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 QS+ R Q(pQ) = h 2(pQ) + R QS+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
Q≥0TC(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 dQTC(Q) = h 2(p)− R Q2S=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 2p √ 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= p1 if q0≤ Q≤ q1 p2 if q1< Q≤ q2 . . . . pN if qN−1 <Q<qN
where we set q0=0 and qN =∞. Prices are monotonically ordered by pn−1≥ pnfor 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≤2 40 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 opposed
to 2 shoes, not only the price of the third shoe but also the price of the first two shoes drop. That is why this is an all-unit discount scheme. If you look at the prices closely, you see that the cost of buying two shoes is the same 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 income
Q:
Tax per dollar of taxable income=
See tax tables if 0 ≤Q≤ 100, 000 28.0% if 100, 000 <Q≤ 183, 250 33.0% if 183, 250 <Q≤ 398, 350 35.0% if 398, 350 <Q≤ 400, 000 39.6% 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 that
the tax on each dollar depends on the total taxable income. In this example, prices (tax rates) are increasing with the taxable income, so we do not have a discount in tax rates but a mark-up. Accountants call this sort of 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 is marked up for all-units (taxable income). In view of this gap, you may question the fairness of an all-unit quantity mark-up tax regime.⋄
When we buy Q units, the purchasing cost is as follows.
Purchasing cost of Q units= p1Q if 0≤Q≤q1 p2Q 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):= h 2(pnQ) + R QS+ R Q(pnQ). (1)
Then the total cost over all regions is obtained by patching all TCntogether, i.e.,
TC(Q) = TC1(Q) if 0≤Q≤q1 TC2(Q) if q1< Q≤q2 . . . . TCN(Q) if qN−1 <Q<qN In the all-units quantity discount problem, we want to solve
min
Q≥0TC(Q).
We propose the ordering algorithm in Table 1 to find the optimal order quantity. Note that this algorithm does 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.
Initialize n :=N and CandidatesComplete= False.
While n≥1 and CandidatesComplete=False do
Find EOQn.
If qn−1< EOQn≤qn
A1: Add EOQnto the set of candidate solutions;
A2: CandidatesComplete= True;
else if EOQn ≤qn−1
B: Add qn−1to the set of candidate solutions;
else if
C: CandidatesComplete=True;
If CandidatesComplete=False
n:=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 theadditionalunits bought on top of Q units:
Price for anadditionalunit
given that Q is already purchased = p1 if 0≤Q≤q1 p2 if q1 <Q≤q2 . . . . pN if qN−1 <Q<qN
Let Vnbe the cost of buying exactly qnunits. 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 are bought but drops the price of the third or more shoes to $40. Then
Price peradditionalshoe= {
60 if 0 ≤Q≤2 40 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 the
third 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 peradditionaldollar of taxable income= 28.0% if 0 < Q≤ 183, 250 33.0% if 183, 250 < Q≤ 398, 350 35.0% if 398, 350 < Q≤ 400, 000 39.6% 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
33%. If your taxable income increases from $183,250 to $183,251, your tax increases by $0.33. Compare this increase 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<q1 V1+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):= h 2(Vn−1+pn(Q−qn−1)) + R QS+ R Q(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<q1 TC2(Q) if q1≤Q<q2 . . . . TCN(Q) if qN−1≤ Q< qN .
In the marginal-units quantity discount problem, we want to solve min
Q≥0TC(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-all-unit discount is TC(Q).
3.
Preliminaries from Probability
3.1 Numerical events and random variablesConsider tossing a coin once. We do not know what the outcome will exactly be but we know that it is either Head (H) or Tail (T). The set of possible outcomes is called a sample space and is denoted by S. For a single toss, S= {H, T}. For two tosses in a row, S= {HH, HT, TH, TT}. A real valued function X from the sample space S to real variables is a random variable. For a single coin toss, X1given as X1(H) = 0 and X1(T) = 1 is
random variable. X2(H) = 1 and X2(T) =0 is another random variable. X1denotes the number of tails and
X2denotes the number of heads in a single toss.
