗).

dΠS(w)

dw= y∗ − pF(y∗)− c

p f (y)

Therefore the optimal order quantity for the retailer when the wholesale price maximizes supplier’s profit, y∗,satisfies:

F(y∗)(

1− y∗ f (y∗)F(y∗)

)=

cp

(3)

Once this y∗ is found than the optimal wholesale price is simply:

w = pF(y∗) (4)

As (4) is a simple equation, we focus on (3). For the existence and uniqueness of y∗, consider f (y)/F(y) insidethe parentheses in (3). This ratio is nothing but the failure rate function of the demand density. For a momentsuppose that this failure rate is increasing in y, than the entire term inside the parentheses decreases in y. Inaddition F(y) decreases in y. Then the left hand side of (3) is decreasing in y while the right hand side isconstant. As y increases from 0 to ∞, the left hand side decreases from 1 to zero and hits c/p at a unique y.These observations yield the following proposition of [2].

Proposition 1. If demand distribution has increasing failure rate, then the unique wholesale price maximizing supplier’srevenue is found by solving (3) and (4).

8. Solved Examples

8.1 Statement of Exercises

1. [All-units quantity discounts] A popular shoe store sells 8000 pairs per year. The fixed cost of orderingshoes from the distribution center is $15 and holding costs are taken as 25% of the shoe costs. The perunit purchase costs from the distribution center is given as

C1 = 60, if 0 ≤ Q ≤ 50C2 = 55, if 50 ≤ Q ≤ 150C3 = 50, if 150 ≤ Q

where Q is the order size. Determine the optimal order quantity.

2. [Marginal units quantity discounts] Refer to the previous exercise, suppose the distribution center givesmarginal unit discount and compute the optimal order quantity.

3. [Harvey’s Speciality Shop] Harvey’s Specialty Shop is a popular spot that specializes in internationalgourmet foods. One of the items that Harvey sells is a popular mustard which be purchases from anEnglish company. The mustard costs Harvey $10 a jar, and requires a six-month lead time for replenish-ment of stock. Harvey uses a 20 percent annual interest rate to compute holding cost, and estimates thatif a customer requests the mustard when he is out of stock the loss-of-goodwill cost is $25 a jar. Book-keeping expenses for placing an order amount to about $50. During the six-month replenishment lead

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time, Harvey estimates that he sells an average of 100 jars, but there is substantial variation from one six-month period to the next. He estimates that the standard deviation of demand during each six-monthperiod is 25, assume that demand is described by a normal distribution. How should Harvey control thereplenishment of the mustard?

4. [Arts and Humanities Coffee] In front of the Arts and Humanities secretarial office coffee is brewed andoffered to general public at a cost of 25 cents per cup. Coffee powder is bought in bags and each bag costs2 dollars and contains enough coffee to prepare 10 cups. Suppose that the coffee demand over every twohours is normally distributed with expected value of 6 and variance of 1. The office is open for 8 hoursevery day and does not provide coffee outside the office hours. Suppose that any coffee drinker whoarrives when the office is out of coffee joins a queue to wait for the next coffee preparation. Thus, atthe beginning of every 2-hour time blocks, there can be a bunch of people who arrived earlier but havewaited for the coffee.a) Assume that demand in every two hour period is independent of other period’s demand, find thedistribution of the coffee demand per day.b) Suppose the office uses 3 coffee bags every day to prepare coffee. First find out how many cups areprepared and then compute the probability of coffee stockout.c) If the office uses 3 coffee bags every day, find the safety sock level and the expected number of peopleper day that cannot get coffee.d) Briefly explain why independence assumption of (a) might be flawed.

5. [UTD Bookstore] UTD Bookstore is trying to determine how many Supply chain books to order for thenext year. The book store buys the books at 50 dollars each and sells at 80 dollars. Any unsold book isbought back by Prentice Hall at 40 dollars. It is estimated that the class size will be some where between(and including) 21 and 32. In this range, the class size can take any integer value with equal probabilities.a) What is an appropriate distribution for book demand, specify the distribution with its parameters.b) Compute how many books should be ordered.c) Compute the marginal cost of ordering one more book beyond the number you found in (b).d) What is the total cost of understocking and overstocking with the number of books found in (b)?e) If you were the manager in charge of ordering books at the bookstore, what steps would you considerto reduce the costs in (d) ?

6. [Light bulbs from Taiwan] Philips manufactures lighting products (various lamps) in Taipei, Taiwanand ships them to Balkan countries for sale. The weekly sales in these countries are independently andNormally distributed with mean demand and standard deviation as below:

Country Mean Standard deviationMacedonia 100 40Romania 300 80

Currently, Philips packs the lamps with manuals in each country’s language in Taiwan. These packagesare sent to DCs in each country to meet that country’s demand. The transportation time from Taiwan tothese countries are 6 weeks.a) How much safety stock is needed in each country for a CSL of 90%?b) Suppose Philips establishes a regional DC in Romania to meet both Romanian and Macedonian de-mand. The manuals are quickly put into packages in this DC after the retailer orders in each country areobserved. Thus, Philips produces to stock and the regional DC orders to stock while the country specificcustomization happens according to retailer orders. What is the distribution of the weekly demand forthe regional DC? Compute the safety stock savings with regard to a) if the CSL of 90% is still maintained.

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7. [Relating overstock and understock] For a better understanding and comparison of overstock and under-stock, let us compare demand D and order quantity Q in a single period (season) context. Suppose that Qis given and is not necessarily optimal for our context. Further suppose that the demand is nonnegativeand is less than 100. By the definitions of understock and overstock, recall that

E(understock) = E(D−Q)+ = E(max{0, D−Q}) and E(overstock) = E(Q−D)+ = E(max{0, Q−D}).

a) Further suppose that Q = 10 units and consider the function y1(D) = (D− 10)+. Note that this is afunction of one variable (i.e. D) and is almost a line. Note that y1(D) can also be written as

y1(D) = max{D− 10, 0}.

Express y1(D) in your words. Draw y1(D) as D varies (x-axis is D, y-axis is y1(D)).b) Keep Q = 10 and define and y2(D) = (10− D)+ and y3(D) = D− 10 . Express y2(D) and y3(D) inyour words. Draw y2(D) and y3(D) onto the same coordinate system used in a). You may use pencils ofdifferent colors or different line styles.c) Now looking at your drawings determine if

i. (D− 10)+ − (10− D)+ = D− 10 or ii. (D− 10)+ + (10− D)+ = D− 10.

is correct. Express the correct equality in your words.d) Now take the expected value of all the terms in the correct equality to obtain either

i′. E(D− 10)+ − E(10− D)+ = E(D− 10) or ii′. E(D− 10)+ + E(10− D)+ = E(D− 10).

Express in your words the correct equality.e) Explain why your conclusion in d) will not change if Q = 10. Then for any order quantity

i′′. E(D−Q)+ − E(Q− D)+ = E(D−Q) or ii′′. E(D−Q)+ + E(Q− D)+ = E(D−Q).

