Simulation Lecture Notes and the Gentle Lentil Case

General Overview of the Case
• What is the decision problem presented in the case? • What are the issues Sanjay must consider in deciding among the alternative choices?
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Accounting Model of Gentle Lentil’s Monthly Earnings
What are the monthly fixed costs?
L = labor costs (between $5,040 and $6,860) U = rent, utilities, other unavoidable costs = $3,995

What are the monthly variable costs?
F = food costs M = number of meals served in month F = $11 x M

What are the monthly total costs?
L + U + F = L + 3,995 + 11 x M
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Gentle Lentil’s Monthly Earnings,cont.
What are the monthly revenues?
R = monthly revenues P = price of meal R = PxM

What are the monthly earnings?
X = monthly earnings = revenues – costs = P x M – ( L + 3,995 + 11 x M ) = (P – 11 ) x M – L – 3,995

Which of these quantities are random variables?
P M L X = = = = price of prix fixe meal number of meals sold labor cost monthly earnings
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(X is a function of random variables, so it is a random variable)

Assumptions Regarding the Behavior of the Random Variables M P = number of meals sold per month
We assume that M obeys a Normal distribution with µ = 3,000 and σ = 1,000

=

price of the prix fixe meal

We assume that P obeys the following discrete probability distribution

Scenario Very healthy market Healthy market Not so healthy market Unhealthy market

Price of Prix Fixe Meal $20.00 $18.50 $16.50 $15.00

Probability 0.25 0.35 0.30 0.10

L

= labor costs per month
We assume that L obeys a uniform distribution with a minimum of $5,040 and maximum of $6,860 4

The Behavior of the Random Variables, cont.
X = earnings per month
We do not know the distribution of X . We assume, however, that

X = (P – 11 ) x M – L – 3,995
Always ask the following questions in any management analysis:

• •

How realistic is this model? How good are the assumptions?
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What question do we want to ask & answer about the random variable X?
• What is the shape of the probability distribution of X ? • What is E(X) ? • What is SD(X) ? • What is P(X ≤ $5,000) ? • What is P(X ≥ $6,667) ? • Other questions to ask?
And at the end:

What would you do if you were Sanjay?
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Simulation of Gentle Lentil Monthly Earnings for “February”
Choose a value of M (number of meals served) that obeys the probability distribution N ( 3,000 , 1,000 ) Choose a value of P (price of prix fixe meal) that obeys the discrete probability distribution for P Choose a value of L (labor cost) that obeys the uniform distribution in the range [ 5,040 , 6,860 ] Compute the monthly earnings X by computing X = (P – 11 ) x M – L – 3,995 Run this n times. We used n = 1,000 This will generate as output n numbers x1, x2, ..., xn.

What shall we do with this output?

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What to do with output created by the simulation?
We have as output the n numbers x1, x2, ..., xn
• Create an estimate of the shape of the underlying probability distribution of X (earnings). • Create an estimate of the mean µ of X • Create an estimate of the standard deviation σ of X • Create an estimate of P(X ≤ $5,000) • Create an estimate of P(X ≥ $6,667)
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Example: Let P denote the price of the prix fixe meal at Gentle Lentil
We assume that P obeys the following discrete probability distribution:
Price of Prix Fixe Meal ($) 20.00 18.50 16.50 15.00 Create the following random number assignments: Price of Prix Fixe Meal ($) 20.00 18.50 16.50 15.00 Probability .25 .35 .30 .10 Random Number .973 .020 .802 .663 .965 .553 Random Number Assignment .00 -- .25 .25 -- .60 .60 -- .90 .90 -- 1.00 Price of Prix Fixe Meal ($) 15.00 20.00 16.50 16.50 15.00 18.50 0.25 0.35 0.30 0.10

Illustration:

Trial 1 2 3 4 5 6

How to Generate Random Numbers from a Continuous Probability Distribution
• Most software programs that perform simulation have the capability to generate random numbers from a variety of standard continuous distributions, such as the Normal distribution, the uniform distribution, etc. • The user need only specify the type of distribution and the parameters (µ and σ for the Normal, a and b for the uniform, etc.)

