isom201 class notes
TRANSCRIPT
ISOM 201 - DATA AND DECISION ANALYSIS
CLASS NOTES AND SUPPLEMENT TO TEXT
Prof. Stephen McDonald
©2011
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Table of Contents
Introduction to the Course 3
Chapter 1 - Introduction to Quantitative Analysis 4
Chapter 2 - Probability and Statistics 8
Chapter 3 - Decision Analysis 18
Chapter 4 - Linear Regression 29
Chapter 5 - Forecasting 38
Chapter 7 - Linear Programming 53
Chapter 8 - Linear Programming Applications 67
Chapter 15 - Simulation 69
Appendix A - Activating the Excel Add-ins 88
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Introduction to the course
ISOM201 falls squarely in the middle of a critical sequence of courses which every Sawyer Business School student must successfully complete. The sequence is:
Math 130 (Math 134 for accounting or finance majors)ISOM120 Stats250 (Stats 240 is also acceptable)ISOM 201ISOM 319SIB429 (formerly MGT429)
The sequence is intended to build a strong quantitative, analytical and strategic skill set in all our students.
ISOM201 may be one of the first courses in your program of study that requires recall of prior course material and synthesis of that material into business problem solving. The three critical areas of prior coursework are:
1. Fundamental math and algebra skills2. Proficiency with Microsoft Excel3. The ability to recall and apply major concepts from statistics
Students who have traditionally had trouble in these areas or who may have earned good grades in these courses without actually retaining much of the material presented will struggle in this course. It is important to understand that while this course uses these important skills, it cannot re-teach them. The emphasis of ISOM201 is to apply these skills to business problems.
Students with a strong foundation in the prior coursework will be able to learn how to assess business problems in a logical and orderly fashion, develop quantitative models to examine alternative solutions and begin to think more critically about the different alternatives business decision makers face.
Doing well in this course may require a greater time commitment than you have become accustomed to in your previous studies. There are very few shortcuts to learning this material. For most students, success in this course is a direct reflection of the time that they are willing to commit to the course and their persistence in overcoming their mistakes and frustrations. Mastering the material demands repetition of many different problems to build comfort, speed and accuracy.
Using the Class Notes
The following class notes are a supplement to the required text, not a substitute. They are a dynamic work in progress and change every semester. Students should use these in conjunction with the text, classroom lectures and homework problems to create their own individual approach to mastering the material.
4.
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Chapter 1 – Introduction to Quantitative Analysis
Chapter 1 provides a broad overview of Quantitative Analysis. The concepts covered in this chapter will be used throughout the course. All of the specific applications of quantitative analysis we study this semester will involve:
Application of the quantitative analysis approach to problem solving Model building, solution and evaluation Computer solutions Business use of the models and techniques Understanding of the specific limitations, strengths and weaknesses of each modeling
technique
One of the more important concepts we will apply throughout this course is the quantitative analysis approach, described in section 1.3 of your text. We’ve all seen different versions of this logical, systematic and consistent process as far back as the very first science courses we ever took. The principles here are exactly the same as the principles we learned then.
Developing your ability to recognize and solve business problems in a structured, logical manner will increase your value in any field you
choose.
Prior to applying the steps outlined in this section, the analysis of any business problem begins with Observation. This requires much more than casual “looking.” It demands a level of concentration and scrutiny that is the product of a trained mind. Observation of a business system, process or problem is critical to the identification and anticipation of problems.
Chapter 1 is a very light warm-up and review of simple quantitative analysis concepts and techniques. This material should not be very challenging. If you find that you are having trouble understanding the concepts of break-even, contribution margin or sensitivity analysis, you might want to go back to earlier coursework and refresh your memory. I know you’ve done this before!
If the algebra is troubling or your Excel skills are weak, now is the time to get to work. Do MORE problems than I have assigned and solve every
problem with algebra AND Excel. There is no magic to these skills. They are acquired and improved through repetition. You may also want to
consider contacting the Ballotti Learning Center now to sign up for study groups or tutoring services.
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1.3 The quantitative analysis approach
1) Defining the Problem. This step follows from our observation of the business situation. The definition of a problem should tell managers what objectives of the firm are not being achieved. Problem Definition should not extend into speculation regarding the causes of the problem or potential solutions.
2) Developing a Model. Model Building is the technique that we will focus on in this course. Models are used to represent existing problems or situations. A well-constructed model will permit us to see the relationships that exist among the variables in the problem.
3) Acquiring Input Data. The accuracy and quality of the data we collect is critical to the success of the model. The techniques learned in statistics will serve us well in collecting fair, random, representative and objective data.
4) Developing a solution. Once the appropriate model is constructed, the model must be manipulated in order to find the optimal or best possible solution to the problem. Often this will involve some trial and error, or other techniques which MS Excel can greatly simplify.
5) Testing the solution. Finding a single solution is not the end, but a new beginning. After a working model has been solved, it needs to be tested to make sure that the goals of the model are consistently applied across a broad array of input data and results. The process of analyzing the results for different input data requires both a thorough knowledge of the operating mechanics of the model and a strong level of common sense to assure that the model is correctly constructed.
6) Analyzing the results. The solution to the model suggests certain management actions and the implication of those actions must be fully understood before implementing the solution. Additionally, since the solution is often based on a single set of data, the behavior of the model and the actions suggested for management must be evaluated for differing input data. The process of analyzing the results as the input data is changed, or a business conditions changes in called Sensitivity Analysis.
7) Implementing the results. The final step, the only real proving ground is implementation. This can be much more difficult than an organization might imagine, as it brings in all the subjective, organizational and emotional issues that mangers must face when implementing change in an organization. The ability to skillfully develop and implement positive changes in organizations requires excellent quantitative and people skills and is a critical skill that separates top quality managers and executives from the “middle of the pack.”
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1.4 Model Development
Many of the mathematical models we construct will be solved in three ways:1. Graphically (using handwritten visual estimates or MS Excel)2. Algebraically (using formulas provided in the text)3. With MS Excel
a. Building our own spreadsheetsb. Using MS Excel Add-ins such as Data Analysis Toolpak and Solverc. Using MS Excel Graphs and their advanced options
Our models will be generally categorized by whether they involve risk or chance. Models that do not contain an element of risk are known as deterministic models. These rely on the assumption that the values in the model are known with certainty. If risk is present, we have a probabilistic model and we will generally use probability to represent the likelihood that different events may occur in our model.
Cost/Volume/Profit and Break-Even Analysis
The first model presented in the text is Break-Even Analysis. I consider this the Fundamental Equation of Business because it forces us to focus on profit and the three basic variables that cause a firm to earn profit or incur losses.
Profit = Revenue – Cost
Now let’s add some algebra to make the model a little more useful:
Revenue (R) is defined as the selling price per unit (s) multiplied by the number of units sold (X). R = sX
Cost (C) is comprised of fixed costs (f) and variable costs (v). Fixed costs remain constant and are independent of the volume of sales or production. Variable costs depend on the number of items produced (X). C = f + vX
Profit (P) is determined by: P = sX - (f + vX). Simplifying with a little more algebra, we derive:
P = sX - (f + vX) P = sX - f - vX P = X(s - v) - f
This formula identifies two key concepts:
1. Contribution Margin per unit = (s – v)
2. Break Even volume (Solve for X with P = 0) : X= fs−v
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Break even Example - Problem 1-15
Given: Variable cost is $20 per unit.Selling price is $50 per unit.Fixed cost is $150.
Find: Number of units required to break even
Long method:
Shorter: X= fs−v
= 15050−20
=15030
=5
5 units must b e sold to break even. At that sales level, the income statement is:
Sales $250
Variable costs 100Fixed costs 150Total costs 250
Net Income $ 0
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P=sX−(vX+ f )0=50 X−(20 X+150 )0=50 X−20 X−1500=30 X−15030 X=150X=5
Chapter 2 – Probability Concepts and Applications
Disclaimer: It is impossible to overemphasize how incomplete this review is, how important a firm command of statistics is for this course and for your future career and how strongly I believe you should find a good resource to make sure your retention of this material is fairly complete. In addition, certain liberties have been taken with proper statistical procedures and the precise interpretation of statistical tests in order to illustrate the application of the core concepts of the tests to this course. While these liberties may be acceptable in general management practice, they do not constitute a complete or completely correct presentation of the concepts.
What follows is the bare minimum necessary for survival in ISOM201 and nothing more.
Chapter 2 in our text presents a good review of many of the key concepts from your statistics course. If other aspects of your statistics course need some “refreshing,” I recommend the following book:
Schaum’s Easy OutlinesBusiness Statistics
Leonard J. Kazmier, Ph. D.ISBN 0-07-139876-7
I generally carry this around with me and have found it to be a quick and effective guide to the essential statistical concepts that all business professionals should know.
Topics: Basic Probability Rules Expected value of a Discrete Random Variable Conditional, Total and Bayesian Probabilities The Normal Distribution Confidence Interval Estimation Hypothesis Testing (including ANOVA)
I have found it to be an effective guide to the essential statistical concepts that all business professionals should know but can’t always remember from prior coursework.
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Basic Probability Rules for your review:
All probabilities values must be between 0.00 and 1.00, also expressed as 0% to 100%.
The sum of all the probabilities for all possible outcomes of an experiment equal 1.00.
Two events are mutually exclusive if only one of the events can occur on any one trial.
An outcome set is defined as collectively exhaustive if the list of outcomes includes every possible outcome.
The Rule of Complements - The probability of event A happening is equal to 1 minus the probability of event A not happening.
P( A )=1−P ( A ) The Union rule – The probability that event A or event B or both event A and event B will
happen is equal to the probability that event A will happen, plus the probability that event B will happen, minus the probability that both event A and event B will happen.
P( A∪B )=P( A or B )=P( A )+P(B )−P ( A and B ) The Intersection Rule for Independent Events - If events A and B are independent, then the
probability of both event A and Event B happening is equal to the product of the probability of event A and event B.
P( A∩B )=P( AB)=P( A and B )=P( A )∗P (B )
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Expected Value of a Discrete Random Variable
A probability distribution is a set of possible events (Xi) that comprises all possible outcomes. The sum of the probabilities in a probability distribution must be 1.00 and the expected value of the distribution is equal to:
E( X )=μ=∑i=1
n
[X i∗P( X i) ]
Example: Demand for white latex paint at Diversity Paint and Supply has always been between 0 and 4 gallons per day, with the probability of each outcome as follows:
Gallons Probability 0 0.20 1 0.40 2 0.25 3 0.10 4 0.05
This calculation can easily be done in Excel using the =SUMPRODUCT function:
4.1)(
2.3.5.4.0)(
05.*41.*325.*24.*12.*0)(
)(*)(*)(*)(*)(*)(
*)(
5544332211
1
XE
XE
XE
XPXXPXXPXXPXXPXXE
XPXXEn
iii
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Conditional Probability
Conditional probabilities are important to decision trees because these decision making situations often have a sequence of probabilistic events in which the posterior (later) probabilities change depending on the prior (earlier) outcomes.
Conditional Probability Rule -
P( A|B)=P( AB)P (B )
or P ( AB )=P( A|B)∗P (B )The first formula above should be read as, “The probability that event A will happen, given that event B has happened, is equal to the probability that both events A and B will happen divided by the probability that even B will happen.”