Express in your words the correct equality. With this example, you have just found an important equalityinvolving understock and overstock. The equality can be used to compare these quantities.

8. [Tailored two product delivery] A retailer buys products A and B from a supplier. Product A is demandeda lot with respect to product B so the retailer receives a truck of product A delivery every Monday. Itis not clear whether B should be in every truck (relative frequency nB = 1). When product B is carriedon a truck, there is a loading/unloading (handling) cost of $4000 for the retailer. Moreover, the weeklydemand for B is 800 units and the holding cost rate per week per unit is $1 at the retailer. The fixed costof handling product B is s = 4000. Let us fix time unit to a single week. Then demand is R = 800 perweek, and the holding cost H = 1 per week per unit.a) Using the relative order frequency nB of product B, express the total B handling and inventory holdingcost over nB weeks.b) Use TC(nB; over nB weeks) above to express the total cost (handling and holding) over 1 week. Thisis the average cost incurred per week.c) When finding the relative frequency n0

B, decide whether we should minimize TC(nB; over nB weeks) orTC(nB; over 1 week). In a sentence, explain the rationale behind your decision. By taking the derivativeof the appropriate objective function and setting equal to zero, compute n0

B.d) The frequency n0

B computed in c) may not be an integer. Suppose that it is n0B = 5/3. Intuitively this

may call for three relative frequencies n1B = 1, n2

B = 1, n3B = 3 and trucking cycles with period of 5 weeks:

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Cargo of a truck on each week5-week cycle 1 5-week cycle 2 5-week cycle 3

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15A A A A A A A A A A A A A A AB B B B B B B B B

Above table indicates that in the 1st, 2nd and the 3rd weeks of the 5-week cycles A and B are received,in the 4th and 5th week only A is received. Express and compute the total cost (handling and holding) Bover 5 weeks that will result from the schedule above. Also compute the “average” total cost incurred perweek. In particular, TC(Schedule in the table; over 5 weeks) = ? TC(Schedule in the table; over 1 week) =?e) Compute TC(nB = 5/3; over 1 week) by using your expression in b).f) Compare TC(Schedule in the table; over 1 week) in d) and TC(nB = 5/3; over 1 week) in e). Explainif they need to be the same.

8.2 Solution of Exercises

ANSWER for Exercise 1There are three ranges for lot sizes in this problem: (0, q1 = 50), (q1 = 50, q2 = 150) and (q2 = 150, ∞).

Holding costs in there ranges of shoe prices are given as h1 = (0.25)60 = 15, h2 = (0.25)55 = 13.75 andh3 = (0.25)50 = 12.5. EOQ quantities in these ranges are

EOQ1 =

√2(15)(8000)

15= 126.5; EOQ2 =

√2(15)(8000)

13.75= 132.1; EOQ3 =

√2(15)(8000)

12.5= 138.6;

Only EOQ2 = 132.1 is in the appropriate range,i.e. q1 ≤ EOQ2 ≤ q2, so it is a candidate solution. SinceEOQ1 > q1, we take q1 = 50 as the candidate solution for the second range. Since EOQ3 < q2, we takeq2 = 150 as the candidate solution for the third range.

It is clear that ordering Q = 50 at the cost C1 = 60 is worse than ordering Q = 50 at the cost C2 = 55.Moreover ordering Q = 132.1 is better than ordering Q = 50 in the second range. Combining the last twostatements, the costs at Q = 132.1and C2 = 55 is smaller than the costs at Q = 50and C2 = 60. Therefore wecan eliminate Q = 50 from the consideration. Evaluating the remaining lot sizes

TC2(Q = 132.1) = 8000(55) + 8000(15)/132.1 + (0.25)(55)(132.1)/2 = 441, 800

TC3(Q = 150) = 8000(50) + 8000(15)/150 + (0.25)(50)(150)/2 = 401, 900

Then Q = 150 is the optimal solution with a cost of 401,900.

ANSWER for Exercise 2V0 = 0, V1 = V0 + (50)60 = 3000 and V2 = V1 + 100(55) = 8500 are cost of buying exactly 0, q1 = 50 and

q2 = 150 units. Then EOQs are

EOQ1 =

√2R(S + V0 − q0c0)

hC1=

√2(8000)(15 + 0− 0)

15= 126.5;

EOQ2 =

√2R(S + V1 − q1c1)

hC2=

√2(8000)(15 + 3000− (50)55

13.75= 555.3;

EOQ3 =

√2R(S + V2 − q2c2)

hC3=

√2(8000)(15 + 8500− (150)50)

12.5= 1139.8;

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For the first and second ranges candidates are Q = 50 and Q = 150. Then TC1(Q) is decreasing over 0 ≤Q ≤ 50 and similarly TC2(Q) is decreasing over 50 ≤ Q ≤ 150. For the third range Q = 1139.8. We do notneed to compute TC1(Q = 50) or TC2(Q = 150) because they are larger than TC3(Q = 1139.8) is decreasing.However we write these costs down for the sake of exercise.

TC1(Q = 50) = (8000/50)15 + (0 + (50)60)0.25/2 + (8000/50)(0 + (50)60)

TC2(Q = 150) = (8000/150)15 + (3000 + (100)55)0.25/2 + (8000/150)(3000 + (100)55)

TC3(Q = 1139.8) = (8000/1139.8)15+(8500+(1139.8− 150)50)0.25/2+(8000/1139.8)(8500+(1139.8− 150)50)

= 421, 266

Then Q = 1139.8 is the optimal solution with a cost of 421,266.

ANSWER for Exercise 3We wish to find the optimal values of the reorder point R and the lot size Q. In order to get the calculation

started we need to find the EOQ. However, this requires knowledge of the annual rate of demand, which doesnot seem to be specified. But notice that if the order lead time is six months and the mean lead time demand is100, that implies that the mean yearly demand is 200,giving a value of R = 200. It follows that

EOQ =√

2SR/hC =√

2 ∗ 50 ∗ 200/(0.2 ∗ 10) = 100.

The next step is to find ROP from Equation

CSL = F(ROP) = P(Demand ≤ ROP) = 1− QhCbR

.

Substituting Q = 100, we obtain

P(Demand ≤ ROP) = 1− QhCbR

= 1− 100 ∗ 0.2 ∗ 1025 ∗ 200

= 0.96.

Since the demand is normally distributed,

ROP = Norminv(0.96, 100, 25) = 144.

As a result, we can run place orders of size Q = 100 whenever inventory level is below ROP = 144.