However, it is worthwhile to point out how the computer accomplishes this task
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The Probability Density Function f(y) of the Random Variable f(y) 0.3

0.2

0.1

0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

y

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Cumulative Distribution Function F(y) of the Random Variable
F(Y) 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

y

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Creating Sample Data Drawn from a Continuous Probability Distribution
1. Use a random number generator to generate a number x that obeys a uniform distribution between 0.0 and 1.0 2. Place the number x on the vertical axis of the graph of the cdf F(y) of the distribution of interest. Then find the point y on the horizontal axis whose cdf value F(y) is equal to x
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Cumulative Distribution Function F(y) of a Random Variable
F(Y) 1.0 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0.0 1.0 2.0 3.0 4.0 5.0 y 6.0 7.0 8.0 9.0 10.0 y = 6.75

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1,000 Trials of Gentle Lentil Monthly Simulation ($)
Trial Number 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 … … 991 992 993 994 995 996 997 998 999 1,000
Sample Mean Sample S.D.

Random Number 0.6616 0.9233 0.3003 0.3774 0.3621 0.4994 0.8688 0.3809 0.3605 0.1199 0.4668 0.0229 0.956 0.9251 0.6625 0.0778 0.9002 0.4938 0.6698 0.669 0.0554 … … 0.8662 0.3161 0.2757 0.6844 0.0572 0.7899 0.5202 0.1155 0.3012 0.9151

Price of Meal 16.50 15.00 18.50 18.50 18.50 18.50 16.50 18.50 18.50 20.00 18.50 20.00 15.00 15.00 16.50 20.00 15.00 18.50 16.50 16.50 20.00 … … 16.50 18.50 18.50 16.50 20.00 16.50 18.50 20.00 18.50 15.00

Random Number 0.1024 0.0971 0.1691 0.2059 0.8036 0.2903 0.6069 0.3227 0.9958 0.8115 0.2747 0.4822 0.8403 0.0298 0.6104 0.9529 0.732 0.6949 0.4064 0.4822 0.2164 … … 0.3053 0.363 0.0123 0.797 0.9871 0.8378 0.3713 0.7075 0.2901 0.292

Number of Meals Served 1,732.13 1,701.95 2,042.38 2,179.43 3,854.52 2,447.60 3,271.18 2,539.71 5,634.33 3,883.56 2,401.48 2,955.35 3,995.79 1,115.85 3,280.36 4,673.74 3,618.72 3,509.67 2,763.19 2,955.45 2,215.43 … … 2,490.73 2,649.49 751.69 3,831.11 5,230.17 3,985.51 2,671.58 3,546.01 2,446.91 2,452.37

Random Number 0.803 0.649 0.262 0.609 0.409 0.208 0.469 0.224 0.557 0.817 0.422 0.91 0.446 0.384 0.459 0.178 0.773 0.187 0.171 0.785 0.288 … … 0.186 0.537 8E-04 0.87 0.197 0.201 0.165 0.754 0.627 0.365

Labor Costs 6,500.94 6,221.59 5,517.08 6,147.86 5,784.60 5,417.72 5,893.94 5,448.50 6,053.06 6,527.21 5,807.71 6,695.67 5,851.35 5,738.51 5,874.85 5,364.04 6,446.92 5,380.98 5,350.51 6,469.55 5,564.93 … … 5,378.26 6,017.32 5,041.42 6,623.90 5,398.59 5,405.81 5,339.57 6,412.91 6,181.58 5,704.71

Monthly Earnings -969.23 -3,408.81 5,805.77 6,202.87 19,129.32 8,944.26 8,102.56 9,604.32 32,209.44 24,429.81 8,208.35 15,907.51 6,136.81 -5,270.09 8,172.15 32,704.61 4,032.97 16,946.53 5,852.02 5,790.42 10,378.95 … … 4,325.72 9,858.83 -3,398.72 10,452.21 37,677.96 12,519.49 10,702.25 21,506.21 8,175.24 109.75
10,303.59 8,491.70

