Simulations

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-2 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Marginal Analysis P = probability that demand > a given supply. 1-P = probability that demand < supply. MP = marginal profit. ML = marginal loss. Optimal decision rule is: P*MP (1-P)*ML or MLMP

MLP

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-3 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Marginal Analysis -Discrete Distributions

Steps using Discrete Distributions:

Determine the value for P.P. Construct a probability table and add a cumulative

probability column. Keep ordering inventory as long as the probability

of selling at least one additional unit is greater than P.P.

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-4 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Café du Donut:Marginal Analysis

Daily Sales

(Cartons)

Probability of Sales

at this Level

Probability that Sales Will

Be at this Level or Greater

4 0.05 1.00

5 0.15 0.95

6 0.15 0. 80

7 0.20 0.65

8 0.25 0.45

9 0.10 0.20

10 0.10 0.10

1.00

Café du Donut sells a dozen donuts for $6. It costs $4 to make each dozen. The following table shows the discrete distribution for Café du Donut sales.

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-5 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Café du Donut: Marginal Analysis Solution

Marginal profit = selling price - cost

= $6 - $4 = $2Marginal loss = cost

Therefore:

667.06

4

24

4

MPML

MLP

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-6 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Café du Donut: Marginal Analysis Solution

Daily Sales

(Cartons)

Probability of Sales

at this Level

Probability that Sales Will

Be at this Level or Greater

4 0.05 1.00 ≥ 0.66

5 0.15 0.95 ≥ 0.66

6 0.15 0. 80 ≥ 0.66

7 0.20 0.65

8 0.25 0.45

9 0.10 0.20

10 0.10 0.10

1.00

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-7 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

In-Class Example 3Let’s practice what we’ve learned. You sell cases of goods for $15/case, the raw materials cost you $4/case, and you pay $1/case commission.

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-8 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

In-Class Example 3:Solution

Daily Sales Cases

Probability of Sales at this Level

Probability that Sales Will Be at this

Level or Greater 4 0.1 1.0 > .286 5 0.1 .9 > .286 6 0.4 .8 > .286 7 0.3 .4 > .286 8 0.1 .1 1.00

MP = $15-$4-$1 = $10 per case ML = $4P>= $4 / $10+$4 = .286

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-9 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Marginal AnalysisNormal Distribution

= average or mean sales = standard deviation of sales MPMP = marginal profit MLML = Marginal loss

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-10 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Marginal Analysis -Discrete Distributions

• Steps using Normal Distributions: Determine the value for P.

Locate P on the normal distribution. For a given area under the curve, we find Z from the standard Normal table.

Using we can now solve for: X

*XZ

MPMLMLP

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-11 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Marginal Analysis -Marginal Analysis -Normal Curve ReviewNormal Curve Review

*X

area = .30

Use table to find Z

area = .70

MPMLML

.3

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-12 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Joe’s Newsstand Example

Joe sells newspapers for $1.00 each. Papers cost him $.40

each. His average daily demand is 50 papers with a

standard deviation of 10 papers. Assuming sales follow a

normal distribution, how many papers should Joe stock?

MLML = $0.40 MPMP = $0.60 = Average demand = 50 papers per day = Standard deviation of demand = 10

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-13 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Joe’s Newsstand Example (continued)

Step 1:

..MPMPMLMLMLML

PP

.

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-14 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Joe’s Newsstand Example (continued)

Step 2: Look on the Normal table for

PP = 0.4 ZZ = 0.25,

and

or:

**XX

XX** = 10 * 0.25 + 50 = 52.5 or 53 newspapers = 10 * 0.25 + 50 = 52.5 or 53 newspapers

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-15 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Joe’s Newsstand Example B Joe also offers his clients the “Times” for $1.00. This paper is flown in

from out of state, which greatly increases its costs. Joe pays $.80 for the “Times.” The “Times” has average daily sales of 100 papers with a standard deviation of 10. Assuming sales follow a normal distribution, how many “Times” papers should Joe stock?

MLML = $0.80 MPMP = $0.20 = Average demand = 100 papers per day = Standard deviation of demand = 10

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-16 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Joe’s Newsstand Example B (continued)

Step 1:

..MPMPMLML

MLMLPP

.

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-17 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Step 2:

Z = 0.80

= -0.84 for an area of 0.80

And

or: X=-8.4+100 or 92 newspapers

**XX

Joe’s Newsstand Example B (continued)

YASAI• GENUNIFORM(a, b): Both arguments are numbers. Normally, it is expected that a

< b. If so, a random number uniformly distributed over the interval [a, b) -- that is, x such that a < x < b -- is returned. If a = b, then the value a (or equivalently b) is returned. If a > b, an error value is returned.