ANSWER for Exercise 4a) Because of independence, the sum of the demands over 4 2-hour blocks remain to be normally distributedand its mean and variance are 24=6+6+6+6, 4=1+1+1+1. The daily demand has N(24, 22) distribution.b) With 3 coffee bags everyday, 30 cups are prepared. The stockout happens when these 30 cups are notenough. That probability is given by P(N(24, 22) ≥ 30) = 1− P(N(24, 22) ≤ 30) = 1− normdist(30, 24, 2, 1) =1− 0.99865.c) With 3 coffee bags, the safety stock is 30-24=6. We now need to compute the expected number of stockoutsper day

ESC = E(max{0, N(24, 22)− 30})= −ss[1− normdist(ss/σ, 0, 1, 1)] + σnormdist(ss/σ, 0, 1, 0)= −6[1− normdist(6/2, 0, 1, 1)︸ ︷︷ ︸

0.99865

] + 2 normdist(6/2, 0, 1, 0)︸ ︷︷ ︸0.004432

= 0.000764

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The expected shortage per day is low as consequence of high safety stock of 6.If we reduce the safety stock down to 4, we can see what happens to the corresponding ESC:

ESC = −4[1− normdist(4/2, 0, 1, 1)︸ ︷︷ ︸0.97725

] + 2 normdist(4/2, 0, 1, 0)︸ ︷︷ ︸0.053991

= 0.016981

With a safety stock of 2, ESC grows further to 0.166631.

ANSWER for Exercise 5a) The demand takes values from {21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32} with equal probability (of 1/12) soit has discrete uniform distribution.b) The overstocking and understocking costs are co = 10 dollars and cu = 30 dollars. We need to solve forP(Demand ≤ Q) = cu/(cu + co) = 0.75 = 9/12. This equality yields Q = 29.c) The marginal cost of ordering the 30th book is coP(Demand ≤ 29) = 10(9/12) = 15/2. If you also wantto compute the marginal benefit of ordering the 30th book, then cuP(Demand ≥ 30) = 30(3/12) = 15/2.In this case of discrete demand, the equivalence of marginal cost and benefit is a coincidence. In the case ofcontinuous demand distribution, this equivalence is actually the optimality condition.d) The total cost of overstocking and understocking is

TC(Q = 29) = coE(max{0, 29−Demand}) + cuE(max{0, Demand− 29})= 10[(29− 28)/12 + (29− 27)/12 + · · ·+ (29− 21)/12]

+30[(32− 29)/12 + (31− 29)/12 + (30− 29)/12]= 10[3] + 30[0.5] = 45.

e) The manager can consider decreasing the cost overstocking (with a higher buyback price), decreasing thecost of understocking (with transshipments from other bookstores) and fast response from Prentice-Hall sothat two or more orders can be placed in a single semester.

ANSWER for Exercise 6a) In Macedonia, the safety stock is given by ssM = norminv(0.9, 0, 1)

√6 40. For Romania, we have ssR =

norminv(0.9, 0, 1)√

6 80.b) The weekly demand is N(100, 402) + N(300, 802), which itself is normal with mean 400 = 100 + 300 andvariance 8000 = 1600 + 6400. Now the safety stock becomes ssM+R = norminv(0.9, 0, 1)

√6√

8000. The safetystock saving is norminv(0.9, 0, 1)

√6(40 + 80−

√8000).

ANSWER for Exercise 7a) y1(D) = max{D− 10, 0} is the random variable for understock (the number of stockouts) if we order 10. Todraw y1(D), put D on x-axis and y1(D) on y-axis. Draw two lines y = D − 10 and y = 0. The maximum ofthese two lines is y1(D).b) y2(D) = (10− D)+ = max{10− D, 0} is the random variable for overstock (the number of leftover in theinventory at the end of the season) if we order 10. y3(D) = D− 10 is the random variable for the negative ofthe safety stock. To draw y2(D), first draw two lines y = 10− D and y = 10. The maximum of these two linesis y2(D). Since y3(D) is a simple line, it is straightforward to draw it. c) If you have the drawings, you can seethat

(D− 10)+ − (10− D)+ = D− 10

or that(D− 10)+ + (10− D) = (10− D)+

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which says that understock plus safety stock is overstock.d) Taking the expected value, we obtain

E(D− 10)+ − E(10− D)+ = E(D− 10),

which says that the expected value of understock plus the expected value of safety stock is the expected valueof overstock.e) The conclusion in d) will not change if Q = 10 because we never used Q = 10 in arriving this conclusion.For example, with Q = 5, understock plus safety stock is still overstock. For any order quantity

E(D−Q)+ − E(Q− D)+ = E(D)−Q.

With this example, you have just found an important equality involving understock and overstock. The equal-ity can be used to compute/compare these quantities.

ANSWER for Exercise 8a) With a relative order frequency of nB, product B is ordered very other nB weeks. Thus, the order size mustsuffice for nB weeks. The order size then is nBR. Since the retailer inventory at maximum is nBR and drops tozero at the uniform rate of R, the average inventory is nBR/2.

TC(nB; over nB weeks) = s︸︷︷︸Fixed ordering cost

+ nBR/2︸ ︷︷ ︸Average inventory

nB︸︷︷︸kept nB weeks

hC︸︷︷︸at the holding cost rate hC︸ ︷︷ ︸

Inventory holding cost over nB weeks

= 4000 +12

800(1)n2B,

where H = hC = 1 is holding cost per item per week.

b)

TC(nB; over 1 week) =TC(nB; over nB weeks)

nB=

snB

+12

RhCnB =4000nB

+12

800(1)nB.

c) When finding the relative frequency, we cannot use an objective that depends on the relative frequency;That is circular logic. In management, our objectives are often computed over fixed and independent (fromthe decisions) time intervals. Such a time interval is 1 week. If you wish you can also use 2 weeks or 5 weeksas your time intervals, in these cases you must multiply the objective below by 2 and 5 respectively and recallthat multiplying an objective by a positive number does not affect the outcome of the optimization.

We minimize TC(nB; over 1 week) to find nB:

ddnb

TC(nB; over 1 week) = −4000n2

B+

12

800(1) = 0.

So,

nB =

√4000400

=√

10.

d)

TC(Schedule in the table; over 5 weeks) =

Week 1︷ ︸︸ ︷4000 +

12

800(1)12︸ ︷︷ ︸n1

B=1

+

Week 1︷ ︸︸ ︷4000 +

12

800(1)12︸ ︷︷ ︸n2

B=1

+

Weeks 3−4−5︷ ︸︸ ︷4000 +

12

800(1)32︸ ︷︷ ︸n3

B=3

= 12000 + 4400 = $16, 400

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TC(Schedule in the table; over 1 week) =TC(Schedule in the table; over 5 weeks)

5= $3280

e)

TC(nB; over 1 week) =40005/3

+12(800)(1)5/3 = 2400 + 2000/3 ≈ $3066.

f) The costs are not the same. Although the average of n1B = 1, n2

B = 1, n3B = 3 is n0

B = 5/3, the con-tribution of longer cycles to the holding cost is not proportional to the length of the cycle. Indeed, thiscontribution is proportional to the square of the cycle length. A longer B-cycle like n3

B = 3 causes dis-proportionately more costs than their lengths. Since the schedule in the table has such a cycle, we expectTC(Schedule in the table; over 1 week) > TC(nB = 5/3; over 1 week). Note that n0

B = 5/3 is fractional andcannot be implemented, TC(Schedule in the table; over 1 week) − TC(nB = 5/3; over 1 week) > 0 can becalled the implementability cost.