Discussion of Simulation Model Output
Let x1, x2, ..., xn denote the values of the monthly earnings obtained for each of the n = 1,000 trials Then x1, x2, ..., xn are each observed values from the distribution of the random variable X

Armed with these numbers, we want to estimate:
• • • • the shape of the probability distribution of X the mean µ of X the standard deviation σ of X the probability that X will lie in a given range, i.e., Pr( a ≤ X ≤ b) for given values of a and b
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Discussion of Simulation Model Output, cont.
Estimating the shape of the distribution of the random variable X:
A histogram of the trial values x1, x2, ..., xn is a good estimate of the shape of the probability distribution of the random variable X

Estimating the probability that X will lie in a given range:

Suppose that we want to estimate Pr(a ≤ X ≤ b) for given values of a and b. Let m denote the number of values among the n observations x1, x2, ..., xn that are in the range between a and b. Let p =

that are in the range between a and b.

m . Note that p is simply the fraction of the observations x1, x2, ..., xn n

Then p is an estimate of the probability Pr(a ≤ X ≤ b).

Probability
0.10 0.02 0.04 0.06 0.08 0.12 0
-6 ,0 00 -4 ,0 00 -2 ,0 00

Monthly Earnings ($)

Histogram of Monthly Earnings from the Gentle Lentil Simulation Model

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0 2, 00 0 4, 00 0 6, 00 0 8, 00 0 10 ,0 00 12 ,0 00 14 ,0 00 16 ,0 00 18 ,0 00 20 ,0 00 22 ,0 00 24 ,0 00 26 ,0 00 28 ,0 00 30 ,0 00 32 ,0 00 34 ,0 00 36 ,0 00 38 ,0 00 40 ,0 00 42 ,0 00 44 ,0 00

Estimating the mean µ and standard deviation σ
Compute the observed sample mean of the random variable X

x=

For example, from the spreadsheet we obtain x= −$969 .23 − $3,408 .81 + ... + $109 .75 = $10 ,303 .59 1,000

x1 + x2 + ... + xn n

Compute the observed sample standard deviation of the random variable X : s = i=1 ∑

n

( xi − x ) n − 1

2

2 2 2 (–969.23 – 10,303.59) + (–3,408.81 – 10,303.59) + ... + (109.75 – 10,303.59) s = 999 = 72,108,968.89 s = $8,491.70 2

For example, from the spreadsheet we obtain

Simulation of Monthly Earnings at Gentle Lentil using Crystal Ball™
A random variable is modeled in Crystal Ball as an "assumption" cell • for each assumption cell, the user must choose and describe the parameters of the probability distribution of the random variable in the cell Suppose we are interested in finding out the unknown probability distribution of one or more quantities in the spreadsheet. In Crystal Ball, if we designate a cell as a "forecast" cell, then Crystal Ball will automatically compute :

• the histogram • sample mean • sample standard deviation • all sorts of other useful information for the cell in the simulation

Crystal Ball Output
™

Monthly Salary
Forecast: Monthly Salary [GL.XLS]Monthly - Cell: H21 Summary: Display Range is from ($15,000) to $35,000 $/Month Entire Range is from ($11,038) to $40,144 $/Month After 1,000 Trials, the Std. Error of the Mean is $272 Statistics: Trials Mean Median (approx.) Mode (approx.) Standard Deviation Variance Skewness Kurtosis Coeff. of Variability Range Minimum Range Maximum Range Width Mean Std. Error Value 1000 $10,526 $9,605 $6,620 $8,602 $73,995,647 0.51 3.06 0.82 ($11,038) $40,144 $51,183 $272.02

Forecast: Monthly Sal ary

Cel l H21
.040 .030 .020 .010 .000 ($15,000) ($2,500)

Frequency Chart

995 Trial s Sho wn
40 30 20 10 0

$10,000 $/Month

$22,500

$35,000

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Interpretation of Simulation Results: Monthly Earnings at Gentle Lentil
• What is the shape of the distribution of monthly earnings? • What is an estimate of the expected monthly earnings? • The standard deviation?

x

= $10,526

s = $8,602

• What is an estimate of the probability that Sanjay will earn less than $5,000 in a given month? 27.6% • What is an estimate of the probability that Sanjay will earn more than in consulting ($6,667 per month) in this particular month? 63.3%

What would you do if you were Sanjay?