Example: A consulting firm is bidding for two projects. The company executives estimate that the probability of winning the bid for project A is 0.45. The executives also believe that if the company wins the bid for project A, then there is a 0.90 probability that they will also win the bid for project B. What is the probability the company will win both projects?
Given: Probability of A = P(A)=0.45Probability of B, given A = P(B|A)=0.90
Equation: P(AB) = P(B|A)*P(A)P(AB) = 0.90 * 0.45P(AB) = 0.405
Answer: There is a 0.405 chance the company will win both projects.
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The Law of Total Probability
This law is an extension of the conditional probability rule and is used to calculate the total probability of an event when only conditional probabilities are known:
P( A )=P( A|B)∗P(B )+P( A|B )∗P (B )This should be read as, “The probability of event A is equal to the Probability of A given B, times the probability of B, plus the probability of A given “not” B, times the probability of “not” B.
Example:
A company is considering the opportunity to conduct product test marketing before deciding to introduce a new product. Based on historical results from similar product introductions, the company estimates that there is a 60% probability that the product will be successful. An analysis of prior test marketing result had also indicated that 80% of all successful products had positive test marketing results and that 50% of all unsuccessful products also had positive test market results. Using this information, the company wants to know the probability that the test marketing results will be positive.
Probabilities:Test marketing results = Event AProduct introduction success = Event B
We want to know the probability that event A will happened and all we are given are the following facts:
Using these probabilities, we can calculate the probability of event A happening as follows:
P( A )=P( A|B)∗P(B )+P( A|B )∗P (B )P( A )=. 8∗. 6+. 5∗. 4P( A )=. 48+. 20P( A )=. 68
Note that the probability of event A differs depending on the outcome of event B. This tells us that these events are conditionally dependent.
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P( B)=.6 P ( B )=. 4P( A|B)= .8P( A|B)= .5
Bayes’ Theorem - The basic principle of Bayesian analysis is that additional information, if available, can sometimes help a decision maker improve the marginal probabilities of the occurrence of an event. The altered probabilities are referred to as revised probabilities.
In general, given two events, A and B that are conditionally dependent, Bayes’ rule can be written as:
P( A|B)=P( B|A )⋅P ( A )
P( B|A )⋅P ( A )+P( B|A )⋅P ( A )
Bayes’ Theorem allows us to “flip” the conditional probabilities and solve for P(A|B) when we are given its opposite, P(B|A).
Let’s consider two possible events, A and B. Event A represents the probability that market demand for a product will be high. Without any market research, this is estimates to be 60%. Event B represents the probability that market research will be predict high demand. The market research firm has provided us with their historical track record of market predictions and it shows that they predict high demand 90% of the time, given that demand later turned out to actually be high. What this means is that they are “right” 90% of the time when demand is high. Their track record also shows that they predict high demand 20% of the time, given that demand later turned out to be low. This means that they incorrectly issue high demand predictions 20% of the time. Notice that these are not complements. They represent pieces of two different conditional probabilities.
Here’s what we were given:
P( A )=. 60 P ( A )= . 40P( B|A )= .90P( B|A )=.20
We would like to know the probability of having high demand for the product, given that we conducted market research and it predicted high demand. Common sense would tell us that the probability of high demand should be better for a product that already has a reliable market research prediction of high demand than it would be for a product with no market research. We use Bayes’ theorem to produce the following:
P( A|B)=P (B|A )⋅P( A )P (B|A )⋅P( A )+P (B|A )⋅P( A )
P( A|B)= .9∗. 6.9∗. 6+. 2∗. 4
=. 54. 54+. 08
=. 54. 62
=.871
There is an 87.1% chance that demand will be high, given that the market research predicted that demand would be high. This is substantially better than the base, 60% probability of high demand.
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Contingency Tables
Contingency tables are very useful tools for solving Bayesian probability problems if you understand how the work.
Let’s look at the previous example using contingency tables.
Here’s what we were given, including the complements:
P( A )=. 60 P ( A )= . 40P( B|A )= .90 P( B|A )=.10P( B|A )=.80 P( B|A )=.20
From this, the Law of Conditional Probability permits the following calculations:
P( AB)=P (B|A )∗P( A )=. 90∗.60=.54P( A B)=P (B|A )∗P( A )=. 10∗.60=. 06P( A B)=P (B|A )∗P( A )=. 20∗. 40=. 08P( AB)=P (B|A )∗P( A )=. 80∗. 40=. 32
The contingency table would look like this:
Total
Total 1.00
A A
B
B
)(ABP )( BAP
)(ABP)( BAP
)(AP )(AP
)(BP
)(BP
Total
0.54 0.08 0.62
0.06 0.32 0.38
Total 0.60 0.40 1.00
A A
B
B
From the table the following calculations can be made:
The total probability of B = P(B) = 0.54 + 0.08 = 0.62
Posterior (revised) probabilities:
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P( A|B)=0. 540. 62
=0 . 871 P( A|B )=0. 080. 62
=0 .129
P( A|B)=0. 060. 38
=0 . 158 P( A|B )=0. 320. 38
=0 .842
The Normal DistributionMany data sets tend to follow a normal, or relatively normal distribution. The classic normal distribution is symmetrical (the mean = the median) and bell-shaped. The empirical rule can be applied to this distribution and tells us about the probability of a data point being close to or far from the mean. You should remember this picture:
The Empirical Rule:
68% of the population lies +/- 1 standard deviation from the mean. 95% of the population lies +/- 2 standard deviations from the mean. 99.7% of the population lies +/- 3 standard deviations from the mean.
Example 1: 100 students take a midterm exam. The results are normally distributed with a mean of 75 and a standard deviation of 9. What range of exam scores should include 95% of the students’ scores?
It can be inferred that 95% of the students scored between 57 and 93.o 57 = 75 - 2*9o 93 = 75 + 2*9
Example 2: What is the probability that a student scored between 65 and 85 on the exam?
Table 2.9, located on page 48 of the text, contains the areas under the curve for the standard normal distribution. This table can be used to calculate the probabilities associated with the following example:
o
6875.15625.84375.)8565(
15625.84375.1)11.1(11.1
84375.)11.1(11.1
11.19
10
9
758511.1
9
10
9
7565
XP
XPZ
XPZ
ZandX
Z
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Confidence Interval Estimation
Continuing with the data above, let’s assume a random selection of 10 exams was selected to get a rough feel for how students performed. The sample mean was 78 and the sample standard deviation was 11.
Note that the sample mean and standard deviation do not match the population mean and standard deviation. Sample statistics are only estimators of the true
population parameters. Oftentimes, we do not know the true population parameters and must use sample statistics to infer the values of these
parameters.
How should we present our estimate of the true mean exam score based solely on the sample statistics? We could use the sample mean and conclude that the population mean will be exactly 78. Does this make sense?
It shouldn’t make sense because we are making a sample statistic sound much too precise when we should know it is only an estimator of the population mean.
A confidence interval is a much better way of presenting our estimate of the population mean. A simple 95% confidence interval can be constructed in two steps:
1. Calculate the standard error of the estimate: s
√n by dividing the standard deviation by
the square root of the sample size. In this case 11
√10= 11
3.162=3.48.
2. Build the confidence interval +/- 2 standard errors from the sample mean.
78 ± (2∗3.48 )=78 ±6.96=[71.04 , 84.96 ]
It is reasonable to state that based on the sample data we are 95% confident the population means score will be between 71 and 85.
Does this range seem too wide to be useful? We often run into this problem when working with small sample sizes. Let’s see what happens to the range when we have a larger sample size, but the same sample mean and standard deviation. With a sample size of 50 instead of 10, we calculate the following standard error and confidence interval:
11
√50= 11
7.071=1.56
78 ± (2∗1.56 )=78 ±3.12=[74.88 , 81.12 ]
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Confidence intervals are important to our study of regression, as they permit us to make sensible and logical statements about the potential value of the dependent variable as projected by the regression equation.Hypothesis Testing - (including ANOVA)
Let’s see if we can simplify the often confusing topic of hypothesis testing. Here’s the basic process for all hypothesis tests:
A statement is made that needs to be tested. This is generally the alternative hypothesis (H1) because an overwhelming amount of evidence is required to support a claim.
The opposite, or all other conditions, then becomes the null hypothesis (H0). This hypothesis will either be rejected, or fail to be rejected based on the sample evidence.
We set the guidelines for the test:o The significance level
5% α = 95% confidence (most commonly used) 1% α = 99% confidence (stricter test, harder to reject H0) 10% α = 90% confidence (looser test, easier to reject H0)
o The distribution to be used for the test statistic
z = normal distribution t = student’s distribution Also F, χ2 and others
Conduct the test Evaluate results
A simplified approach:
For our purposes, always assume a 5% level of significance.o We will not reject the null hypothesis unless there is less than a 5% probability
that it is true. Remember, the null hypothesis gets the “benefit of the doubt.” It takes a lot of evidence to reject the null hypothesis.
In lieu of evaluating the test statistic, which can be both confusing and tedious, we should always convert the test statistic to its p-value.
o Fortunately, this is done for us by Excel in all the areas in which we need to
evaluate hypothesis test. Always (in this class) evaluate the results as follows:
o If p-value < 0.05, reject the null hypothesis, accept the alternative as true.
o If p-value > 0.05, do not reject the null hypothesis, there is not enough evidence
to accept the alternative as true.
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Chapter 3 – Decision Analysis
Decision Analysis
Our objective in studying decision making is to bring a strong element of analysis and objectivity to the decision making process. With a basic understanding of probability, an organized approach to sorting through the issues a decision maker faces and a commitment to logical analysis, we can construct models of decision making alternatives and use these models to better analyze the potential outcomes and consequences of our decisions.
Components of Decision Making
The two basic elements of decision making are the decision alternatives, the choices we can make, and the events (states of nature), that may occur in the future. It is important to distinguish carefully between these two.
We choose what to do when we are faced with a decision.
Probability governs the occurrence of an event.
Six steps of Decision Making:1. Define the problem – What’s going on?2. List the alternatives – What choices (decisions) do I have?3. Identify possible outcomes or events – What might happen?4. List the payoff for each “end point” – Monetary Outcomes5. Select an appropriate model – The right tool for the job6. Apply the model and analyze the results
We evaluate our decision alternatives, the monetary outcomes of our decisions and the possible events by constructing a payoff table. A payoff table helps us organize all the different outcomes possible and consider their relative desirability and/or probability.
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Let’s consider the following payoff table for a situation with 3 decision alternatives and four possible events:
A manufacturer is trying to decide on the appropriate production level of its product for the next month. It has three levels of production available and is trying to match production to unknown market demand for the product. Historically, the company has been able to classify demand into 4 categories and has forecast the following financial outcomes for each combination of production level and market demand.
Event:Low
DemandModerate Demand
Good Demand
High Demand
Decisions:Minimum Production 7,000$ 8,000$ 10,000$ 10,000$ Average Production (10,000)$ -$ 10,000$ 30,000$ Maximum Production (25,000)$ (10,000)$ 15,000$ 50,000$
How should we decide what to do?