9. Solved Exercises from Operations Management

9.1 Statement of Exercises

1. [Walton Bookstore] In August, Walton Bookstore must decide how many of next year’s nature calendarsshould be ordered. Each calendar costs the bookstore $2 and is sold for $4.50. After January 1, anyunsold calendars are returned to the publisher for a refund of $0.75 per calendar. Walton believes thatthe number of calendars sold by January 1 follows the probability distribution shown in following table.

Number of Calendars Sold 100 150 200 250 300Probability 0.3 0.2 0.3 0.15 0.05

Walton wants to maximize the expected net profit from calendar sales. How many calendars should thebookstore order in August?

2. [Childcare] UT Dallas allows an employee to put an amount into an account at the beginning of eachyear, to be used for child-care expenses. This amount is not subject to federal income tax. Assume thatall other income is taxed by the federal government at a 40% rate. If this amount can only be usedfor childcare expenses. If the amount is more than the childcare expenses, the difference is lost. If thechildcare expenses are more than the amount, the employee must pay for the excess out of his/her ownpocket. This excess payment can be claimed in an income tax return to receive tax credit at the rate of25% of the excess payment. Prof. Cakanyıldırım believes that his childcare expenses for his son for thecoming year will be $3000, $4000, $5000, $6000, or $7000 with equal probabilities. At the beginning of theyear, how much money should he place in the child-care account?

3. [Elevators] At the western side of the SOM, there are two elevators. Whenever one calls for an ele-vator from a certain floor, it seems like the elevators are at a different floor. An OM student team iscommissioned to decide on the optimal floor for both elevators. That is, after carrying people to theirdestination floor, the elevators will return to their optimal floor to wait for forthcoming people unlessthey are already called by some people. The OM student team found out the distribution of the elevatorcalls coming from the four floors of SOM:

P(Call from the 1st floor) = 0.4, P(Call from the 2nd floor) = 0.1,P(Call from the 3rd floor) = 0.2, P(Call from the 4th floor) = 0.3.

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Suppose that the elevators, while moving, has a constant speed of 1 floor every 4 seconds. If the elevatorsare both placed at the 2nd floor, the expected waiting time for an elevator would be

E[Waiting time if elevators at the 2nd floor] = 4[0.40|2− 1|+ 0.1|2− 2|+ 0.2|2− 3|+ 0.3|2− 4|]= 4[0.4 + 0.2 + 0.6] = 4.8 seconds.

The OM team aims to minimize the waiting time for the forthcoming elevator calls. Cast this problem asa newsvendor problem and find the optimal location of the elevators.

4. [2 Elevators] In Exercise 3, we addressed the optimal location by assuming that both elevators will wait atthe same floor. Now we relax this assumption by saying that elevator A can wait at a certain floor whileelevator B waits at another floor. It is easy to convince ourselves that we can obtain a better solution byhaving elevators wait at different floors. What are the optimal floor locations for the elevators?

5. [Hotel Reservations] A hotel near a university always fills up on the evening before football games.History has shown that when the hotel is fully booked, the number of last-minute cancellations has amean of 5 and standard derivation of 3. The average room rate is $80. When the hotel is overbooked,policy is to find a room in a nearby hotel any to pay for the room for the customer. This usually coststhe hotel approximately $200 since rooms booked on such late notice are expensive. How many roomsshould the hotel overbook?

9.2 Solutions

ANSWER for Exercise 1:Let q be the number of calendars ordered in August and D be the number of calendars demanded by

January 1. If D ≤ q, the costs shown in following table are incurred where the revenue is negative cost.

For D ≤ q costBuy q calendars at $2/calendar 2qSell D calendars at $4.50/calendar -4.50DReturn q− D calendars at $0.75/calendar -0.75(q− D)Total cost 1.25q− 3.75D

Thus, the overage cost is co = 1.25 which is the multiplier of q in the total cost above.

If D ≥ q + 1, the costs are:

For D ≥ q + 1 costBuy q calendars at $2/calendar 2qSell D calendars at $4.50/calendar -4.50qTotal cost -2.50q

From the multiplier of q, the underage cost is cu = 2.5. Then

cu

co + cu=

2.503.75

=23

.

Walton should order q∗ calendars, where q∗ is the smallest number for which P(D ≤ q∗) ≥ 2/3. Note that

P(D ≤ 100) = .30, P(D ≤ 150) = 0.50, P(D ≤ 200) = 0.80.

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Hence, q∗ = 200 calendars.

ANSWER for Exercise 2:Let q be the amount put aside for the childcare and d be the childcare expense. If d ≤ q, the costs shown in

following table are incurred where the revenue is negative cost.

For d ≤ q costTax avoided -0.4qLost money q-dTotal after-tax cost 0.6q-d

Thus, the overage cost is co = 0.6 which is the multiplier of q in the total cost above.

If d ≥ q, the costs are:

For d ≥ q costTax avoided -0.4qTax credit -0.25(d-q)Total after-tax cost -0.25d-0.15q

From the multiplier of q, the underage cost is cu = 0.15. Then

cu

co + cu=

0.150.75

=15

.

Since P(Childcare expense = 3000) = 0.2, Prof. Cakanyıldırım should put aside 3000 for childcare.

ANSWER for Exercise 3:To cast the elevator problem as a newsvendor problem, we first let Q be the optimal floor and let D be the

floor the next call comes from. The empirical distribution of D is given in the problem statement. If Q > D,the elevator is located above where the call happens. In this case, the elevator travels Q− D in 4(Q− D) sec-onds. By increasing Q, we can reduce the waiting time by 4 seconds, so the underage cost is cu = 4 seconds.Similarly, the overage cost is co = 4 seconds. Then the critical ratio is 0.5. In view of the empirical distributionof D, we have P(D ≤ Q) = 0.5 when Q = 2. In other words, the elevator must go to the second floor to waitfor forthcoming calls.

ANSWER for Exercise 4:Let QA and QB the waiting location of the elevators. It is safe to assume that when one calls for an elevator,

the closest elevator goes to pick that person up. If two elevators are equidistant from the person, one of theelevators can be sent arbitrarily. For (QA = 1, QB = 2), we have the waiting time W(QA = 1, QB = 2), whoseexpected value can be computed (in units of 4 seconds) as follows.

EW(QA = 1, QB = 2) = 0.2 ∗ 1 + 0.3 ∗ 2 = 0.8.

In general,

EW(QA, QB) =4

∑i=1

P(D = i)min{|i−QA|, |i−QB|}

= 0.4 ∗min{|1−QA|, |1−QB|}+ 0.1 ∗min{|2−QA|, |2−QB|}+ 0.2 ∗min{|3−QA|, |3−QB|}+ 0.3 ∗min{|4−QA|, |4−QB|}.

