Simulations
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
ML
P 
or
MP ML
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 Discrete Distributions
Steps using Discrete Distributions:
 Determine the value for 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.
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
3-3
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Café du Donut:
Marginal Analysis
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.
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
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 Solution
Marginal profit = selling price
- cost
= $6 - $4 = $2
Marginal loss = cost
Therefore:
ML
P
ML  MP
4
4


 0.667
42
6
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
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
3-6
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
In-Class Example 3
Let’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.
Daily
Sales
Cases
4
5
6
7
8
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
Probability that
Probability of
Sales at this Level Sales Will Be at this
Level or Greater
0.1
0.1
0.4
0.3
0.1
1.00
3-7
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
In-Class Example 3:
Solution
MP = $15-$4-$1 = $10 per case
P>= $4 / $10+$4 = .286
Daily
Sales
Cases
4
5
6
7
8
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
ML = $4
Probability of
Probability that
Sales at this Level Sales Will Be at this
Level or Greater
0.1
1.0 > .286
0.1
.9 > .286
0.4
.8 > .286
0.3
.4 > .286
0.1
.1
1.00
3-8
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Marginal Analysis
Normal Distribution




 = average or mean sales
 = standard deviation of sales
MP = marginal profit
ML = Marginal loss
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
3-9
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Marginal Analysis Discrete Distributions
• Steps using Normal Distributions:
 Determine the value for P.
ML
P
ML MP
 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
Z 
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
X -
*

3-10
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Marginal Analysis Normal Curve Review
area = .30
area = .70
ML
.3 
ML MP

X
*
Use table to find Z
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
3-11
© 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?
 ML = $0.40
 MP = $0.60
  = Average demand = 50 papers per day
  = Standard deviation of demand = 10
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
3-12
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Joe’s Newsstand Example (continued)
.40 
ML

0 .40
P
ML  MP .40  .60
Step 1:
.
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 2: Look on the Normal table for
P = 0.4  Z = 0.25,
and
or:
0 25 
X - 50
*
10
X* = 10 * 0.25 + 50 = 52.5 or 53 newspapers
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 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?




ML = $0.80
MP = $0.20
 = Average demand = 100 papers per day
 = Standard deviation of demand = 10
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
3-15
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Joe’s Newsstand
Example B (continued)
.8
ML

 0. 80
P
ML  MP .8  .2
Step 1:
.
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 2:
Z = 0.80
= -0.84 for an area of 0.80
And
- 0 .84 
X
*
- 100
10
or: X=-8.4+100 or 92 newspapers
To accompany Quantitative
Analysis for Management, 9e
by Render/Stair/Hanna
3-17
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
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.
Newsvendor Problem
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
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
To accompany Quantitative Analysis for
Management, 9e
by Render/Stair/Hanna
3-22
© 2006 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
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?