First, we must understand that our decision is the production level and that the market demand is an event that we do not control.
Next, we must assess the amount of information available to us about the likelihood of each even happening. This assessment leads us into one of three decision-making environments: certainty, uncertainty and risk.
Certainty – the decision makers know which event will occur.
Uncertainty – the decision maker does not know what will happen and does not know enough to reliably assess the probability of occurrence of each event.
Risk – the decision maker does not know what will happen, but does know enough to reliably assess the probability of occurrence of each event.
4 Events (States of Nature)
3 Decision Alternative
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Decision Making under Certainty
When the decision maker in this example know which event will occur, it is easy to choose the alternative that produces the best possible outcome.
If Low Demand is known, then choose Minimum Production, resulting in $7,000 profit.
If Moderate Demand is known, then choose Minimum Production, resulting in $8,000 profit.
If Good demand is known, then choose Maximum Production, resulting in $15,000 profit.
If High Demand is known, then choose Maximum Production, resulting in $50,000 profit.
Decision Making Under Uncertainty
The text presents 5 techniques for decision making without probabilities. Decision makers are often faced with situations for which the probabilities of events occurring cannot be reasonably estimated. In these situations, more subjective decision analysis should be applied to the payoff table. Although the following techniques are often used by decision makers, they are rarely presented as formally as we have here.
The Maximax Criterion
The optimist uses this criterion. The maximax decision maker searches for the single best outcome and makes the decision that could achieve that outcome. In this case it would be Maximum Production.
Note that this criterion ignores all lesser outcomes from that decision, even though they still could occur.
The Maximin Criterion
Here’s the pessimist’s choice. Believing that the worst will happen, the maximin decision makes locates the worst possible outcome for each decision:
Minimum Production $7,000Average Production $(10,000)Maximum Production $(25,000)
Then, the decision alternative with the “best” worst outcome is selected. Here, the Maximin decision maker would select Minimum Production.
Note that this criterion ignores all possible “better” outcomes, even though they could occur.
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Event:Best
OutcomeWorst
OutcomeWeighted Average
Coefficient of Optimism 0.65 0.35
Decisions:Minimum Production 10,000$
7,000$
8,950$
Average Production 30,000$
(10,000)$
16,000$
Maximum Production 50,000$
(25,000)$
23,750$
Event:Low
DemandModerate Demand
Good Demand
High Demand
Weighted Average
Likelihood 0.25 0.25 0.25 0.25
Decisions:Minimum Production 7,000$
8,000$
10,000$
10,000$
8,750$
Average Production
(10,000)$
-$
10,000$
30,000$
7,500$
Maximum Production (25,000)$
(10,000)$
15,000$
50,000$
7,500$
The Criterion of Realism (Hurwicz)
This method introduces weighted average calculations. The decision maker using this criterion must assign a value to their level of optimism. The value called the coefficient of realism, α, and can range from 0 to 1, representing 0% to 100% optimism.
In this example, we will assume the decision maker is 65% optimistic.
The decision maker constructs a table showing only the best and worst outcomes for each decision. All other potential outcomes other than the best and worst are ignored in this method.
The coefficient of optimism is applied to each best outcome and its complement (1 - .65) is applied to each worst outcome, creating a weighted average outcome. The decision maker chooses the decision alternative with the best weighted Average outcome, in this case Maximum Production.
The Equal Likelihood Criterion (LaPlace)
This method treats all of the events as if they have an equal chance of occurring, creating the following payoff table:
Once again, the decision maker would find the highest weighted average outcome, selecting Minimum Production based on this method.
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The Minimax Regret Criterion
This method requires the construction of an opportunity loss or regret table. A regret table begins with the best decision for each event state and measures the amount of monetary regret from the best outcome to each of the other outcomes for that event state. The regret table for this example is as follows:
Event:Low
DemandModerate Demand
Good Demand
High Demand
Decisions:Minimum Production -$ -$ 5,000$ 40,000$ Average Production 17,000$ 8,000$ 5,000$ 20,000$ Maximum Production 32,000$ 18,000$ -$ -$
This criterion next looks at the maximum regret for each decision:
Minimum production: $40,000Average production: $20,000Maximum production: $32,000
The decision maker using this method then selects the decision with the minimum of all the maximum regrets, in this case Average Production.
Summary of preferred decisions using decision making without probability:
Criterion: Preferred decision: Key outcome value:
Maximax criterion Maximum production $50,000Maximin criterion Minimum production $ 7,000Hurwicz criterion (α = .65) Maximum production $23,750Equal likelihood criterion Minimum production $ 8,750Minimax regret criterion Average production $20,000
Note that the key outcome values do not represent the expected result of the decision. They are merely the value used by the decision maker to select the preferred decision using the respective criterion.
The actual outcome of each decision would still be dependent upon the state of market demand that occurred after the production decision was made.
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Decision Making Under Risk
Most of the problems we will work with have probabilities assigned for each event. In practice this is often done to help construct a payoff table and evaluate decision alternatives. It is always useful to question the probabilities assigned to events when you are outside the classroom. The decision analysis model is only as good as the accuracy of the event probabilities being used in the model. The key concept for these problems is the Expected Monetary Value for each decision alternative. This is calculated exactly the same way we calculated the expected value of a discrete probability distribution in chapter 2.
In our example problem, suppose we assigned the following event probabilities:Low demand .15Moderate demand .25Good demand .40High demand .20
The full payoff table would now look like this:
Event:Low
DemandModerate Demand
Good Demand
High Demand
Expected Value
Probability 0.15 0.25 0.40 0.20
Decisions:Minimum Production 7,000$ 8,000$ 10,000$ 10,000$ 9,050$ Average Production (10,000)$ -$ 10,000$ 30,000$ 8,500$ Maximum Production (25,000)$ (10,000)$ 15,000$ 50,000$ 9,750$
The calculation of the expected value for the Minimum Production alternative would be:
(7,000 * .15) + (8,000 * .25) + (10,000 * .40) + (10,000 * .20) = 9,050
Based on these event probabilities, we would prefer the Maximum Production alternative, as it has the highest expected value ($9,750).
When working with probabilities, we also can consider the concept of regret, or opportunity loss, and construct an opportunity loss table as follows:
Event:Low
DemandModerate Demand
Good Demand
High Demand
Expected Opportunity
LossProbability 0.15 0.25 0.40 0.20
Decisions:Minimum Production -$ -$ 5,000$ 40,000$ 10,000$ Average Production 17,000$ 8,000$ 5,000$ 20,000$ 10,550$ Maximum Production 32,000$ 18,000$ -$ -$ 9,300$
Again, the preferred decision would be maximum production, as it has the lowest expected opportunity loss ($9,300).
Must add to 1
23
Expected Value of Perfect Information
Decision makers are often presented with an opportunity to acquire information to help them make better decisions. In order to understand the value of the information, there must be a cost/benefit analysis performed on the information. One technique for doing this is the calculation of the Expected Value of Perfect Information (EVPI).
What decision would be chosen if the decision maker knew which event would occur with absolute certainty?
Clearly, the best decision for each event could be chosen. In this case:
If the event ________ was known… Choose…. Outcome…
Low demand Minimum production $ 7,000Moderate demand Minimum production $ 8,000Good demand Maximum production $15,000High demand Maximum production $50,000
The Expected Value WITH Perfect information is the sum of the products of each of these outcomes and their respective event probabilities:
Event Outcome Probability ProductLow demand $ 7,000 .15 $ 1,050Moderate demand $ 8,000 .25 $ 2,000Good demand $15,000 .40 $ 6,000High demand $50,000 .20 $10,000
Expected Value WITH Perfect Information $19,050
If a decision maker could achieve an outcome of $19,050 WITH perfect knowledge of the future, then the value of knowing the future would be the difference between the expected value WITH knowledge of the future and the expected value WITHOUT knowledge of the future:
Expected Value WITH Perfect Information: $19,050Expected Value WITHOUT Perfect information: $ 9,750
Expected Value OF Perfect Information (EVPI): $ 9,300
24
Decision Trees
Decision trees display the time sequence of decisions and events and are important when the decision making situation involves more than simply one decision followed by one event.
Decision trees are all constructed from a series of two symbols:
Decisions are represented square “decision nodes”
States of Nature (Events) are represented by circular “event nodes
Consider the example beginning on page 83 of the text. The decision facing Mr. Thompson is to choose among the construction of a large plant, construction of a small plant, or to do nothing.
This is represented as:
After he makes a decision, an event will happen. In this case the event, which is beyond his control, is either a favorable or unfavorable market. The tree is expanded to reflect this as follows:
Large
Small
Nothing
Large
Small
Nothing
favorable
favorable
unfavorable
unfavorable
25
The next step is to assign probabilities to each event node. The probabilities must sum to 1 at each node.
Next, all the payoffs must be recorded in the tree and the terminal outcome calculated for each branch of the tree.
At this point, the tree is fully constructed, but it is not solved.
200
-180
200
-180
0
Solving a decision tree requires two processes:
At each event node, calculate the EMV of the event. (SUMPRODUCT)At each decision node, choose the best financial result.
200
-180
100
-20
0
The EMV of construction a large plant is $10,000 (200,000*.5) + (-180,000*.5).The EMV of construction a small plant is $40,000 (100,000*.5) + (-20,000*.5).
The highest value is associated with the decision to construct a small plant. This is the best decision.
Large
Small
Nothing
Unfavorable (.5)
Favorable (.5)
Favorable (.5)
Unfavorable (.5)
40
Large
Small
Nothing
10
40
Unfavorable (.5)
Favorable (.5)
Unfavorable (.5)
Favorable (.5)
26
Note that, at this point, a payoff table could be used as easily as a decision tree for this problem. Decision trees become mandatory when a sequence of decisions and events develops as illustrated in the second part of the problem.
The opportunity to conduct a market survey creates a new branch off the initial decision node and expands the tree significantly.
200
-180
100
-20
0
190
-190
90
-30
-10
190
-190
90
-30
-10
49.2
Large
Small
Nothing
10
40
Unfavorable (.5)
Favorable (.5)
Unfavorable (.5)
106.4
Large
Small
Nothing
106.4
63.6
Unf (.22)(-$180k)
Fav (.78)($100k)
Unf (.22)(-$20k)
49.2
Fav (.78)($200k)
2.4
Large
Small
Nothing
-87.4
2.4
Unf (.73)(-$180k)
Fav (.27)($100k)
Unf (.73)(-$20k)
Fav (.27)($200k)
Conduct Survey
Survey Fav. (.45)(-$10k).
Survey Unf. (.55)(-$10k).
Favorable (.5)
27
Reviewing this expanded tree, there are several important points to consider.
The outcome of the survey is an event. Mr. Thompson does not control it. After receiving the report, Mr. Thompson must make a decision. The alternatives available
to Mr. Thompson are the same (Large, Small, Nothing) but the probabilities are different depending on a favorable or unfavorable report.
The cost of the survey is subtracted from each branch and is reflected in the final payoff at each endpoint.
The tree now has three separate decision points. The answer to the problem must address Mr. Thompson’s best alternative at each point.