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Specializing this formula to the specific QA and QB values, we obtain:

EW(1, 3) = 0.1 ∗ 1 + 0.3 ∗ 1 = 0.4, EW(1, 4) = 0.1 ∗ 1 + 0.2 ∗ 1 = 0.3,EW(2, 3) = 0.4 ∗ 1 + 0.3 ∗ 1 = 0.7, EW(2, 4) = 0.4 ∗ 1 + 0.2 ∗ 1 = 0.6,EW(3, 4) = 0.4 ∗ 2 + 0.1 ∗ 1 = 0.9.

Among all the floors to place the elevators, (QA = 1, QB = 4) yields the lowest waiting time. Thus, one ofthe elevators should wait on the first floor, the other should wait on the fourth floor.

ANSWER for Exercise 5:The cost of underestimating the number of the cancellations is $80 and the cost of overestimating cancella-

tions is $200.

P(Cancellations ≤ Overbooking) =cu

co + cu=

80200 + 80

= 0.2857.

Using Normsinv(0.2857) in Excel gives a z-value of -0.56599. The negative value indicates that we shouldoverbook by a value less than the average of 5. Number of overbooked rooms = 5 + 3(-0.566) = 3.302.

Another common method for analyzing this type of problem is with a discrete probability distributionfound using actual data and marginal analysis. For our hotel, consider that we have collected data and ourdistribution of no-shows is as follows.

Cancellations 0 1 2 3 4 5 6 7 8 9 10Probability 0.05 0.08 0.10 0.15 0.20 0.15 0.11 0.06 0.05 0.04 0.01

Using these data, a table showing the impact of overbooking is created. Total expected cost of each over-booking option is the calculated by multiplying each possible outcome by its probability and summing theweighted costs. The best overbooking strategy is the one with minimum cost.

Number Proba- Number of OverbookingsCancelled bility 0 1 2 3 4 5 6 7 8 9 10

0 0.05 0 200 400 600 800 1000 1200 1400 1600 1800 20001 0.08 80 0 200 400 600 800 1000 1200 1400 1600 18002 0.10 160 80 0 200 400 600 800 1000 1200 1400 16003 0.15 240 160 80 0 200 400 600 800 1000 1200 14004 0.20 320 240 160 80 0 200 400 600 800 1000 12005 0.15 400 320 240 160 80 0 200 400 600 800 10006 0.11 480 400 320 240 160 80 0 200 400 600 8007 0.06 560 480 400 320 240 160 80 0 200 400 6008 0.05 640 560 480 400 320 240 160 80 0 200 4009 0.04 720 640 560 480 400 320 240 160 80 0 200

10 0.01 800 720 640 560 480 400 320 240 160 80 0TotalCost 337.6 271.6 228 212.4 238.8 321.2 445.6 600.8 772.8 958.8 1156

From the table, the minimum total cost is 212.4 when 3 extra reservations are taken. This approach using dis-crete probability is useful when valid historic data are available.

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10. Exercises

1. [Cost sensitivity] Let the total annual holding and ordering cost of an inventory system be TC(Q)

TC(Q) = (R/Q)S + (Q/2)hC

which is minimized by Q∗ = EOQ.a) Compute TC(2Q∗)/TC(Q∗).b) Compute TC(Q∗/2)/TC(Q∗).c) In (a) and (b) you computed the sensitivity of total costs to using a nonoptimal lot size. In the first caseit was 2Q∗ and in the second Q∗/2. According to your results, is TC(Q) sensitive to Q?

2. Solve Chopra exercise 4 of Chapter 10 on p.297.

3. Solve Chopra exercise 5 of Chapter 10 on p.297.

4. List and compare the motives of Kraft and American Airlines for trade promotions/discounting.

5. Consider an all-unit discounting scheme with two prices p1 and p2, p1 ≥ p2 where

Price per unit ={

p1 if Q < qp2 if Q ≥ q

}Clearly q is measured in units and the price drops to p2 from p1 at q. Suppose that EOQ1 and EOQ2refer to the EOQ quantities for regions Q < q and Q ≥ q respectively. Further suppose that EOQ1 ≥ q.Argue in English or by a figure that the optimal order quantity with the discount would be EOQ2. Hint:Compare the cost at Q < q to cost at Q = q. Then compare the cost at Q = q to the cost at Q = EOQ2.

6. Consider an all-unit discounting scheme with N prices, {pn : 1 ≤ n ≤ N}, which are monotonicallyordered by pn−1 ≥ pn for 1 < n ≤ N. Suppose that EOQn refers to the optimal order quantity with pricepn. Prove that EOQn is nondecreasing in n: EOQn−1 ≤ EOQn for 1 < n ≤ N.

7. Consider an all-unit discounting scheme, prove that TCn(Q) ≥ TCn+1(Q) for any lot size Q. Express thedifference TCn+1(Q)− TCn(Q) in terms of the cost parameters.

8. PhD Validation of the ordering algorithm in Table 1. In order for us to prove the validity of the algorithm,we need to argue that no candidates need to be considered after these steps.a) Validation of step A2. Suppose that the loop stops with n = nA in step A2. Prove that the total costTC(EOQnA) cannot be larger than any one of the costs TC(EOQn), TC(qn−1), TC(qn) for n < nA. Hint:Use the last exercise.b) Validation of step C. Suppose that the loop stops with n = nC in step C. Prove that the total costTC(qnC) cannot be larger than any one of the costs TC(EOQn), TC(qn−1), TC(qn) for n < nC. Alsoprove that TC(qnC) cannot be smaller than the total cost evaluated at the candidate qnC picked up forn = nC + 1.

9. [Discounting to Coordinate] Consider the online retailer DO which faces a customer demand of 10,000boxes of vitamins per month. DO incurs a fixed order placement (with the manufacturer) cost of 100dollars and incurs a holding cost rate of 20% over the purchase price of each per year. DO buys a boxof vitamins at 3 dollars from the manufacturer. Each time DO places an order, the manufacturer has toprocess the order which costs a 250 to the manufacturer. The manufacturer also incurs a 20% holdingcost rate and manufactures a box of vitamins for 2 dollars.a) Find the EOQ quantity for DO. Compute DO’s and the manufacturer’s holding and ordering costs per

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year.b) Now consider an all units discount scheme offered by the manufacturer to the DO: DO will pay theregular price of 3 dollars for purchases in quantities below 9165 and will pay 2.9978 dollars for purchasesin larger quantities. Find the optimal order quantity for DO under the new pricing scheme.c) Compute DO’s holding, ordering and purchase costs when the order quantity in b) is used. Alsocompute the manufacturer’s profit (= revenue made from DO - production costs - holding cost - orderingcost) when DO orders in the quantities computed in b).d) Is DO or the manufacturer better off or worse off with the discounts? If both are better off where arethe savings coming from?