GENNORMAL(m, s): Both arguments are numbers. If s < 0, an error value is returned. If s is zero, the return value is m. Otherwise, a random value with a normal distribution with mean m and standard deviation s is returned.

GENBINOMIAL(n, p): The first argument n must be a nonnegative integer, and the second argument p must be a number in the range [0, 1]. Otherwise, an error value is returned. If these conditions are met, then the return value is an integer drawn randomly from a binomial distribution with n trials and probability p of success at each trial. Note that if n = 0, then the return value is 0. The implementation is efficient even when n is large.

• GENPOISSON(m): The argument m is a nonnegative number. A negative argument causes an error value to be returned. A zero argument causes zero to be returned. Otherwise, the return value is randomly chosen from a Poisson distribution with mean value m. The implementation is efficient even when m is large.

• GENTABLE(V, P): The argument V and P are blocks of cells or lists (for example, "{1,3,7}") having the same number of cells. Essentially, the function returns each value in V with the probability specified by the corresponding element in P. If the two arguments have the same number of cells but differing numbers of rows and columns, the correspondence is determined by scanning first across the first row, then across the second row, and so forth. Non-numeric entries in P are treated as if they were zero. If the two arguments do not have the same number of cells, or if P contains any negative numbers, or if P contains only zeroes, an error value is returned. If the values in P do not sum to 1, they are rescaled proportionally so that they do. For example, GENTABLE({1,2,3},{.2,.5,.3}) returns 1 with probability 0.2, 2 with probability 0.5, and 3 with probability 0.3.

• GENEXPON(a): The argument must be a positive number, or an error value is returned. If so, the return value is randomly chosen from an exponential distribution with mean value 1/a.

• GENGEOMETRIC(p): Returns a geometric random variables with a probability p of being 1. This variable is equal to the number of trials of a mean p Bernoulli (or equivalently, GENBINOMIAL(1,p)) variable until the value 1 is obtained. The value of p must be greater than 0, and less than or equal to 1, or an error value is returned.

• GENTRIANGULAR(a, b, c): Returns a value from a triangular distribution with minimum a, mode b, and maximum c. The arguments must be numbers with the property a < b < c, or an error value is returned.

Specifying Output

• To specify an output of the simulation, use the formula SIMOUTPUT(x, name):

• For example a cell containing =SIMOUTPUT(A4+B7,"profit") defines an output called "profit" whose value is A4+B7.

• Running the Simulation Once you have built your model, specified scenarios (if any), and specified outputs, you can run your simulation. To do so, select "YASAI Simulation" from the Tools menu.

To accompany Quantitative Analysis for Management, 9e by Render/Stair/Hanna

3-22 © 2006 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Newsvendor Problem

Daily Sales

(Cartons)

Probability of Sales

at this Level

Probability that Sales Will

Be at this Level or Greater

100 0.30 1.00

150 0.20 0.70

200 0.30 0. 50

250 0.15 0.20

300 0.05 0.05

1.00

Calendar is sold for $4.5. Each calendar costs $2. The following table shows the discrete distribution for sales.Any unsold calendar are returned for a $0.75 refund

Overbooking Problem• You are taking reservations for an airline flight. This particular flight uses

an aircraft with 50 first-class seats and 190 economy-class seats. • First-class tickets on the flight cost $600, with demand to purchase them

distributed like a Poisson random variable with mean 50. Each passenger who buys a first-class ticket has a 93% chance of showing up for the flight. If a first-class passenger does not show up, he or she can return their unused ticket for a full refund. Any first class passengers who show up for the flight with tickets but are denied boarding are entitled to a full refund plus a $500 inconvenience penalty.

• Economy tickets cost $300. Demand for them is Poisson distributed with a mean of 200, and is independent of the demand for first-class tickets. Each ticket holder has a 96% chance of showing up for the flight, and "no shows" are not entitled to any refund. If an economy ticket holder shows up and is denied a seat, however, they get a full refund plus a $200 penalty. If there are free seats in first class and economy is full, economy ticket holders can be seated in first class.

Overbooking Problem• The airline allows itself to sell somewhat more tickets than

it has seats. This is a common practice called "overbooking". The firm is considering the 18 possible polices obtained through all possible combinations of

• Allowing overbooking of up to 0, 5, or 10 first-class seats

• Allowing overbooking of up to 0, 5, 10, 15, 20, or 25 economy seats

• Which option gives the highest average profit? What are the average numbers of first-class and economy passengers denied seating under this policy. If no overbooking of first class is allowed, what is the best policy?