The complete answer to this problem is,
Mr. Thompson should decide to conduct the survey. If it is favorable, he should build a large plant. If it is unfavorable, he should build a small plant. The expected value of this decision is $49,200.
Additionally, the maximum amount Mr. Thompson should be willing to pay for the survey is $19,200 ($10,000 + ($49,200 - $40,000)).
28
Chapter 4 – Linear Regression
Linear Regression
Objective: Linear regression is a tool used to analyze the relationship between multiple variables when we suspect that one of the variables is dependent upon a single or multiple independent variables.
The Simple Linear Regression Model
Predicting one thing (Y) from another (X).
The linear relationship is defined by the Population intercept (b0) and the Population slope (b1).
The form of the equation is: Y = intercept + (slope)(X) or, Y = b0 + b1X.
Y is the dependent variable, the variable we wish to explain or predict.
X is the independent variable or the predictor variable.
The model has the following assumptions: The relationship between X and Y is a straight-line relationship. The values of X are fixed, not random.
What we want to know…
1. What is the best possible line (model) for the data?2. Is there a statistically significant relationship between the dependent variable and any
of the independent variables in the model?3. How much of the variability in the dependent variable is explained by the model?4. Which independent variable (in a multiple regression) has a stronger or weaker
relationship with the dependent variable?5. How accurate will the model be as an estimator or predictor of the dependent
variable?
What we want to do…
1. Calculate point estimates for the predicted value of the dependent variable (Y) given value(s) for the independent variable(s) (X).
2. Construct a “rough” 95% interval for the predicted value of the dependent variable (Y) given value(s) for the independent variable(s) (X).
3. Understand the predicted change in the dependent variable that would result from a change to any of the independent variable values.
29
What is the best possible line (model) for the data?
The example beginning on page 118 of your text provides the following data regarding company sales for Triple A Construction and total local area payroll (a general indicator of the region’s economy).
Local Payroll ($100,000,000's
)
Sales ($100,000'
s3 64 86 94 52 4.55 9.5
Because we believe that the local payroll affects sales, we will use payroll as the independent variable (X) and sales as the dependent variable (Y).
Approach #1 – An Excel scatter plot with a trend line, equation and coefficient of determination.
Excel 2007 Instructions:
Simple Linear Regression Charts
1. Highlight the data, making sure X is in the first column and Y is in the Second.
2. Select Insert and Scatter.
30
3. Select “Scatter with only marker.”
4. Right click on any data point and select Add Trendline. Left click a check mark in Display Equation on chart and Display R-squared value on chart.
31
5. The following is produced.
6. Clean up and format to improve presentation.
y = 1.25x + 2R² = 0.6944
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7
Sales
Local Payroll
Triple A Construction
32
Approach #2 - Algebraic Solutions
The least squares method is used to calculate the estimated regression line. By minimizing the sum of the squared differences between each data point and the predicted Y for that point, we are applying the same concepts used in computing variance to create a line that “best fits” the data.
Least squares = Best fit
We will now solve Triple A using the algebraic method using the formulas in the text.
Triple A ConstructionX = Local Payroll ($100,000,000s) (Independent Variable)Y = Tiple A's Sales ($100,000's) (Dependent Variable)
X Y
3 6 -1 1 -1 1 1 5.75 0.0625 1.5625
4 8 0 0 1 0 1 7 1 0
6 9 2 4 2 4 4 9.5 0.25 6.25
4 5 0 0 -2 0 4 7 4 0
2 4.5 -2 4 -2.5 5 6.25 4.5 0 6.25
5 9.5 1 1 2.5 2.5 6.25 8.25 1.5625 1.5625
4 7 0 10 0 12.5 22.5 6.875 15.625
slope denominator slope numerator SST SSE SSR
Slope (b1) 1.25Intercept (b0) 2.00
Determination (r2) 0.6944
)( XX 2)( XX )( YY ))(( YYXX
2)( XX ))(( YYXX
2)( YY
2)( YY
Y 2)ˆ( YY
2)ˆ( YYX Y
2)ˆ( YY
2ˆ YY
The correct regression equation is Y = 2 + 1.25X and the coefficient of determination is .6944.
This approach can be used for both simple linear regression and multiple linear regression. However, the calculations for a multiple linear regression are beyond the scope of this course and have been made less important because of the ease with which Excel and other tools can conduct this analysis.
33
Approach #3 – Regression using Excel
Linear Regression on MS Excel 2007
1. Select Data
2. Select Data Analysis. If this does not appear, see the instructions for installing the data analysis add-in.
3. Select Regression from alphabetical list and click on “OK”.
34
4. Complete the dialog box as shown.
5. Select “OK” and the following output will be produced. Format the column widths, number of decimal places and other aspects to make the presentation clear and professional.
SUMMARY OUTPUT
Regression StatisticsMultiple R 0.833333R Square 0.694444Adjusted R Square0.618056Standard Error1.311011Observations 6
ANOVAdf SS MS F Significance F
Regression 1 15.625 15.625 9.090909 0.039352Residual 4 6.875 1.71875Total 5 22.5
CoefficientsStandard Error t Stat P-value Lower 95% Upper 95%Lower 95.0%Upper 95.0%Intercept 2 1.742544 1.147747 0.31505 -2.83808 6.838077 -2.83808 6.838077X 1.25 0.414578 3.015113 0.039352 0.098947 2.401053 0.098947 2.401053
The Output Range can be either the first cell that should contain the output in the current worksheet or a new worksheet.
35
If you review the 3 approaches carefully, you will find that they all produce the same results, including the slope, intercepts and coefficient of determination values for the model. However, only the Excel output above produces the additional statistical information that will permit us to fully understand the regression.
Interpreting the Regression Output
Is there a statistically significant relationship between the dependent variable and any of the independent variables in the model?
This is analyzed be looking to the Significance F value above. It is the p-value of the hypothesis test conducted on the model to determine whether X and Y are related. The null hypothesis in a regression is that there is NO statistical relationship between Y and any X variable. The p-value represents the probability that this statement may be true. In this case there is a .0394, or about 4% probability that there is NO relationship between Local Payroll and Triple A’s Sales.
How much of the variability in the dependent variable is explained by the model?
The coefficient of determination represents the percentage of total variability in Y that is explained by the regression equation. It is our best measure of the fit of the regression model. Determination is written as r2 and its value is between 0 and 1. 0 indicates that there is no relationship between X and Y and 1 indicates that the regression equation PERFECTLY explains Y.
Which independent variable (in a multiple regression) has a stronger or weaker relationship with the dependent variable?
This question is only important in a multiple linear regression containing more than 1 independent variable, but the principles can be applied here. The p-value associated with the individual X variable, .0394 is once again the result of a hypothesis test. But this test compares this single X to Y and again carries the null hypothesis that there is no relationship between X and Y. Since this is a simple (1X) linear regression, this value is the same as the Significance F. In a multiple regression, each X will have a different p-value. P-values closer to 0 indicate a stronger relationship with Y and values closer to 1 indicate a weaker relationship with Y.
How accurate will the model be as an estimator or predictor of the dependent variable?
The Standard Error of the estimate, 1.31, is a statistical measure of the deviation in predicted values of Y. Using the principals learned in statistics, a 95% interval may be developed for the predicted value of Y by calculating the value of Y, given X’s). This is the point estimate and is the starting point for the interval. Since approximately 95% of normally distributed population values should fall within 2 standard deviation of the mean, we construct this “rough” interval as follows:
95% interval for predicted Y = Y ± (2 ) ( Standard Error )
36
Calculate point estimates for the predicted value of the dependent variable (Y) given value(s) for the independent variable(s) (X).
Point Estimate: A single value estimate of Y for a given X.
The equation for this example is Y=2+1.25 X. Therefore if we believe that Local Payroll (X) next year will be $550,000,000, we would predict the Triple A’ Sales for next year would be:
Y=2+1.25 (5.5 )=2+6.875=8.875=$ 887,500
Construct a “rough” 95% interval for the predicted value of the dependent variable (Y) given value(s) for the independent variable(s) (X).
Using the same value for X, the 95% interval for the predicted value of Y would be:
8.875 ± (2 ) (1.31 )=8.875 ±2.62={6.255 ,11.495}
If Local Payroll next year is $550,000,000, then Triple A’s Sales should be between $625,500 and $1,149,500.
Understand the predicted change in the dependent variable that would result from a change to any of the independent variable values.
The slope, 1.25 indicates that the sales for Triple A will change by 1,25 for each 1 unit change in local payroll. Each $100,000,000 increase in local payroll should result in a $125,000 increase in Triple A’s sales. However, this is once again only an estimate derived from sample data. Accordingly, we would prefer to express this potential change as an interval. The Lower 95% and Upper 95% values for X help us with this.
The lower 95% value is 0.10 and the upper 95% value is 2.40. The 95% interval for the predicted CHANGE in Y that would result from a 1 unit change in X would be {0.10, 2.40} of between $10,000 and $240,000.
37
Chapter 5 – Forecasting
Forecasting Part 1 – A Basic Introduction
Why Forecast?
The future is uncertain and yet we must make decisions today. The results of our decisions are affected by that uncertainty. We worked with these concepts in decision analysis and tried to understand how to make better decisions by applying a logical evaluation of the possible outcomes and their respective probabilities.
Forecasting is a technique that helps us develop possible future outcomes. Accurate and well-designed forecasts can significantly improve the quality of our decision-making and lead to improved results in all areas of business.
In essence, forecasting uses data from the past to help see potential future results in an organized and logical framework.
If you are interested in seeing the role of forecasting “outside the classroom,” take a look at this website http://www.forecastingprinciples.com.
Forecasting - Steps for Success
The emphasis on taking a step-by-step, logical approach continues to be a major aspect of our course. Once again, the general guidelines of the quantitative analysis approach presented in Chapter 1 have been slightly re-worked to fit the specific needs of forecasting while still maintaining and emphasizing the benefits of a structured approach to building analytical models.
1. How will the forecast be used?
2. What are we trying to forecast?
3. What is the time horizon for our forecast?
4. What forecasting model is most appropriate?
5. Gather data.
6. Test the validity of the model.
7. Create the forecast.
8. Implement the results.
9. Analyze and refine the forecast over time.
38
Basic Types of Forecasts
There are three basic “families” of forecasting models. Each group contains a set of modeling methods that tend to be suitable for specific types of forecast and specific types of data. The text has a useful chart describing the three basic types on page 151.
MovingAverage
Exponential Smoothing
Trend Projections
Time-Series Methods: include historical data over a time interval
Forecasting Techniques
No single method is superior
DelphiMethods
Jury of ExecutiveOpinion
Sales ForceComposite
ConsumerMarket Survey
Qualitative Models: attempt to include subjective factors
Causal Methods: include a
variety of factors
Regression Analysis
Multiple Regression
Decomposition
Time-Series Models are one of the most frequently used tools in business forecasting because we are always interested in predicting the future based on historical data. Time-Series models use time as the most critical factor in preparing forecasts by relying on the assumption that the future will at least partially a function of what happened in the past. The specific tools used in time-series models range from very simple moving averages to weighted moving averages and exponential smoothing. These models then advance to include the use of regression principles for trend projection and very detailed decomposition techniques for complex data.