10. [Order quantity with a promotion] Refer to the textbook where order quantity under promotion (one-time discounting) is studied. Suppose that promotion opportunity is available only now and Qd is aninteger multiple of EOQ. Consider ordering in the lot size Qd only once when the promotion is availableinstead of ordering EOQ in (Qd/EOQ) cycles. The cost over (Qd/EOQ) cycles with order quantity EOQis

TC(EOQ in Qd/EOQ cycles) = TC(EOQ in 1 cycle purchasing at C− d)

+

(Qd

EOQ− 1

)TC(EOQ in 1 cycle purchasing at C)

=EOQ2

2Rh(C− d) + S + (C− d)(EOQ)

+

(Qd

EOQ− 1

)(EOQ2

2RhC + S + C(EOQ)

).

Moreover the cost over 1 cycle with order quantity Qd is

TC(Qd in 1 cycle) =(Qd)2

2Rh(C− d) + S + (C− d)(Qd).

Note that by ordering EOQ in Qd/EOQ cycles and by ordering Qd in 1 cycle, we cover the demand of Qd

during a duration of Qd/R. Thus, TC(EOQ in Qd/EOQ cycles) and TC(Qd in 1 cycle) happen duringexactly the same duration, so they can be compared.a) Let us use the inventory notation to express the cost savings achieved by switching from EOQ to Qd

Savings in material input cost = (Qd − EOQ)d

Savings in ordering cost = (Qd − EOQ)S/EOQ

Explain these expressions using one sentence for each.b) What is Savings in holding cost due to switching from EOQ to Qd? This number will be negative but letus be consistent and work with savings.c) Sum all the savings to construct the total savings TS(EOQ ← Qd) achieved by switching from EOQto Qd. Then check that

∂Qd TS(EOQ← Qd) = d +S

EOQ+

(hC)EOQ2R

− Qd(C− d)hR

Set this derivative equal to zero and simplify by using the definition of EOQ, i.e., S/EOQ = (hC)EOQ/(2R).d) Find the optimal Qd that maximizes total cost savings in (c) and show that Qd ≥ EOQ.

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11. [Demand management with credit] When a supplier sells inventory to a buyer, the supplier does notusually receive the payments immediately. The supplier can actually offer a payment option to the buyerwhere the buyer gets 3% discount if the full payment is made in 10 days or else the full amount will bedue in 30 days. In industry’s lingo, this option is called 3/10 net 30.a) What does 2/20 net 60 means?b) With 3/10 net 30, the supplier is actually discounting the price. Is this an all-unit quantity discount?Can this be used as a promotion scheme to boost the demand? Explain.

12. [Locating RFID readers] We are going to locate two RFID readers on each side of a moving conveyorbelt. Our identical RFID readers can read any RFID tag within their range. But their range (depending oninterference around and in the product packaging materials) is a normal random variable with expectedvalue of 1 meter and standard deviation of 0.2 meter. Use normdist functions to answer the questionsbelow.a) We assume that RFID tags remain at the middle of the conveyor belt (i.e., equidistant from the readerslocated on the left and right-hand sides of the conveyor) when the products pass by the readers. Then,as the belt becomes wider, the read rate (probability of reading a single RFID tag) by two readers drops.How wide the conveyor can be to ensure a read rate of at least 0.91.b) Now suppose that the only available conveyor is 2 meter wide. Moreover, half of the RFID tags passaway from 0.5 meter from the reader on the left and the other half pass away from 1.5 meter (resp. 0.5meter) away from the reader on the left (resp. right). What is the read rate now?

13. Whole Foods offers a good selection of cheeses. One of the cheeses it sells is Bulgarian Feta Cheese.Bulgarian Feta contains more fat than Greek Feta and is more tasty to eat for breakfast. Bulgarian cheesesare purchased in jars from a Bulgarian farm. The replenishment lead time from Bulgaria is 6 months.During the replenishment lead time Whole Foods sell 100 jars of cheese. The standard deviation of thedemand during the lead time is 25 jars. Whole Foods assumes that the demand is Normally distributed.Whole Foods uses an ROP level of 150 jars.a) Find safety stock level.b) Find the proportion of order cycles in which stock outs occur.

14. Whole Foods offers a good selection of cheeses. One of the cheeses it sells is Bulgarian Feta Cheese.Bulgarian Feta contains more fat than Greek Feta and is more tasty to eat for breakfast. Bulgarian cheesesare purchased in jars from a Bulgarian farm. Each jar costs 10 dollars. The replenishment lead timefrom Bulgaria is 6 months. During the replenishment lead time Whole Foods sell 100 jars of cheese.The standard deviation of the demand during the lead time is 25 jars. Whole Foods assumes that thedemand is Normally distributed. Whole Foods uses an interest rate of 20% for its inventory holding costcomputations. When Whole Foods out of the stock for the cheese, it incurs a stockout cost of 25 dollarsper jar. Placing an order costs about 50 dollars, independent of its size.a) Find optimal lot sizes and reorder point for Whole Foods.b) Find safety stock level.c) Find the proportion of order cycles in which stock outs occur.

15. Repeat b) of light bulb exercise 6 if Macedonian and Romanian demands have a correlation of 0.5.

16. [Capacitated Newsvendor Problem] First download the associated Excel file from the web site. Considera restaurant allocating its tables to customer groups A and B which ask for reservation by calling inadvance. If a table is allocated to group but sufficient people do not show from that group, the table hasto be given to walk-in customers. Per table revenue from walk-in customers is $30 and is smaller thanrevenues of groups making in-advance reservations. Actually $30 revenue does not even cover the costof providing service for a table.

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a) According to the parameters in the Excel file, what is the cost of providing service for a table for groupsA and B? What is the revenue per table expected from groups A and B? How many tables are there inthe restaurant?b) How much should the restaurant owner be willing to pay to add 1 more table?c) If there were only 8 tables in the restaurant, how many of these should be allocated to groups A andB?d) If there were 14 tables in the restaurant, how many of these should be allocated to groups A and B?

17. [Beer Game] Play the Beer Game the real version for 30 periods. Save your results (retailer, distributioncenter, warehouse, factory) onto a single floppy. Submit them along with your homework.

18. [Lead Time as a Real Option] Looking at the safety stock computations, we realize that safety stocksare very sensitive to the length of the lead times. This is because, safety stocks are used to cover for thevariabilities during the lead times. If the length of the lead times can be controlled even slightly, thechances of a stockout or overstock during lead time can be decreased. Suppose that the lead times havetwo parts:

Lead Time (LT) = Manufacturing Lead Time (MLT) + Delivery Lead Time (DLT)

The orders are first manufactured in MLT time at a manufacturer and then delivered in DLT time to theretailer. DLT can be decreased (increased) by using a faster (slower) transportation mechanism or mode.Suppose that the retailer has the option of modifying the DLT when the manufacturing is completedat the manufacturer. Should the retailer check its inventory level when making this modification to theDLT, why? When would the retailer ask for a faster transportation mechanism? Can you give a real-lifeexample or make up an example of this kind of DLT modifications? In general, options in supply chainsare on “real” (tangible) aspects so they are called Real Options to differentiate from Financial Options.