Example: Consider tossing a coin twice. The sample space is S ={HT, TH, HH, TT}. Let X1be the
num-ber of tails. Then X1(HT) =1, X1(TH) = 1, X1(HH) =0, X1(TT) = 2. Let X2be the number of heads. Then
X2(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 X3in English. Suppose that X4denotes the earnings in a
betting 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 X14 and X24 be the earnings in the first and the second toss then
X41(HT) = 5, X41(TH) = 10, X41(HH) = 5, X14(TT) = 10 and X24(HT) = 10, X42(TH) = 5, X42(HH) = 5,
X2
4(TT) = 10. Note that X4 = X41+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 the
of a discrete random variable with countably infinite values: Suppose that we are tossing a coin until H comes up, 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 but they 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 a constant. P(X = a)is the probability that the random variable X would assume value a. P(X = a)is called the probability mass function (pmf for short).
Example: Consider tossing a coin twice. Let X be the number of Tails coming up. The sample space, the random variable and associated probabilities are:
S X Probability HT 1 1/4 TH 1 1/4 TT 2 1/4 HH 0 1/4 P(X= a) = 1/4 if a=0 1/2 if a=1 1/4 if a=2 0 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 can stick two of you together and treat you as a super human. Then we have 3! = 6 ways of ordering the super human and the remaining two people. These 6 ways do not consider whether you are the first or your friend is. Taking who is first between you two into account, we obtain 12=2*6 ways. So in 12 ways out of 24, you will be 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 expected value 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 the center of the gravity. No such intuitive reasoning can be provided for variance, except for saying that it is the expected value of the square of the variation from the mean. Surely other measures than variance can be built to measure variation.
Example: Consider tossing a coin twice. Let X be the number of Tails coming up.
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 want to pass a full glass of water to your friend and 1/4 of the water in the glass spills over at every pass. What is the expected percentage of water your friend receives. Let us construct g first, let g(X)be the percentage of the water left in the glass after X passes. g(X =1) =3/4, g(X=2) = (3/4)2and 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 numnum-ber 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 bank teller in a day. There are various other useful discrete densities such as geometric and multinomial that are used 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 the
elapsed time until you have a flat tire are examples of continuous random variables. Let X denote any (discrete or 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 derivative of 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 of the 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
a f
(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 ranran-dom variable,
E(X) =
∫ ∞
−∞x· f(x)dx Var(X) =E(X−E(X))
2 E(g(X)) =∫ ∞
−∞g(x)·f(x)dx
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 ≤1 0 otherwise }
You may first want to check that this density is valid, i.e. it sums up to one:
∫ ∞ −∞ f(x)dx = ∫ 1 0 3x 2/2+xdx= (1/2)x3+ (1/2)x2|x=1 x=0=1.
Now the expected value:
E(X) = ∫ ∞ −∞x· f(x)dx= ∫ 1 0 x (3x2/2+x)dx = (3/8)x4+ (1/3)x3|xx=1=0 =17/24.
Two common continuous distributions are Uniform and Normal. We say that X has a uniform distribution over 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≤b
0 otherwise
}
Also note that
F(x) = 0 if x< a x/(b−a) if a≤ x≤b 1 if x>b
Uniform distribution arises in practice if the knowledge about the random variable is very limited to the level that 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 forecasts are 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 other distributions 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 of
this, 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 property is extensively used in statistics.
Let us elucidate why we are interested in the sums of random variables. Suppose we are modelling the weekly demand for a supplier. One approach is to figure out the weekly demands of each customer of the supplier and sum them up. This is demand aggregation over customers. Another approach is to find daily demands for the supplier and sum them up over a week. This is temporal aggregation over days. Clearly one can aggregate the demand over both customers and days. Important point is that we often deal with aggregated numbers coming from different sources. Then assuming a normal distribution for sum/average of numbers is very
justifiable as such sum/average converges to a normal distribution.
If X is a normal random variable with meanµ 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 but with 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−µ σ a−µ σ ϕ(z)dz
where Z is a standard normal variable, i.e. Z ∼ N(0, 1)andϕ(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 normal densities to compute probabilities for any normal random variable. This is the reason why books tabulate only standard 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−µ σ −∞ 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 cdfΦ, 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 x du f (x)dx= ∫ ∞ ROP ∫ ∞ u f (x)dxdu= ∫ ∞ ROP1−F (u)du
where we change the order of integration to obtain the second equality. However, these manipulations cannot save us from numerical integration. In the case of Normal demand, we can avoid integrals, see the Formulas on 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 integration and all modern software -including Excel- has built-in functions to compute these probabilities.