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Simulation of Annual Earnings at Gentle Lentil
Instead of looking just at monthly earnings at Gentle Lentil, it would be better to look at annual earnings What should the annual model assume? All assumptions for the monthly model, plus: • there are twelve distinct random variables M1, ... , M12 for the twelve distinct number of meals served in each month • these random variables will be independent and identically distributed (i.i.d.) • the random variable P (price of prix fixed meal) remains the same over the entire year • the random variable L (labor cost in month) remains the same over the entire year
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Simulation of Annual Earnings at Gentle Lentil, cont.
Use a random number generator to generate random values of P , L , and M1, ... , M12 according to their respective distributions Xi = (P – 11 ) x Mi – L – 3,995 for i = 1, ..., 12 A = annual earnings = X1 + X2 + X3 +... + X12 How is this model different from the monthly model?
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Simulation Output from Crystal Ball Annual Proprietor Salary
Forecast: Annual Proprietor Salary Summary: Display Range is from ($50,000) to $300,000 $/Year Entire Range is from ($15,853) to $292,519 $/Year After 1,000 Trials, the Std. Error of the Mean is $2,053 Statistics: Trials Mean Median (approx.) Mode (approx.) Standard Deviation Variance Skewness Kurtosis Coeff. of Variability Range Minimum Range Maximum Range Width Mean Std. Error
Fo r e c a s t: A n n u a l P r o p ri e t o r S a l a ry Cell D14
.0 35

™

[GL.XLS]Annual-Poca - Cell: D14

Value 1000 $124,585 $121,030 $81,284 $64,937 $4,216,752,610 0.13 2.05 0.52 ($15,853) $292,519 $308,372 $2,053.47

Fr e q u e n c y C h a rt

1 , 0 0 0 T ri a l s
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.0 26 .0 18 .0 09 .0 00 ( $5 0 ,00 0) $ 37 ,500 $ 12 5 ,0 0 0 $ /Y e ar $ 21 2,5 0 0 $ 30 0 ,00 0

2 6.2 1 7.5 8 .7 5 0

Interpretation of Simultaneous Results: Annual Earnings at Gentle Lentil/Proprietorship
• What is the shape of the distribution of annual earnings? • What is an estimate of the expected annual earnings? The standard deviation? x = $124, 585, s = $64, 937 • How do these annual statistics compare to the monthly statistics? • What is an estimate of the probability that Sanjay will earn less than $60,000?

16.2%
• What is an estimate of the probability that Sanjay will earn more than $80,000?

69.3%

What would you do if you were Sanjay? 26

Summary Table of Some Relevant Estimates
Choice Consulting Estimate of Mean $80,000 Estimate of Std.Dev. $ 0 95% Confidence Estimate of Estimate of Interval of Mean Pr(Salary > $80.000) Pr(Salary≤ $60,000) $80,000 ± 0.0 100% 0%

Proprietorship

$124,585

$64,937

$124,585± 4,025

69.3%

16.2%

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Simulation of Annual Earnings at Gentle Lentil Under a Partnership
What are the terms of the partnership?
Restaurant Earnings in month X $3,500 ≤ Sanjay’s Earnings in month $3,500

$3,500≤ X ≤ $9,000

X

X> $9,000

$9,000 + 0.10 * ( X – 9,000 )

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Sanjay's Monthly Salary
Under Financial Partnership ($)
11,000 10,000

Sanjay's Monthly Salary ($)

9,000 8,000 7,000 6,000 5,000 4,000 3,000 2,000 1,000 0 0 2000 4000 6000 8000 10000 12000 14000
16000

Gentle Lentil Restaurant Monthly Earnings ($)

Simulation Output from Crystal Ball Annual Partnership Salary
Forecast: Annual Partnership Salary Summary: Display Range is from ($50,000) to $300,000 $/Year Entire Range is from $42,000 to $126,452 $/Year After 1,000 Trials, the Std. Error of the Mean is $637 Statistics: Trials Mean Median (approx.) Mode (approx.) Standard Deviation Variance Skewness Kurtosis Coeff. of Variability Range Minimum Range Maximum Range Width Mean Std. Error Value 1000 $89,887 $92,530 $99,005 $20,154 $406,203,025 -0.44 2.25 0.22 $42,000 $126,452 $84,452 $637.34