19. [Quick response] When would manufacturers implement quick response? In a manufacturer and a re-tailer setting, consider quick response; manufacturer’s effort to decrease its lead times to quickly respondto retailer orders. Explain how key retailer inventory characteristics (inventory levels and order sizes)can change with quick response. You are advised to study the quick response example in the textbook.Now consider quick response from the manufacturer’s standpoint. Is it easy for a manufacturer to de-crease its lead times? Can you give 1-2 examples of how such a decrease can be achieved? What doesmanufacturer gain from quick response? For example, does the manufacturer sell more or less to the sin-gle retailer? How would your answer change if the manufacturer is competing with other manufacturersto supply the retailer?

20. PhD For each of the following demand distributions decide if there is a unique wholesale w price thatmaximizes the supplier’s profit ΠS(w): Normal, exponential, uniform, gamma, logistic, Weibull.

21. [BestBuy in November 2008] BestBuy is a specialty retailer of consumer electronics, home office prod-ucts, entertainment software, appliances and related services. It has about 21% market share in the USand $40 billion annual revenue as of March 1, 2008 according the the latest annual report filed on May13, 2008 1. The major portion (≈ 40%) of revenues are obtained from consumer electronics segmentincluding digital cameras.

Despite its good performance in fiscal year 2008 (ended on March 1, 2008) and 2007 (ended on March1, 2007), the company is worried about its performance in fiscal year 2009. Quoting from the BetsBuy’slatest annual report p.56: “As we consider the macroeconomic pressures on the consumer and evaluate

1Annual report can be reached via www.bestbuy.com → For Our Investors → Annual Reports and is available athttp://media.corporate-ir.net/media files/IROL/83/83192/08AR/assets/shared/BestBuy Fiscal08 Annual Report.pdf.

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the industry business trends, we believe it is prudent to plan for a soft consumer environment in thenear-term . . . looking forward to fiscal [year] 2009”. Moreover, p.15 has: “General economic conditions,a decline in consumer discretionary spending or other conditions may materially adversely impact oursales in a disproportionate fashion”.

Losing 1-2 customers does not affect the sales much, because BestBuy has a wide customer base. Quotingfrom p.12 of the annual report: “No single customer has accounted for 10% or more of our [BestBuy’s]. . . revenue”. However, when all the customers are affected by the same economic conditions, they mayall reduce/change their spending.

Since March 1, eight months passed and they confirmed with certainty a strong decline in consumerspending. The consumer spending has dropped so extremely that Circuit City, a major competitor ofBestBuy, declared bankruptcy in early November, 2008. BestBuy’s CEO Brad Anderson commented afterOctober sales reports by saying that “Since mid-September, rapid, seismic changes in consumer behaviorhave created the most difficult climate we have ever seen” [1]. The same-store sales were down by 7.6%.According to [1], “as such, Best Buy now expects same-store sales for the year . . . to fall by 1% to 8%, withresults for the last four months of the fiscal year – which encompass the all-important holiday-shoppingseason – tumbling by between 5% to 15%”. President and COO Brian Dunn has recently said: “In 42years of retailing, we have never seen such difficult times for the consumer. People are making dramaticchanges in how much they spend, and we are not immune from those forces”. In sum, BestBuy stronglybelieves that its sales and hence revenues will drop in fiscal year 2009. In order to stabilize the earnings,BestBuy considers its costs.a) A portion of its costs is due to servicing its debt. According to p.47 of the annual report, the debt hasincreased in 2008:

2008 2007Debt (including current portion) $816 $650Capitalized operating lease obligations (8 times rental expense) $5,902 $5,401Total debt (incldng capitalized operating lease obligations) $6,718 $6,051Debt (incldng current portion) $816 $650Capitalized operating lease obligations (8 times rental expense) $5,902 $5,401Total shareholders equity $4,484 $6,201Adjusted capitalization $11,202 $12,252Debt-to-capitalization ratio 15% 9%Adjusted debt-to-capitalization ratio (incldng capitalized operating lease obligations) 60% 49%

In addition, the report says: “. . . our short-term and long-term debt was comprised primarily of creditfacilities and convertible debentures 2. We do not manage the interest rate risk on our debt throughthe use of derivative instruments. Our credit facilities are not subject to material interest rate risk 3.The credit facilities interest rates may be reset due to fluctuations in a market-based index, such asthe federal funds rate, the London Interbank Offered Rate (LIBOR), or the base rate or prime rate ofour lenders. A hypothetical 100-basis-point change [increase] in the interest rates of our credit facilitieswould change [decrease] our annual pre-tax earnings by $2 million”. The LIBOR from March to October,2008, is given below.

Month Mar Apr May Jun Jul Aug Sep Oct3-Month (Debt Expiration) LIBORMonthly Average in % 2.78 2.79 2.69 2.77 2.79 2.81 3.12 4.01

2Convertible debenture is a debt instrument, issued by a large company to raise capital from investors, that can be converted intostock at the option of the investor. BestBuy investors can sell all or a portion of their debentures on January 15, 2012 and January 15,2017 to BestBuy at a price equal to 100% of the principal amount of the debentures plus accrued and unpaid interest.

3One of those financial facilities went bankrupt by November 2008 [1].

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One of the reasons why increasing interest rates can decrease BestBuy earnings is through inventoryfinancing in place at BestBuy. According to p.70 of the report: “We have inventory financing facilitiesthrough which certain suppliers receive payments from a designated finance company on invoices weowe them. Amounts due under the facilities are collateralized by a security interest in certain merchan-dise inventories. The amounts extended [by the finance company] bear interest, if we exceed certainterms, at rates specified in the agreements. We impute interest based on our borrowing rate where thereis an average balance outstanding”. Using the estimates above and the increase in LIBOR that happenedfrom March to November, compute the drop in pre-tax BestBuy earnings. (Hint: Use the quote abovethat relates increase rate increases to pre-tax earnings.)

b) Another portion of BestBuy costs is due to inventory holding. Most of the inventory is sent frommanufacturers to eight BestBuy distribution centers (DC) in the US: Dinuba, California; Findlay, Ohio;Nichols, New York; Ardmore, Oklahoma; Franklin, Indiana; Staunton, Virginia; Dublin, Georgia; Bloom-ington, Minnesota; Whittier, California (ordered from the largest square-footage to the smallest). TheseDCs then forward the inventory to 923 BestBuy stores in the US.

Ardmore, Oklahoma is about 60 miles from the UTD campus and houses the closest BestBuy DC. ThisDC ships digital cameras to many BestBuy stores, including two that are closest to the UTD campus:

• Store # 202 (Plano Store) is at 2800 North Central Expressway (Route 75).

• Store # 256 (Addison Store) is at 4255 Lyndon B. Johnson Freeway (Interstate 635).