Computing probabilities for X∼ N(µ, σ2)
Functions Matlab Excel
pd 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)
For a given p (0≤ p≤1), inverse cdf finds the outcome value x such that the probability of observing smaller outcomes than x is exactly p. Note that if p = F(x)then x = F−1(p). Note that Excel uses a single function to 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:
ℜ
ϵαδ the textbook
4.1 Cycle Service Level vs. Fill RateCycle service level is a probability while fill rate is a ratio. Unfortunately, the third edition of Chopra & Meindl says: “Fill rate is equivalent to the probability that product demand is supplied from available inventory” on p.307. This sentence is not accurate as it gives the impression that the fill rate is a probability. Please ignore this 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, see chapter SC03. By postponing, it is possible to learn more information about demand which effectively reduces demand forecast variability. As we know from the above discussion, safety stocks are used to counter the variability 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 split supply chain into two before the customization and after the customization. Inventories before the customiza-tion have two significant properties. First, before the customizacustomiza-tion, raw materials or semi-finished assemblies have 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 the customization, 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 variability of 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 of uncertainty and inventory pooling. Many works in the literature fail to identify the distinction between these two factors. They go ahead with a partial analysis of studying only inventory pooling effects on the aggregated demand. Such a partial analysis could be defended by saying that resolution of uncertainty is not measured in almost all of the instances in practice. However, this defense points out the need for resolution of uncertainty data 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 customization stage. 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 to laptops. If the stage where displays are assembled is the customization stage, savings due to postpone-ment will be higher.
3. Demand is highly random and is fluctuating so forecasts are very variable. Therefore, safety stocks are significant 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 customer demands determine where to customize products in a product family. Customization should happen after much of the uncertainty in the demand is resolved. The timing of the resolution of uncertainty can be different. In Figure 1, uncertainty resolution in product family A happened before that in product family C. Thus, product family A can be customized earlier while the customization of product family C needs to postponed further. Product in family C should be kept in a generic form and be customized right before (or perhaps after, which is make-to stock policy) the customer demand. If the supply chain is not flexible enough to postpone the customization of family C until after the last production stage, then we cannot reap all the potential benefits of postponement.
Family A Family B Family C Due Date Time Demand Variability Resolution of Uncertainty in the Demand of Family A Resolution of Uncertainty in the Demand of Family B Resolution of Uncertainty in the Demand of Family C Customize Family A Customize
Family B CustomizeFamily C Make-to-orderFor Family C
ƵƐƚŽŵĞƌ
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 supplier flexibility, e.g. the extent to which supplier can quickly deliver orders. Since some operations are delayed purposefully with the postponement strategy, the risk of missing customer deadlines is higher. It may be necessary and also wise to incur some extra costs to obtain a better supplier service, such as frequent deliveries 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 its warehouses 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
stretches out SC, sometimes to the point of breaking. A functional IT infrastructure is also needed to satisfy 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 postpost-ponement to inefficiencies introduced into SC to achieve component commonality and to training cost of the personnel. Reader should not get the impression that postponement always reduces total costs. Rather it is the proper analysis of the postponement, trading off cost savings against extra costs due to postponement, which has the potential to reduce total costs.
6.
Product Availability:
ℜ
ϵαδ the textbook
7.
Optimal Wholesale Pricing
PhDConsider a supplier(S) and a retailer(R) subject to random demand D with pdf f , cdf F and ¯F :=1−F. Let us
define the following cost parameters:
• c: Supplier’s cost of supplying 1 unit. If supplier manufactures this unit, c is the marginal manufacturing
cost. 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 determine w. We will study finding optimal w to maximize the supplier’s profits. Note supplier’s profits depend on how
many 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, if the wholesale price is large, the retailer will buy less, i.e. y(w)decreases in w. Let us quantify this observation, first letΠR(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=0 dx f (D)dD−wy = p ∫ y 0 ∫ ∞ x f (D)dDdx−wy= p ∫ y 0 ¯ F(D)dD−wy.
It is easy to show thatΠR(y) is concave, so it is maximized at y∗ = F¯−1(w/p). In order to emphasize the dependence of the retailer’s order size, we use the notation y∗(w) = F¯−1(w/p).
We now study supplier’s profitΠS(w), ΠS(w) = (w−c) ( arg max y Π R(y) ) = (w−c)y∗(w) = (w−c)F¯−1(w/p)
The first order condition for a critical w is:
dΠS(w)
dw = F¯
−1(w/p) + w−c
p ¯F′(F¯−1(w/p))
Now remember that ¯F′ =−f , y∗ = F¯−1(w/p)and w= p ¯F(y∗).
dΠS(w)
dw =y
∗− p ¯F(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∗) ) = c p (3)
Once this y∗is found than the optimal wholesale price is simply:
w= p ¯F(y∗) (4)
As (4) is a simple equation, we focus on (3). For the existence and uniqueness of y∗, consider f(y)/ ¯F(y)inside the parentheses in (3). This ratio is nothing but the failure rate function of the demand density. For a moment suppose that this failure rate is increasing in y, than the entire term inside the parentheses decreases in y. In addition ¯F(y) decreases in y. Then the left hand side of (3) is decreasing in y while the right hand side is constant. 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’s revenue is found by solving (3) and (4).