™

[GL.XLS]Annual-Poca - Cell: D15

Forecast: An nual Partne rship Sal ary Cel l D15
.082 .061 .041 .020 .000 ($50,000) $37,500 $125,000 $/Year $212,500 $300,000

Frequency Chart

1,000 T ri als
82 61.5 41 20.5 0

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Interpretation of Simulation Results: Annual Earnings at Gentle Lentil / Partnership
What are the expected annual earnings and standard deviation under the partnership? x = $ 89 , 887 s = $ 20 ,154

In the partnership, what is an estimate of the probability that Sanjay will earn less than $60,000? 10.2% In the partnership, what is an estimate of the probability that Sanjay will earn more than $80,000? 68.0%
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Summary Table of Some Relevant Estimates
Choice Consulting Proprietorship Partnership Estimate of Mean $80,000 $124,585 $89,887 Estimate of Std. Dev. $ 0 $64,937 $20,154 95% Confidence Interval of Mean $80,000 ± 0.0 $124,585 ± 4,025 $89,887 ± 1,249 Estimate of Pr(Salary ≥ $80,000) 100% 69.3% 68.0% Estimate of Pr(Salary ≤ $60,000) 0% 16.2% 10.2%

Estimate of Pr(Proprietorship outperforms Partnership) = 71.7%

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Comparison: Partnership versus Proprietorship

What is the likelihood that the proprietorship is better than the partnership?

71.7%
Should Sanjay go for Gentle Lentil alone, or in financial partnership with his aunt?
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Simulation Output from Crystal Ball Annual Proprietor Premium
Forecast: Annual Proprietor Premium Summary: Display Range is from ($75,000) to $175,000 $/Year Entire Range is from ($57,853) to $166,067 $/Year After 1,000 Trials, the Std. Error of the Mean is $1,448 Statistics: Trials Mean Median (approx.) Mode (approx.) Standard Deviation Variance Skewness Kurtosis Coeff. of Variability Range Minimum Range Maximum Range Width Mean Std. Error Value 1000 $34,698 $27,610 ($753) $45,787 $2,096,427,661 0.38 2.15 1.32 ($57,853) $166,067 $223,920 $1,447.90

™

[GL.XLS]Annual-Poca - Ce ll: D16

Cel l D16
.037 .028 .019 .009 .000 ($75,000)

Forecast: An nual Propri etor Premi um Frequency Chart

1,000 T ri als
37 27.7 18.5 9.25 0

($12,500)

$50,000 $/Year

$112,500

$175,000

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Discussion of the Estimates
How good are these estimates?
• Intuition suggests that when the number of trials is very large, then the estimates are very reliable • However, to give a more precise answer to this question, we will need to learn sampling theory
35

Some Lessons of Simulation
• Simulation attempts to measure things that average case analysis and simple formulas cannot • The successful application of a simulation model depends on the ability to generate random variables that obey a variety of discrete and continuous probability distributions • Simulation can demonstrate effects in a system that cannot otherwise be “derived”

36

Some Lessons of Simulation, cont.
• The results that one can obtain in a simulation are not precise, due to the inherent randomness in a simulation. Care must be used in interpreting simulation results • The typical conclusions that one can draw from a simulation model are:
- estimates of the distributions of particular quantities of interest - means and standard deviations of these distributions

• From these distributions, one can derive confidence intervals and other inferences of statistical sampling
37

Some Lessons of Simulation, cont,
• The question of how many trials or runs of a simulation can become a complex statistical issue • Fortunately, with today’s computing power, this is not a paramount issue for most problems • In practice, one should recognize that gaining managerial confidence in a simulation model will depend on at least three factors:
1. A good understanding of the underlying management problem 2. One’s ability to use the concepts of probability and statistics correctly 3. One’s ability to communicate these concepts effectively
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