According to p.17 of the annual report, BestBuy makes more than half of its earnings in the quarter thatincludes the months of November and December – known as the holiday season in the US. BestBuy isexpecting a dismal holiday season and is focusing on cutting costs. To cut operating costs, BestBuy isinvestigating aggregation of the high-end camera inventory and holding it only at the Addison Storewhich will then serve the demand that Plano Store is serving now. BestBuy first gathers the holidayseason high-end Cannon and Samsung camera sales at Addison and Plano Stores from 2000 to 2007:

Holiday season salesin number of cameras 2000 2001 2002 2003 2004 2005 2006 2007Cannon at

Addison Store 46 64 51 55 47 47 41 66Plano Store 113 112 88 101 108 114 93 86

Samsung atAddison Store 42 44 47 51 38 34 37 62Plano Store 69 78 64 67 74 81 63 52

These sales data are available in bestbuy question.xls from the course webpage under BestBuy Sheet onthe left panel. Download bestbuy question.xls and see how safety stock saving of aggregation is com-puted for Cannon assuming that the market trends of last 8 years will continue in 2008. Replicate thosecomputations for Samsung camera to find safety stock saving of aggregation. Compare the correlationof Cannon sales and Samsung sales, comment if the correlations play a role in safety stock saving ofaggregation.

c) Suppose that each Cannon and Samsung camera has an inventory holding cost of $40 per camera permonth during the months of November and December. Compute the dollar value of safety stock savingsin b). These savings can scaled up easily if BetsBuy implements aggregation throughout the US. If BetBuyaggregates Cannon and Samsung Camera inventories of every two stores in a single store, the savingsshould be multiplied by the number of stores divided by 2. Compute the US-wide saving of aggregationand compare this saving to drop in the earnings in a).

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22. [BestBuy in November 2008 Continued] In part b) above, we assume that the market trends will con-tinue for Cannon. This is clearly not the case when BestBuy talks of seismic changes in consumer behav-ior.a) Suppose that Cannon sales will drop by 20 units at each of the Addison and Plano stores as in ”NextCannon-20” worksheet of bestbuy question.xls. Compute the safety stock saving of aggregation.b) Suppose that Cannon sales will drop by 15% at each of the Addison and Plano stores as in ”NextCannon 85%” worksheet of bestbuy question.xls. Compute the safety stock saving of aggregation.

23. [BestBuy in November 2008 Continued] Suppose that Cannon sales will drop by 15% at each of theAddison and Plano stores as in ”Next Cannon 85%” worksheet of bestbuy question.xls. Furthermore,BestBuy wants to provide 95% CSL.a) Fit a Normal distribution to sales in ”Next Cannon 85%” worksheet of bestbuy question.xls and usenormdist and/or norminv functions to compute the inventory levels.b) Go back to the quotations made from the annual report and identify which of them can be used tojustify a Normal distribution for BetsBuy’s demand.

24. [BestBuy in November 2008 Continued] Suppose that you drove to the Plano store to buy your fa-vorite Cannon camera which costs $172 plus tax. But that camera is not available at the Plano store but itis at the Addison store. You are called an overflowing customer if you drive to Addison store to buy thecamera. Are you an overflowing customer?

References

[1] M. Barris (2008). Sales slump pounds Best Buy. Wall Street Journal, November 12 issue.

[2] M.A. Lariviere and E.L. Porteus (2001). Selling to the Newsvendor: An Analysis of Price Only Contracts. Work-ing paper, Kellogg School of Management, Northwestern University.

[3] W.C. Rinaman. (1993). Foundations of Probability and Statistics published by Saunders College Publishing.ISBN 0-03-071806-6.

[4] S.M. Ross. (1993). Introduction to Probability Models 5th edition published by Academic Press, San Diego,CA. ISBN 0-12-598455-3.

[5] D.D. Wackerly, W. Mendenhall and R.L. Scheaffer. Mathematical Statistics with Applications 5th edition pub-lished by Wadsworth Publishing Company, Belmont, CA. ISBN 0-534-20916-5.

Appendix: Beer Game

SC Costs

• Holding costs incurred at F, D, W, R:

– 0.25 dollar per keg per day at F

– 0.50 dollar per keg per day at D

– 0.75 dollar per keg per day at W

– 1 dollar per unit per day at R

• Backordering costs incurred only at R:

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– 10 dollars per unit per day at R

• Minimize sum of these over 30 days

Playing role and style

• Play as R or W or D or F.

• Play as a team

• Without transferring demand information

• Play the training version first, then the real version

Demand distribution

• Random demand discretized

• Truncated

• With mean 50 and standard deviation 20

Order fulfillment

• Observe the incoming shipment from the upstream partner, add this to the inventory.

• Observe the incoming order from the downstream partner, take out this from the inventory.

If the result is nonnegative hold the inventory, otherwise backorder.

Order of daily events at the Retailer(R)

1. Inform upstream partner of the size of order placed by you Info-LT days ago. In day 4, R informs W ofR’s order in day 2.

2. Upstream partner informs you of the size of incoming shipment sent by the upstream partner Mater-LTdays ago. In day 4, W informs R of the shipment made by W in day 2. Enter.

3. Observe the customer demand.

4. Order fulfillment.

5. Outstanding orders advance 1 day.

6. Decision: Order from W. Enter.

Order of daily events at the Warehouse(W)

1. Downstream partner informs you of the size of the order it placed Info-LT days ago. W learns aboutwhat R ordered Info-LT days ago. In day 4, W receives R’s incoming order placed by R in day 2. Enter.

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2. Inform upstream partner of the size of order placed by you Info-LT days ago. In day 4, W informs D ofW’s order in day 2.

3. Upstream partner informs you of the size of incoming shipment sent by the upstream partner Mater-LTdays ago. In day 4, D informs W of the shipment made by D in day 2. Enter.

4. Inform downstream partner about how many kegs it is receiving on this day.

5. Shipments advance 1 day.

6. Order fulfillment.

7. Outstanding orders advance 1 day.

8. Decision: Order from D. Enter.

Order of daily events at the Distributor(D)

1. Downstream partner informs you of the size of the order it placed Info-LT days ago. D learns about whatW ordered Info-LT days ago. In day 4, D receives W’s incoming order placed by W in day 2. Enter.

2. Inform upstream partner of the size of order placed by you Info-LT days ago. In day 4, D informs F ofD’s order in day 2.

3. Upstream partner informs you of the size of incoming shipment sent by the upstream partner Mater-LTdays ago. In day 4, F informs D of the shipment made by F in day 2. Enter.

4. Inform downstream partner about how many kegs it is receiving on this day.

5. Shipments advance 1 day.

6. Order fulfillment.

7. Outstanding orders advance 1 day.

8. Decision: Order from F. Enter.

Order of daily events at the Factory(F)

1. Downstream partner informs you of the size of the order it placed Info-LT days ago. F learns about whatD ordered Info-LT days ago. In day 4, F receives D’s incoming order placed by D in day 2. Enter.

2. Inform downstream partner about how many kegs it is receiving on this day.

3. Shipments advance 1 day.

4. Order fulfillment.

5. Factory brews scheduled production.

6. Decision: Schedule production to be received in 2 days. Enter.

Reporting

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• After you finished playing the game, you can graph several performance measures, get a summary ofyour actions. Click on “Show History” for this.

• Results can be saved to A drive by first clicking on “End Game” and then saying “Yes”.

• You can copy results and paste into another file.

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