8.
Solved Examples
8.1 Statement of Exercises1. [All-units quantity discounts] A popular shoe store sells 8000 pairs per year. The fixed cost of ordering shoes from the distribution center is $15 and holding costs are taken as 25% of the shoe costs. The per unit purchase costs from the distribution center is given as
C1=60, if 0≤Q≤50 C2=55, if 50≤ Q≤150 C3=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 gives marginal unit discount and compute the optimal order quantity.
3. [Harvey’s Speciality Shop] Harvey’s Specialty Shop is a popular spot that specializes in international gourmet foods. One of the items that Harvey sells is a popular mustard which be purchases from an English 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 that if 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
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-six-month period is 25, assume that demand is described by a normal distribution. How should Harvey control the replenishment of the mustard?
4. [Arts and Humanities Coffee] In front of the Arts and Humanities secretarial office coffee is brewed and offered to general public at a cost of 25 cents per cup. Coffee powder is bought in bags and each bag costs 2 dollars and contains enough coffee to prepare 10 cups. Suppose that the coffee demand over every two hours is normally distributed with expected value of 6 and variance of 1. The office is open for 8 hours every day and does not provide coffee outside the office hours. Suppose that any coffee drinker who arrives when the office is out of coffee joins a queue to wait for the next coffee preparation. Thus, at the beginning of every 2-hour time blocks, there can be a bunch of people who arrived earlier but have waited for the coffee.
a) Assume that demand in every two hour period is independent of other period’s demand, find the distribution 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 are prepared 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 people per 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 the next year. The book store buys the books at 50 dollars each and sells at 80 dollars. Any unsold book is bought 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 consider to reduce the costs in (d) ?
6. [Light bulbs from Taiwan] Philips manufactures lighting products (various lamps) in Taipei, Taiwan and ships them to Balkan countries for sale. The weekly sales in these countries are independently and Normally distributed with mean demand and standard deviation as below:
Country Mean Standard deviation
Macedonia 100 40
Romania 300 80
Currently, Philips packs the lamps with manuals in each country’s language in Taiwan. These packages are sent to DCs in each country to meet that country’s demand. The transportation time from Taiwan to these 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 are observed. Thus, Philips produces to stock and the regional DC orders to stock while the country specific customization happens according to retailer orders. What is the distribution of the weekly demand for the regional DC? Compute the safety stock savings with regard to a) if the CSL of 90% is still maintained.
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 Q is given and is not necessarily optimal for our context. Further suppose that the demand is nonnegative and 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 a
function 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)in
your words. Draw y2(D)and y3(D)onto the same coordinate system used in a). You may use pencils of
different 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 Q̸=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 equality involving 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 demanded a lot with respect to product B so the retailer receives a truck of product A delivery every Monday. It is not clear whether B should be in every truck (relative frequency nB = 1). When product B is carried
on a truck, there is a loading/unloading (handling) cost of $4000 for the retailer. Moreover, the weekly demand for B is 800 units and the holding cost rate per week per unit is $1 at the retailer. The fixed cost of handling product B is s = 4000. Let us fix time unit to a single week. Then demand is R = 800 per week, and the holding cost H=1 per week per unit.
a) Using the relative order frequency nBof product B, express the total B handling and inventory holding
cost over nBweeks.