Causal Models rely upon the assumption that we can identify factors that cause or influence the behavior of the data we wish to forecast. This is basically the same as the relationship between the independent variables and the dependent variable that we recently worked with in regression analysis. The causal model approaches we use in forecasting will be identical to the work we did in regression.
Qualitative Models recognize that there are often forecasting situations that do not lend themselves to a purely quantitative forecasting tool. These models often combine judgment with some data that is not quite as quantitative as historical information. The specific approaches that fall into this category often include the use of outside experts and survey data from internal or external sources. These approaches are usually iterative, requiring several rounds of discussion and refinement to produce a forecast that is useful.
Using Scatter diagrams and Times Series
39
The most basic tool for examining time series data is a scatter diagram. This diagram is constructed and used exactly as it was done for simple linear regression analysis. When constructing a time series scatter diagram, the data we wish to forecast is identified as the dependent variable and plotted on the y-axis and the time series is the independent variable plotted along the x-axis.
To keep this simple, we will ignore the regression possibilities at this point and simply use our eyes or the chart wizard to help fit a line to the data and then forecast that line into the future.
The data and scatter diagrams provided on pages 152 and 153 of the text demonstrate the application of simple plots to time series data.
Scatter Diagrams
0
50
100
150
200
250
300
350
400
450
0 2 4 6 8 10 12
Tim e (Ye a rs)
An
nu
al
Sa
les Radios
Televisions
Compact Discs
Scatter diagrams are helpful when forecasting time-series data because they depict the relationship between variables.
40
Which forecast is likely to be more accurate?
With all of the different forecasting tools available to us, how can we begin to evaluate the potential future accuracy of any method compared to another? Once again, we will use historical data to consider the potential future reliability of the forecast. The basic technique is called mean absolute deviation and requires that we compare each actual value to the value that would have been forecast for that period using the chosen forecasting model. By averaging (mean) the absolute values of the deviations between the forecast and actual amounts, we can compare the relative accuracy of different forecasting models.
Let’s look at this calculation and consider the forecast for sales of CD players using the data on page 152. A scatter diagram of the data, including a forecasting line, looks like this:
CD Players
90
100
110
120
130
140
150
160
170
180
190
200
210
1 2 3 4 5 6 7 8 9 10
Year
Sale
s
From this forecasting line we can estimate the forecast sales levels for the 10 years and compare them to the actual sales levels. We can also compare the relative accuracy of this forecasting method to the accuracy of the naïve forecast described on page 154 that uses the prior year’s sales level for the next year’s forecast. The resulting calculations are as follows:
Year Actual Sales Naïve Forecast Absolute DeviationScatter Diagram
Forecast Absolute Deviation1 110 104 62 100 110 10 115 153 120 100 20 126 64 140 120 20 137 35 170 140 30 148 226 150 170 20 158 87 160 150 10 169 98 190 160 30 180 109 200 190 10 191 9
10 190 200 10 202 12
Total 160 100Count 9 10M.A.D. 17.78 10.00
The lower M.A.D. value for the scatter diagram forecast indicates that this forecasting method is more accurate than the naïve forecast for this data.
41
Forecasting Part 2 – The Real Deal
Decomposition
Decomposition refers to the practice of analyzing historical data in a way that separates out the four possible components of any data stream.
Data = Trend * Seasonality * Cyclicality * Random Events
We will focus on separating trend from seasonality, the two most frequent and important aspects of decomposition. The specific forecasting tools we will use to accomplish this are the deseasonalization of data using centered moving averages and seasonal indexes and the application of multiple linear regression techniques to data that contains both seasonality and trend.
Moving Averages / Weighted Moving Averages
Some of the more basic forecasting tools available include the moving and weighted moving average methods. These methods forecast based on a selected number of prior periods and may also assign varying weights to each period. The selection of both the number of periods and the weighting of each period is judgmental and should be made with the goal of minimizing the M.A.D. for the forecast. The example in the text calculates a three-month moving average for Wallace Garden supply as well as a three-month weighted moving average. The math behind these calculations is illustrated on pages 158 and 159 and below:
Problem 5-12 Weight 1.002.00
Wallace Garden Supply 3.006.00
MonthActual Shed
Sales
3 month Moving Average
Mean Absolute Deviation
3 month Weighted Moving Average
Mean Absolute Deviation
January 10.00February 12.00March 13.00April 16.00 11.67 4.33 12.17 3.83May 19.00 13.67 5.33 14.33 4.67June 23.00 16.00 7.00 17.00 6.00July 26.00 19.33 6.67 20.50 5.50August 30.00 22.67 7.33 23.83 6.17September 28.00 26.33 1.67 27.50 0.50October 18.00 28.00 10.00 28.33 10.33November 16.00 25.33 9.33 23.33 7.33December 14.00 20.67 6.67 18.67 4.67
M.A.D. 6.48 5.44
10 + 12 + 13 = 35
35 / 3 = 11.67| 16 - 11.67 | = 4.33 Compute the average of
this column.
[(10*1)+(12*2)+(13*3)]/6 = 12.17
42
Exponential Smoothing
This forecasting method employs the use of a smoothing constant to adjust the forecast for each period for some portion of the prior period’s forecasting error. The choice of the appropriate smoothing constant is judgmental and should be design to minimize the M.A.D. of the forecast. The basic formula for is as follows:
Forecast for period 2 = F2 = F1 + ((A1 – F1)*C)
The basic spreadsheet format for an exponential smoothing forecast will look like this:
Exponential Smoothing - Step by Step
Smoothing constant 0.3Year 1 Forecast 5000
YearActual
Demand
Previous Forecast
ErrorSmoothing Constant
Forecast Adjusment
Current Rounded Forecast
Rounded Forecast MAD
1 4,000 5,000 5,000.00 1,000.002 6,000 -1,000 0.30 -300 4,700 4,700.00 1,300.003 4,000 1,300 0.30 390 5,090 5,090.00 1,090.004 5,000 -1,090 0.30 -327 4,763 4,763.00 237.005 10,000 237 0.30 71 4,834 4,834.10 5,165.906 8,000 5,166 0.30 1,550 6,384 6,383.87 1,616.137 7,000 1,616 0.30 485 6,869 6,868.71 131.298 9,000 131 0.30 39 6,908 6,908.10 2,091.909 12,000 2,092 0.30 628 7,536 7,535.67 4,464.33
10 14,000 4,464 0.30 1,339 8,875 8,874.97 5,125.0311 15,000 5,125 0.30 1,538 10,412 10,412.48 4,587.52
2,437.19
Previous forecast error = Previous actual – Previous forecastCurrent rounded forecast = Prev. forecast + (Prev. forecast error * smoothing constant)
43
Linear Trend Forecasting
This method brings us back to simple linear regression. The dependent variable, y, is the data we wish to forecast and the independent variable, x, is time.
In general, it makes sense to label the time series starting with “1” even if the series is presented in “real years” (2002).
Using the data from Wallace Garden Supply (5-14), a linear trend forecast would look like this:
YearActual
Demand1 4,0002 6,0003 4,0004 5,0005 10,0006 8,0007 7,0008 9,0009 12,000
10 14,00011 15,000
Wallace Garden Supply y = 1054.5x + 2218.2
R2 = 0.8225
02,0004,0006,0008,000
10,00012,00014,00016,000
1 2 3 4 5 6 7 8 9 10 11 12
Years
Dem
and
The accuracy of this forecast can be measured by using the regression equation to compute predicted demand for each year and computing M.A.D. as in previous examples.
YearActual
DemandForecast Demand M.A.D.
1 4,000 3,272.70 727.30 Slope 1,054.502 6,000 4,327.20 1,672.80 Intercept 2,218.203 4,000 5,381.70 1,381.704 5,000 6,436.20 1,436.205 10,000 7,490.70 2,509.306 8,000 8,545.20 545.207 7,000 9,599.70 2,599.708 9,000 10,654.20 1,654.209 12,000 11,708.70 291.30
10 14,000 12,763.20 1,236.8011 15,000 13,817.70 1,182.30
1,385.16
44
Forecasting Basics Wrap-up:
Let’s examine the relative accuracy of the three main methods we have used thus far. We will continue to use the Wallace Garden Supply data from problem 5-14 and we will compare the accuracy of a three-year weighted moving average forecast, an exponential smoothing forecast and a linear trend forecast. For simplicity, we will use weights of 1, 2, and 3 and a smoothing constant of 0.5 as used in the examples on pages 159 – 162.
Wallace Garden Supply
YearActual
Demand
3 Year Weighted Moving Average M.A.D.
Exponential Smoothing M.A.D.
Linear Trend M.A.D.
1 4,000 5,000 1,000 3,273 7272 6,000 4,500 1,500 4,327 1,6733 4,000 5,250 1,250 5,382 1,3824 5,000 4,667 333 4,625 375 6,436 1,4365 10,000 4,833 5,167 4,813 5,188 7,491 2,5096 8,000 7,333 667 7,406 594 8,545 5457 7,000 8,167 1,167 7,703 703 9,600 2,6008 9,000 7,833 1,167 7,352 1,648 10,655 1,6559 12,000 8,167 3,833 8,176 3,824 11,709 291
10 14,000 10,167 3,833 10,088 3,912 12,764 1,23611 15,000 12,500 2,500 12,044 2,956 13,818 1,182
2,333 2,086 1,385
Weights 1 Smoothing Slope 1,054.552 Constant 0.5 Intercept 2,218.18
36
We can improve upon this by adjusting the weighting and the smoothing constant to optimize the accuracy of those forecasts. Doing so produces the following:
Wallace Garden Supply
YearActual
Demand
3 Year Weighted Moving Average M.A.D.
Exponential Smoothing M.A.D.
Linear Trend M.A.D.
1 4,000 5,000 1,000 3,273 7272 6,000 4,300 1,700 4,327 1,6733 4,000 5,490 1,490 5,382 1,3824 5,000 4,333 667 4,447 553 6,436 1,4365 10,000 5,000 5,000 4,834 5,166 7,491 2,5096 8,000 8,167 167 8,450 450 8,545 5457 7,000 7,833 833 8,135 1,135 9,600 2,6008 9,000 7,667 1,333 7,341 1,659 10,655 1,6559 12,000 8,500 3,500 8,502 3,498 11,709 291
10 14,000 10,667 3,333 10,951 3,049 12,764 1,23611 15,000 12,833 2,167 13,085 1,915 13,818 1,182
2,125 1,965 1,385
Weights 1.00 Smoothing Slope 1,054.551.00 Constant 0.7 Intercept 2,218.18
4.006.00
45
Finally, we can graphically represent the actual demand and each of the 3 forecasts and look at how each forecast fits the actual historical data.
Wallace Garden Supply
0
2,000
4,000
6,000
8,000
10,000
12,000
14,000
16,000
1 2 3 4 5 6 7 8 9 10 11 12
Years
Dem
an
d
Actual 3yr WMA Exp. Smooth. Linear Trend
The conclusion that should be reached by our review of M.A.D. and from our review of this chart is that, for this data, a linear trend forecast would be the best choice.