b) Use TC(nB; over nBweeks)above to express the total cost (handling and holding) over 1 week. This
is the average cost incurred per week. c) When finding the relative frequency n0
B, decide whether we should minimize TC(nB; over nBweeks)or TC(nB; over 1 week). In a sentence, explain the rationale behind your decision. By taking the derivative
of the appropriate objective function and setting equal to zero, compute n0B. d) The frequency n0
B computed in c) may not be an integer. Suppose that it is n0B = 5/3. Intuitively this
Cargo of a truck on each week
5-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 15
A A A A A A A A A A A A A A A
B 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) B over 5 weeks that will result from the schedule above. Also compute the “average” total cost incurred per week. 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). Explain
if they need to be the same. 8.2 Solution of Exercises
ANSWER for Exercise 1
There 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 and
h3= (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. Since
EOQ1 > q1, we take q1 = 50 as the candidate solution for the second range. Since EOQ3 < q2, we take
q2=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 two statements, the costs at Q = 132.1and C2 = 55 is smaller than the costs at Q = 50and C2 = 60. Therefore we
can 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 2
V0 = 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;
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 not
need 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 3
We 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 does not seem to be specified. But notice that if the order lead time is six months and the mean lead time demand is 100, 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− QhC
bR . Substituting Q=100, we obtain P(Demand≤ROP) =1− QhC bR =1− 100∗0.2∗10 25∗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 4
a) Because of independence, the sum of the demands over 4 2-hour blocks remain to be normally distributed and 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 not enough. 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 stockouts per 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) | {z } 0.99865 ] +2 normdist(6/2, 0, 1, 0) | {z } 0.004432 = 0.000764
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) | {z } 0.97725 ] +2 normdist(4/2, 0, 1, 0) | {z } 0.053991 = 0.016981
With a safety stock of 2, ESC grows further to 0.166631.
ANSWER for Exercise 5
a) The demand takes values from{21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32}with equal probability (of 1/12) so it has discrete uniform distribution.
b) The overstocking and understocking costs are co = 10 dollars and cu = 30 dollars. We need to solve for
P(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 want
to 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 of continuous 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 the cost of understocking (with transshipments from other bookstores) and fast response from Prentice-Hall so that two or more orders can be placed in a single semester.
ANSWER for Exercise 6
a) 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 and variance 8000= 1600+6400. Now the safety stock becomes ssM+R= norminv(0.9, 0, 1)
√
6√8000. The safety stock saving is norminv(0.9, 0, 1)√6(40+80−√8000).
ANSWER for Exercise 7
a) y1(D) =max{D−10, 0}is the random variable for understock (the number of stockouts) if we order 10. To
draw y1(D), put D on x-axis and y1(D)on y-axis. Draw two lines y = D−10 and y = 0. The maximum of
these 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 the
inventory at the end of the season) if we order 10. y3(D) =D−10 is the random variable for the negative of
the safety stock. To draw y2(D), first draw two lines y=10−D and y=10. The maximum of these two lines
is y2(D). Since y3(D)is a simple line, it is straightforward to draw it. c) If you have the drawings, you can see
that
(D−10)+− (10−D)+ =D−10 or that
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 value of 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 8
a) With a relative order frequency of nB, product B is ordered very other nB weeks. Thus, the order size must
suffice for nB weeks. The order size then is nBR. Since the retailer inventory at maximum is nBR and drops to
zero at the uniform rate of R, the average inventory is nBR/2. TC(nB; over nBweeks) = |{z}s
Fixed ordering cost
+ n| {z }BR/2 Average inventory nB |{z} kept nB weeks hC |{z}
at the holding cost rate hC
| {z }
Inventory holding cost over nB weeks
= 4000+1
2800(1)n
2
B,
where H=hC =1 is holding cost per item per week. b) TC(nB; over 1 week) = TC (nB; over nB weeks) nB = s nB +1 2RhCnB = 4000 nB + 1 2800(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 (from the decisions) time intervals. Such a time interval is 1 week. If you wish you can also use 2 weeks or 5 weeks as your time intervals, in these cases you must multiply the objective below by 2 and 5 respectively and recall that 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: d dnb TC(nB; over 1 week) =− 4000 n2 B +1 2800(1) =0. So, nB = √ 4000 400 = √ 10. d)
TC(Schedule in the table; over 5 weeks) =
Week 1 z }| { 4000+ 1 2800(1)1 2 | {z } n1 B=1 + Week 1 z }| { 4000+1 2800(1)1 2 | {z } n2 B=1 + Weeks 3−4−5 z }| { 4000+1 2800(1)3 2 | {z } n3 B=3 = 12000+4400 = $16, 400
TC(Schedule in the table; over 1 week) = TC(Schedule in the table; over 5 weeks) 5 = $3280 e) TC(nB; over 1 week) = 4000 5/3 + 1 2(800)(1)5/3 = 2400+2000/3 ≈ $3066. f) The costs are not the same. Although the average of n1
B = 1, n2B = 1, n3B = 3 is n0B = 5/3, the
con-tribution of longer cycles to the holding cost is not proportional to the length of the cycle. Indeed, this contribution is proportional to the square of the cycle length. A longer B-cycle like n3B = 3 causes dis-proportionately more costs than their lengths. Since the schedule in the table has such a cycle, we expect
TC(Schedule in the table; over 1 week) > TC(nB = 5/3; over 1 week). Note that n0B = 5/3 is fractional and
cannot be implemented, TC(Schedule in the table; over 1 week)−TC(nB = 5/3; over 1 week) > 0 can be
called the implementability cost.