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Forecasting TechniquesA Simplified Approach to Seasonality
The general objective when trying to create a forecast of seasonal data is to separate the seasonality from the underlying trend and develop an accurate trend forecast which can then be adjusted for seasonality.
Both decomposition and multiple regression can be effective tools for this, but neither provides a simple or intuitive approach that clearly explains the process of separating seasonality from trend.
This example is my attempt to more clearly explain these concepts. I hope it helps.
We are presented with data for the last 4 years of quarterly sales tax revenues collected by the State of Texas.
Quarter Year 1 Year 2 Year 3 Year 41 218 225 234 2502 247 254 265 2833 243 255 264 2894 292 299 327 356
Total 1000 1033 1090 1178
The very first thing we should do is put together a quick graph of the data to see what story it tells. In order to do that, we will need to set up the data as a time series of 16 data points. The raw data and graph would be as follows:
Quarter Number
Tax Revenue
1 2182 2473 2434 2925 2256 2547 2558 2999 234
10 26511 26412 32713 25014 28315 28916 356
Texas Sales Tax Revenues
200
250
300
350
400
1 5 9 13
Quarter
Rev
enu
e
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Our first observation from the chart should be that the data appears to have consistent, seasonal patterns in which the first quarter of the year is low and the fourth quarter is high. The pronounced seasonality in this data will make the basic forecasting tools ineffective as shown in the following linear regression of the raw data.
Texas Sales Tax Revenues y = 5.01x + 226.23
R2 = 0.41
200
250
300
350
400
1 5 9 13
Quarter
Rev
enu
e
A linear regression of this data only explains 41% of the variability in revenues and would clearly do a poor job of forecasting future quarterly revenue because it ignores the seasonality within the data. In order to do a better job, we must deal with seasonality and trend separately.
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The easiest way to eliminate seasonality from data is to eliminate the seasons! Let’s look at the annual totals and a linear regression of the annual data.
Year Number
Tax Revenue
1 10002 10333 10904 1178
Texas Sales Tax Revenues y = 59.10x + 927.50
R2 = 0.96
800850900950
10001050110011501200
1 2 3 4
Year
Rev
enu
e
It should be evident that the annual data has a linear trend and that simple linear regression does an excellent job of describing this trend. We should feel comfortable that we can accurately predict annual sales tax revenues using this model. Accordingly, we can predict that annual revenues for year 5 will be:
59.10(5) + 927.5 = 1,223
Now that we have a good projection for next year’s revenues, we must develop a good way of allocating it among the 4 quarters of the next year.
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An intuitive approach to this revolves around a simple calculation of the average revenues collected in each quarter over the 4 years in the data set.
Texas Sales Tax Revenues ($millions)
Quarter Year 1 Year 2 Year 3 Year 4Quarterly Average
Average Percent
1 218 225 234 250 231.75 21.55%2 247 254 265 283 262.25 24.39%3 243 255 264 289 262.75 24.44%4 292 299 327 356 318.50 29.62%
1075.25 100.00%
Based on simple averages, we can allocate our annual projection of $1,223 to each quarter as follows:
Texas Sales Tax Revenues ($millions)
QuarterAverage Percent
Annual Projection
Quarterly Projection
1 21.55% 1223.00 263.592 24.39% 1223.00 298.293 24.44% 1223.00 298.854 29.62% 1223.00 362.27
100.00% 1223.00
We now have a simplified projection of quarterly sales tax revenues for next year.
But how might it compare to a more sophisticated multiple regression model?
Let’s look.
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Multiple Regression for Forecasting Seasonal Data
Here’s a full picture of the data setup and results of a multiple regression analysis that employs categorical variables for each season.
Year Quarter Number Q1 Q2 Q3 Amount1 1 1 1 0 0 2181 2 2 0 1 0 2471 3 3 0 0 1 2431 4 4 0 0 0 2922 1 5 1 0 0 2252 2 6 0 1 0 2542 3 7 0 0 1 2552 4 8 0 0 0 2993 1 9 1 0 0 2343 2 10 0 1 0 2653 3 11 0 0 1 2643 4 12 0 0 0 3274 1 13 1 0 0 2504 2 14 0 1 0 2834 3 15 0 0 1 2894 4 16 0 0 0 356
SUMMARY OUTPUT
Regression StatisticsMultiple R 0.9842R Square 0.9687Adjusted R Square 0.9574Standard Error 7.67Observations 16.00
ANOVAdf SS MS F Significance F
Regression 4.00 20055.20 5013.80 85.21 0.00Residual 11.00 647.24 58.84Total 15.00 20702.44
CoefficientsStandard Error t Stat P-value Lower 95% Upper 95%Intercept 281.56 5.75 48.94 0.0000 268.90 294.22Number 3.69 0.43 8.61 0.0000 2.75 4.64Q1 -75.67 5.57 -13.57 0.0000 -87.94 -63.40Q2 -48.86 5.49 -8.90 0.0000 -60.95 -36.78Q3 -52.06 5.44 -9.57 0.0000 -64.03 -40.08
Quarterly Projection for Year 5
281.56Coefficients 3.69 -75.67 -48.86 -52.06
Quarter Quarter # Q1 Q2 Q3 Projected Revenue1 17 1 0 0 268.692 18 0 1 0 299.193 19 0 0 1 299.694 20 0 0 0 355.44
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Let’s compare.
Texas Sales Tax Revenues ($millions)
QuarterAverage Percent
Annual Projection
Quarterly Projection
Multiple Regression
ForecastPercentage Difference
1 21.55% 1223.00 263.59 268.69 -1.90%2 24.39% 1223.00 298.29 299.19 -0.30%3 24.44% 1223.00 298.85 299.69 -0.28%4 29.62% 1223.00 362.27 355.44 1.92%
100.00% 1223.00 1223.00
While these projections match up very well and I am reasonably confident that the differences between them would not be significant in most situations, this is not a repudiation of multiple regression or other sophisticated forecasting techniques. Rather, it is a reminder that we should seek to keep our work simple and clear when the data allows us to do so.
There are many situations requiring more powerful forecasting tools and more complex models, but we can often gain a better understanding of data with simple tools and then proceed to tackle the problem with more sophistication because we understand what needs to be done and not simply because we memorized a specific tool.
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Chapter 7 – Linear Programming
Linear Programming is a problem-solving approach that has been developed to help managers make better decisions.
All linear programming models we will study are “Constrained Optimization” models. In each problem, the firm will seek to optimize the value of some desired financial outcome (profit, revenue or cost). However, the firm’s ability to optimize is constrained by limitations on the inputs needed to produce the desired outcome. Examples of common constraints include:
Limited production resources (materials, labor, capital)Limited customer demandMinimum production requirements (Orders already accepted)Required product mix levels
The successful application of linear programming to solve business problems requires an ability to construct a mental model of each business and then translate that mental model into a set of linear equations that describe all the factors that will influence the firm’s decision.
Our study of linear programming will concentrate on the development of five critical skills:
1. Understanding the business problem.
2. Formulating the business problem as a set of equations.
3. Graphical solutions to two-variable problems.
4. MS Excel solutions to multi-variable problems.
5. Management interpretation of results.
Using the Problem Solving Methodology on the next page will help organize your work and guide you through these problems. As always, there is simply no substitute for hard work and repetition in order to develop these skills.
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Linear ProgrammingProblem Solving Methodology
1) Read and understand the basic business problem.What does the firm do?What financial objective does it want to achieve?What decisions must it make?
2) Describe the business OBJECTIVE desired by the firm.Optimization is always the goal.
Maximize profitMaximize revenueMinimize cost
3) Identify and name the DECISION VARIABLES.What decision does the firm control?Label each decision variable.
4) Identify and describe each CONSTRAINT.What are the limits, or requirements that the firm must adhere to?Dissect every sentence in the problem.Describe the impact of every constraint on the decision variables.
5) FORMULATE the OBJECTIVE FUNCTION as a linear equation using the DECISION VARIABLES.Find the coefficient for each decision variable.Convert the objective into an equation. (From the Flair Furniture example in section 7.3.)
MAX Profit: Z=70T + 50C
6) FORMULATE each CONSTRAINT as a linear inequality using the DECISION VARIABLES.Find the coefficients for each decision variable in each constraint.Convert each constraint into an inequality. (From the Flair Furniture example in section 7.3.)
Carpentry: 4T + 3C ≤ 240Painting and varnishing: 2T + 1C ≤ 100
7) Choose and appropriate solution method.Two variable problems can be solved graphically.All problems can be solved using MS Excel Solver.
8) Solve the problem for the optimal quantities of the DECISION VARIABLES.Identify the quantity of each decision variable that provides the optimal solution to the objective function.
9) Calculate the value of the OBJECTIVE FUNCTION at the optimal solution point.Identify the optimal profit, revenue or cost that is produced by the optimal solution.
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Linear ProgrammingSolution Approaches to 2-variable problems
All problems containing two decision variables can be solved graphically. The graphical solution approach follows the following steps:
1.) Graph EACH constraint by solving for its intercepts.
For the equation 7X + 3Y ≤ 210 we would,
Substitute 0 for X, producing7(0) + 3Y ≤ 2103Y ≤ 210Y ≤ 70
Substitute 0 for Y, producing7X + 3(0) ≤ 2107X ≤ 210X ≤ 30
This equation is defined by the line extending from (0, 70) to (30, 0).
2.) Locate the feasible region defined by the constraint. The feasible region is that portion of the graph that contains solutions which do not violate the constraint equation.
Because the equation is a “≤” the feasible region is below the line, heading toward (0,0).
3.) Repeat the above steps for each constraint equation.
4.) Locate the feasible region for the problem by eliminating all areas of the graph that violate one or more constraints. The feasible region will contain the set of all possible solutions that satisfy all of the constraints.
5.) The optimal solution will always lie at one of the extreme points along the frontier, or border formed by the intersections of the constraints and/or the X and Y axes. Find the (X, Y) coordinates for each of the intersecting points along the frontier.
6.) Solve the objective function equation for each of the coordinate sets. The coordinate set with the optimal result will be the optimal solution to the problem.
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Linear ProgrammingISO-profit line solutions.
The ISO-profit line method for solving 2 variable problems is helpful when there are multiple possible solutions. It also provides an excellent visual reference for understanding the problem. However, this solution can prove difficult for some students because it requires visualization skills and facility with graphing.
Start with a graph that has all constraints and the feasible region identified. This is where you are at the end of step 4 above.
Choose a small value for the solution to the objective function that lies somewhere in the feasible region.
If the objective function is 10X + 8Y = Z (max profit) and the coordinates (4, 5) lie in the feasible region, it would make sense to select 80 [10(4) + 8(5)] as the starting point for the ISO-profit line.
Using 10X + 8Y = 40, we then graph this line the same way we graphed the constraint lines:
Substitute 0 for X, producing10(0) + 8Y = 808Y = 80Y = 10
Substitute 0 for Y, producing10X + 8(0) = 8010X = 80X = 8
The starting ISO=profit line is defined by the points (0,10) and (8,0).
This line can then be moved toward the frontier of the feasible region by maintaining a constant slope for the line. Continue moving the ISO-profit line until it is touching the very last point in the feasible region. That point is the optimal solution to the problem.