9.
Solved Exercises from Operations Management
9.1 Statement of Exercises1. [Walton Bookstore] In August, Walton Bookstore must decide how many of next year’s nature calendars should be ordered. Each calendar costs the bookstore $2 and is sold for $4.50. After January 1, any unsold calendars are returned to the publisher for a refund of $0.75 per calendar. Walton believes that the number of calendars sold by January 1 follows the probability distribution shown in following table.
Number of Calendars Sold 100 150 200 250 300 Probability 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 the bookstore order in August?
2. [Childcare] UT Dallas allows an employee to put an amount into an account at the beginning of each year, to be used for child-care expenses. This amount is not subject to federal income tax. Assume that all other income is taxed by the federal government at a 40% rate. If this amount can only be used for childcare expenses. If the amount is more than the childcare expenses, the difference is lost. If the childcare expenses are more than the amount, the employee must pay for the excess out of his/her own pocket. This excess payment can be claimed in an income tax return to receive tax credit at the rate of 25% of the excess payment. Prof. C¸ akanyıldırım believes that his childcare expenses for his son for the coming year will be $3000, $4000, $5000, $6000, or $7000 with equal probabilities. At the beginning of the year, 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 eleele-vators are at a different floor. An OM student team is commissioned to decide on the optimal floor for both elevators. That is, after carrying people to their destination floor, the elevators will return to their optimal floor to wait for forthcoming people unless they are already called by some people. The OM student team found out the distribution of the elevator calls 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.
Suppose that the elevators, while moving, has a constant speed of 1 floor every 4 seconds. If the elevators are 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 as a 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 at the same floor. Now we relax this assumption by saying that elevator A can wait at a certain floor while elevator B waits at another floor. It is easy to convince ourselves that we can obtain a better solution by having 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 a mean 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 costs the hotel approximately $200 since rooms booked on such late notice are expensive. How many rooms should 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 cost
Buy q calendars at $2/calendar 2q Sell D calendars at $4.50/calendar -4.50D Return 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 cost
Buy q calendars at $2/calendar 2q Sell D calendars at $4.50/calendar -4.50q
Total cost -2.50q
From the multiplier of q, the underage cost is cu =2.5. Then cu co+cu = 2.50 3.75 = 2 3.
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.
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 cost
Tax avoided -0.4q
Lost money q-d
Total 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 cost
Tax avoided -0.4q
Tax 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 = 0.15 0.75 = 1 5.
Since P(Childcare expense=3000) =0.2, Prof. C¸ akanyıldırım 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 distribution
of D, we have P(D≤ Q) = 0.5 when Q =2. In other words, the elevator must go to the second floor to wait for forthcoming calls.
ANSWER for Exercise 4:
Let QAand QBthe 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 the elevators can be sent arbitrarily. For(QA =1, QB =2), we have the waiting time W(QA =1, QB =2), whose
expected 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|}.Specializing this formula to the specific QAand QBvalues, 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 of
the 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
= 80
200+80 =0.2857.
Using Normsinv(0.2857) in Excel gives a z-value of -0.56599. The negative value indicates that we should overbook 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 distribution found using actual data and marginal analysis. For our hotel, consider that we have collected data and our distribution of no-shows is as follows.
Cancellations 0 1 2 3 4 5 6 7 8 9 10
Probability 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 the weighted costs. The best overbooking strategy is the one with minimum cost.
Number Proba- Number of Overbookings
Cancelled 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 2000 1 0.08 80 0 200 400 600 800 1000 1200 1400 1600 1800 2 0.10 160 80 0 200 400 600 800 1000 1200 1400 1600 3 0.15 240 160 80 0 200 400 600 800 1000 1200 1400 4 0.20 320 240 160 80 0 200 400 600 800 1000 1200 5 0.15 400 320 240 160 80 0 200 400 600 800 1000 6 0.11 480 400 320 240 160 80 0 200 400 600 800 7 0.06 560 480 400 320 240 160 80 0 200 400 600 8 0.05 640 560 480 400 320 240 160 80 0 200 400 9 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 0 Total Cost 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.