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Linear ProgrammingSample Problem – Graphical Solution
The Pinewood Furniture Company produces chairs and tables from two resources – labor and wood. The company has 80 hours of labor and 36 pounds of wood available each day. Demand for the chairs is limited to 6 per day. Each chair requires 8 hours of labor and 2 pounds of wood to produce, whereas a table requires 10 hours of labor and 6 pounds of wood. The profit derived from each chair is $400 and from each table $100. The company wants to determine the number of chairs and tables to produce each day in order to maximize profit.
1.) The business objective is to maximize overall profit by finding the optimal production quantity for each product, without violating any of the constraints.
2.) The decision variables are the two products, Chairs “C” and Tables “T”.
3.) The firm’s ability to maximize profit is constrained by the limited availability of labor and wood as well as the limited demand for chairs. There are only 80 labor hours available per day. There are only 36 pounds of wood available per day. The demand for chairs is limited to a maximum of 6 per day.
4.) The objective function for this problem is:Maximize Profit: Z = $400(C) + $100(T)
5.) The linear inequalities that represent each of the constraints are:CONSTRAINTSLabor: 8(C) + 10(T) ≤ 80 hoursWood: 2(C) + 6(T) ≤ 36 poundsDemand: 1(C) ≤ 6 units
The complete formulation for this problem is:
Maximize: Z = $400(C) + $100(T)Subject to: 8(C) + 10(T) ≤ 80 hours
2(C) + 6(T) ≤ 36 pounds1(C) ≤ 6 units
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6.) Graphical Solutiona. Plot the line for the labor constraint.b. Plot the line for the wood constraint.c. Plot the line for the demand constraint.d. Identify the feasible region.e. List the coordinates of the possible solution points.f. Solve the objective function for the possible solution points.g. The highest profit achieved is the optimal solution.
0
1
2
3
4
5
6
7
8
9
10
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
Ta
ble
s
Chairs
PinewoodLabor Wood Demand C
The optimal profit would be achieved by producing 6 Chairs and 3.2 Tables.
Profit: Z = $400(6) + $100(3.2) = 2400 + 320 = 2720
Labor: 8(6) + 10(3.2) ≤ 80 48 + 32 ≤ 80 80 ≤ 80 There is no slack.
Wood: 2(6) + 6(3) ≤ 36 12 + 18 ≤ 36 30 ≤ 36 There are 6 units remaining.
Demand: 1(6) ≤ 6 6 ≤ 6 There is no slack.
The optimal solution fully utilizes both labor and demand.
Feasible
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Objective Function: Max Profit: 9A + 12B + 15C + 11D = Z
Constraints: Wiring: .5A + 1.5B + 1.5C + 1D ≤ 15,000 Drilling: .3A + 1B + 2C + 3D ≤ 17,000 Assembly: .2A + 4B + 1C + 2D ≤ 26,000 Inspection: .5A + 1B + .5C + .5D ≤ 12,000 Min Dem A: 1A + 0B + 0C + 0D ≥ 150 Min Dem B: 0A + 1B + 0C + 0D ≥ 100 Min Dem B: 0A + 0B + 1C + 0D ≥ 300 Min Dem C: 0A + 0B + 0C + 1D ≥ 400
Linear ProgrammingMS Excel Solutions
We will be generally following the method shown in section 7.5 for the development of Excel solutions to linear programming problems. I recommend that all students use the following approach:
The development of linear programming models and solutions using MS Excel begins when the algebraic formulation of the problem is complete.
As an example, let’s consider problem 7-40 in our text. The complete formulation of this problem is as follows:
Using Excel Solver for Linear Programming problems:
Section 7.5 in our text does a very good job of explaining the use of this add-in on pages 271 - 275. Please understand that there are an infinite number of ways to set up a linear programming problem to properly work with the Solver. The approach I am providing below will be used in class. Although slightly different from the text, it addresses all of the same concepts and issues and also allows us to work directly from the Formulation of any problem.
This may require a lot of practice, so please make good decisions about the number of problems you need to solve in order to build up your strength, speed and accuracy with this technique.
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Our spreadsheet is divided into 3 sections. At the top is a description of each decision variable and its objective function coefficient. Remember that the objective function coefficient may relate to profit, revenue or cost, depending on the appropriate business objective for the problem. In this case, the spreadsheet is formatted as follows:
The top section of the spreadsheet matches the objective function. The second section of the spreadsheet deals with all of the constraints in the problem. The third section of the spreadsheet leaves space for the solution that Excel will develop. The optimal production quantities of each decision variable, as well as the value of the objective function achieved at the optimal solution are recorded here. The format is as follows:
Note that the “LHS” and “SLACK” cells are blank. These cells will require formulas that we will create later.
Note that the cells are left empty. The “Profit” cell will require a formula and the “Optimal Quantity” cells will be filled in by the Excel Solver and will display the optimal production quantity of each product.
At this point, the spreadsheet for 7-38 should look as follows:
A B C D E F G H I1 OBJECTIVE A B C D2 Profit per unit 9.00$ 12.00$ 15.00$ 11.00$ 34 CONSTRAINTS A B C D LHS SIGN RHS SLACK5 Wiring 0.5 1.5 1.5 1.0 ≤ 150006 Drilling 0.3 1.0 2.0 3.0 ≤ 170007 Assembly 0.2 4.0 1.0 2.0 ≤ 260008 Inspection 0.5 1.0 0.5 0.5 ≤ 120009
10 MIN XJ210 1.0 ≥ 15011 MIN XM897 1.0 ≥ 10012 MIN TR29 1.0 ≥ 30013 MIN BR788 1.0 ≥ 4001415 Optimal Quantity1617 Profit
As an overall guide, the LHS and profit cells, highlighted in yellow, will require SUMPRODUCT formulas referencing the optimal solution quantities in all cases and the relevant coefficient ranges for each formula.
The SLACK cells, highlighted in green, will require simple (=RHS – LHS) formulas.
The Optimal Quantity cells, highlighted in blue, are left blank. The Solver will fill these cells with the optimal solution quantities and the other formulas will react automatically to these quantities.
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A B C D E F G H I1 OBJECTIVE A B C D2 Profit per unit 9.00$ 12.00$ 15.00$ 11.00$ 34 CONSTRAINTS A B C D LHS SIGN RHS SLACK5 Wiring 0.5 1.5 1.5 1.0 ≤ 150006 Drilling 0.3 1.0 2.0 3.0 ≤ 170007 Assembly 0.2 4.0 1.0 2.0 ≤ 260008 Inspection 0.5 1.0 0.5 0.5 ≤ 120009
10 MIN XJ210 1.0 ≥ 15011 MIN XM897 1.0 ≥ 10012 MIN TR29 1.0 ≥ 30013 MIN BR788 1.0 ≥ 4001415 Optimal Quantity1617 Profit
Writing the Formulas: There are 2 basic formulas that we need to create in this model. First, we need a formula that will calculate the result of the objective function. This formula will be located in the “Objective Function Result” cell in the third section of the spreadsheet. This formula will always calculate the result of the objective function, even as the production quantities or profit/revenue/cost per unit may change. It is critical to use cell references and not values in this formula.
Cell B17 = (b2 * b15) + (c2 * c15) + (d2 * d15) + (e2 * e15) = Objective Function Result
This formula can be simplified by using the SUMPRODUCT function in Excel. Cell B17 should contain the formula =SUMPRODUCT(B15:E15, B2:E2)
The next formulas calculate the actual usage (LHS) for each of the constraints in the problem. Once again, we can develop the Excel formulas from the formulation of the problem. For example,
Cell F5 = (b5 * b15) + (c5 * c15) + (d5 * d15) + (e5 * e15) = Actual Wiring Hours Used
This is an example of just one of the constraints in the problem. The SUMPRODUCT function can again be used to greatly simplify the formula and also to permit us to write it one time and copy it properly to all of the other “LHS” cells.
Cell F5 should contain the formula =SUMPRODUCT($B$15:$E$15, B5:E5)
Note that the first array, corresponding to the optimal quantity cells, has been set as an absolute cell reference (the F4 key). This permits us to copy this formula to all the other LHS cells and retain all the correct cell references. After this formula is properly constructed for the first constraint, it should be copied to all of the remaining LHS cells.
Finally, we need a simple formula to calculate the Slack for each constraint.
Cell I5 should contain the formula =H5-F5.
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When all the formulas are written, the worksheet will look like this:
A B C D E F G H I1 OBJECTIVE A B C D2 Profit per unit 9.00$ 12.00$ 15.00$ 11.00$ 34 CONSTRAINTS A B C D LHS SIGN RHS SLACK5 Wiring 0.5 1.5 1.5 1.0 0 ≤ 15000 15,0006 Drilling 0.3 1.0 2.0 3.0 0 ≤ 17000 17,0007 Assembly 0.2 4.0 1.0 2.0 0 ≤ 26000 26,0008 Inspection 0.5 1.0 0.5 0.5 0 ≤ 12000 12,0009
10 MIN XJ210 1.0 0 ≥ 150 15011 MIN XM897 1.0 0 ≥ 100 10012 MIN TR29 1.0 0 ≥ 300 30013 MIN BR788 1.0 0 ≥ 400 4001415 Optimal Quantity1617 Profit 0
Applying MS Excel Solver:
To open the solver dialog box, “Data” and look for “Solver” in the far right of the ribbon. Click on it and the dialog box below.
The dialog box begins with “Set Target Cell:” This is the “Profit” cell in the spreadsheet, cell B17 in our example.
Next, select the appropriate button (Max, Min or Value of): for 7-40 select Max. Solver has now been instructed to make sure the value of the objective function result cell for 7-40 is maximized. It now understands our measurable business objective!
Click in the “By Changing Cells:” box. In this box, we will specify the cell range that represents Optimal Production Quantities of our decision variables. Solver will change the value of these cells automatically when we run the Solver until the value of the objective function result is Optimized. This should contain cells $B$15:$E$15.
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Finally, we must tell Solver about our problem constraints. To do so, click on the “Add” button. A smaller dialog box will open and we will put each constraint into the model individually. The new dialog box looks like this:
The “Cell Reference” box will refer to the “LHS” cell for each constraint.
The inequality box must show the correct inequality. Use the drop-down button to select the appropriate inequality.
The “Constraint:” box will be the “RHS” cell for each constraint:
Each constraint can be entered individually, or they can be entered as groups as long as all constraints in the group have the same inequality.
The Wiring constraint in our example would be entered like this:
The completed dialog box for our example looks like this:
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After all the constraints are entered and have been CAREFULLY checked for accuracy, click on “Options” and place a mark in the boxes labeled “Assume Linear Model” and “Assume Non-Negative”.
Click on “OK” and you will return to the initial dialog box. You are now ready to click on “Solve” and see what the Solver produces.
Running the Solver successfully causes a new dialog box to pop up. If everything is perfect, you will get a message telling you that solver has found a solution.
It is possible to get this message and not have the correct answer because Solver only does what you tell it to do. You can enter faulty instructions or bad data and Solver will be able to solve the problem, but it will not be solving it correctly.
Remember, Solver brings speed and accuracy to the process. You must bring judgment, accuracy, logic and common sense.
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If there are critical problems in the worksheet that prevent Solver from running correctly, you will get a variety of messages telling you that Solver could not work.
I assure you that your particular version of Solver is not faulty. These messages occur because the worksheet is incorrect or because the instructions that you gave to the Solver are incorrect.
Do not get frustrated at the tool. It is doing EXACTLY what you told it to do!
Once the Solver has run properly, you will see a dialog box that asks you to specify the supplementary reports that you would like Solver to produce. Always select the Answer and Sensitivity reports from this list. They will automatically appear on separate spreadsheet tabs. Make sure you always print out these reports with your answers, as these reports are very important to our study of sensitivity analysis in linear programming.
Clicking on “OK” at this point will cause the Answer and Sensitivity reports to appear as additional pages in your worksheet. Your worksheet should now look like this.
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Please look carefully at the bottom of the worksheet to see the tabs for the Answer and Sensitivity reports. Your final printed output for the problem will consist of the basic worksheet, the Answer report and the Sensitivity report. They should look like this:
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Chapter 8 – Linear Programming Applications
Chapter 8 presents specific applications of the linear programming techniques from chapter 7. It does not present any new material in terms of the solution process or the interpretation of results, but it does present a more challenging series of problems. These notes will briefly review each type of problem we will encounter in this chapter.
8.2 – Media Selection Problems
These are exceptions to the “Max Profit, Max revenue, Min cost” rule, as the objective is generally to reach as large an audience as possible. In this case, “money” is a budgetary constraint. This problem can also be set up to minimize media costs, while reaching an acceptable minimum audience size.
Preview problem 8-6 fits this type and homework problem 8-11 is combined pricing strategy and media problem. Problem 8-11 is a truly critical problem to test your real grasp of business issues explored in this course. It is not difficult to set up, but you must think clearly when you review the solution.
8.3 – Production Mix Problems
These are the most like the problems in chapter 7, but the complexity is greater. The typical problem seeks maximization of profit by finding the optimal production quantities of a set of products. Homework problem 8-13 is our example. It will require you to calculate the profit per unit of each product and you should be both careful and thorough in your calculation.
8.4 – Labor Planning Problems
These problems minimize the staffing levels (number of employees) or staffing cost while meeting all the work requirements given in the problem. Preview problem 8-3 and homework problem 8-14 fit this type. 8-3 is quite straightforward, while 8-14 is a good test of your ability to put the pieces together and formulate a realistic hiring strategy over a 5 month period.
8.5 - Portfolio Selection Problems
These problems always generated the most student interest, at least until the stock market collapsed, because they involve maximizing return on investment and use exactly the same techniques that are the foundation of most investment management decisions. In these problems, we must find the optimal allocation of money across a series of potential investments. Preview problem 8-2 and homework problem 8-24 are examples of this type. These problems are usually easy to set up, although care should be taken in the formulating and understanding of the constraints. These problems also work very well for learning to interpret the sensitivity report.
8.6 - Shipping Problems
These problems represent the original application of linear programming; the logistics of getting material from one place to another efficiently. The objective is generally to minimize cost (or
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distance as a proxy for cost) and the nature of these problems leads to a large number of decision variables. This can makes them tedious to solve in exactly the same Excel template used for all other problems, and an alternative set up is provided for preview problems 8-10 and 8-23. These problems also illustrate that there are an infinite number of ways to set up a linear programming problem and that our “universal” template is really only for ease of instruction and not necessarily the best format for each problem.
8.8 – Diet Problems
How do we feed people, or animals, and make sure they receive proper nutrition, while minimizing cost? This is the point of the diet problem and it is a model in wide use in many institutions. The setup of these problems is very simple, although a little time consuming. Homework problem 8-12 is a classic diet problem and the solutions to these problems also work well for studying the interpretation of the sensitivity report.
Summary
Linear programming is the basis for many business decision models. Additionally, the logic and discipline these problems require is widely applicable to the analysis of business situations. The purpose of these two chapters is to expose you to the process, build some solid Excel solution development skills and demonstrate just a few of the specific applications of this technique.
This is the one topic we cover, above all others, where students should be able to see relevance and applicability in both their future careers and in their current daily lives. At its core, almost every decision is a constrained optimization decision in which we try to do “the best we can” while satisfying a series of internal (self-imposed) or external constraints.
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Chapter 15 – Simulation Modeling
Simulation models are used in many business situations to study the impact of various decisions on company performance. Simulation models can be as simple as the example we will cover in this guide, or a complex and detailed as any real-world situation.
Whenever we are interested in comparing the effect of decision alternatives, asking “what-if” questions or examining the relationships between individual variables or factors in a business situation or process, simulation can be a useful approach.
Simulation also permits us to model many periods of time instantly and allows us to consider the impact of our decisions over time without disrupting the actual business process currently in place.
The goal of simulation modeling is to help mangers make better decisions by providing them with practical information about the likely impact of and results obtained from their decisions BEFORE the decision is actually implemented in the organization.
Our study of simulation will focus first on the basic concepts of building a simulation model manually and in Excel. We will then examine the application of simulation models in three business processes:
Inventory management, stocking levels and reordering policy Queuing, or waiting line processes Repair and maintenance planning
In addition, we will consider the important differences between two general types of simulation models:
Fixed time increment models use data that reflects the total demand on the system and the number of services performed within a fixed period of time such as an hour, day, week or month. The example that follows, the Simkin’s inventory problem and the queuing problem for the Port of New Orleans all represent fixed time increment models.
Next event increment models require that we update the model each time demand occurs. In these models, demand happens at variable intervals of time and we measure the time between each occurrence instead of the total number of occurrences within a fixed period of time. The Three Hills Power Company example is a next event increment model.
While each type of model does use a similar framework for the basic construction of the model, the logic required within the model is quite different and it is important to identify which type of model would work best for each problem or business situation you face.
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Typical Simulation Problems
Basic one element of a business process is subject to variability a probability distribution of the variable is developed from observation a Monte Carlo simulation is constructed to model the problem decision alternatives are evaluated within the simulation
Inventory Analysis Variables to be simulated
o Customer demando Delivery lead time
Decisionso re-order point – when to reordero re-order quantity – how many to reorder
Monetary implicationso cost of storing inventoryo cost of placing orderso opportunity cost of a stock out
Queuing (Waiting Lines) Variables to be simulated
o Arrivals into the system (demand)o Service time
Decisionso Service / staffing levelo Acceptable customer waiting time
Monetary implicationso Cost of additional staffingo Cost of customer waiting time (lost goodwill, etc.)
Repair and Maintenance Policy Variable to be simulated
o Breakdown frequencyo Repair time
Decisionso Staffing level of repair crewo Acceptable downtime (machine out of service)
Monetary implicationso Cost of additional repair staffingo Cost of downtime (lost production, profits, etc.)
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A quick guide to building a basic Monte Carlo Simulation
Billy Matthews has a part time job selling copies of the Sunday newspaper out in front of the local Dunkin Donuts. Once a quarter, he must place an order for the number of newspapers he wants to buy from the publisher for the next 13 weeks. After he places the order, the quantity cannot change. Billy is locked into that quantity for all 13 weeks. Billy pays $1.00 for each newspaper and sells them for $2.50. Any newspapers that are unsold are waste, but there is no penalty for having a shortage. He also pays the owner of the Dunkin Donuts a flat fee of $50 per week in order to use the space in front of the store.
The challenge Billy faces is that demand for newspapers can vary greatly from week to week depending on weather and other variables that are beyond his control. He has been studying data analysis and decision modeling in his math class (Billy is in the 10th grade) and he knows that he must develop a model that permits him to set his order at the right level to maximize profit. He has decided to construct a Monte Carlo simulation for the next 13 weeks and then place his order according to the results of that simulation.
Step 1: Establishing a Probability Distribution
Billy needs some information about the demand for newspapers. Because he is meticulous about his business, Billy has kept accurate records of the demand for newspapers for the last 50 weeks and developed the following frequency distribution.
Demand Frequency50 275 6
100 20125 15150 7
Weeks = 50
Because Billy is comfortable with probability and knows that a probability distribution is the key to a Monte Carlo simulation, he converts the frequency distribution into a probability distribution. And calculates the mean, or expected value of the probability distribution.
Demand Frequency Probability50 2 0.0475 6 0.12
100 20 0.40125 15 0.30150 7 0.14
Weeks = 50 1.00
μ=E ( x )=∑ [ x∗P( x ) ]=109 .5
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Step 2: Building a Cumulative Probability Distribution for Each Variable
Understanding how his MS Excel prefers to handle probability distributions, Billy extends his work to create a cumulative probability distribution for the data.
Demand Frequency ProbabilityCumulative Probabiltiy
50 2 0.04 0.0475 6 0.12 0.16
100 20 0.40 0.56125 15 0.30 0.86150 7 0.14 1.00
Weeks = 50 1.00
Step 3: Setting Random Number Intervals
Billy’s last step in preparing his probability distribution to be used in the simulation is to set up the interval ranges for the random numbers and organize it in a format that he knows MS Excel will like.
Demand Frequency ProbabilityCumulative Probabiltiy
Lower Random
Upper Random Demand
50 2 0.04 0.04 1 4 5075 6 0.12 0.16 5 16 75
100 20 0.40 0.56 17 56 100125 15 0.30 0.86 57 86 125150 7 0.14 1.00 87 100 150
Weeks = 50 1.00
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Step 4: Generating Random Numbers
Setting aside the probability interval table above, Billy must now generate a random number for each of the 13 weeks. He remembers the first time he discussed this solution with his dad, who excitedly ran up to the attic and came back down flushed with pride and holding a dusty copy of a mysterious book that seemed to consist entirely of pages filled with numbers. Although Billy loves his dad, he couldn’t help but laugh when dad said, “This is so much cooler that watching the Red Sox, we can spend the afternoon with this book of random number tables and we can enjoy a few hours of good father-son bonding over the wonders of these tables.”
Billy managed to stop laughing only long enough to reply, “Random number tables from a book? What is this, the 70’s? Dad, I appreciate the offer of help, but I’ll let MS Excel generate my random numbers for me. My simulation will be finished before the first pitch and we can still watch the game together as long as you promise not to try to calculate OPS in your head after every at bat.”
Billy turns to Excel and sets up the following basic template for the simulation.
WeekRandom Number Demand
12345678910111213
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In the Random Number column, Billy wants Excel to generate random numbers. Because he has set up his interval table using integers from 1 to 100, he wants Excel to generate random numbers that match his intervals. Therefore the random numbers must be between 1 and 100. Accordingly, Billy chooses the RANDBETWEEN function.
Note: If your version of Excel does not contain the RANDBETWEEN function, the following should be used in its place and will work exactly the same way. Please remember that getting this formula into excel EXACTLY as it is shown is critical to it working correctly.
=ROUND(RAND(),2)*100+1
Finally, because he doesn’t want the random numbers to change every time the model is changed, he converts the group of cells in the Random Number column to fixed values by using the following commands:
Copy (select entire column)Edit – Paste Special (select “Values” from the dialog box)
When all this is done Billy has the following random numbers in the simulation template.
WeekRandom Number Demand
1 742 463 904 155 356 547 868 359 8910 3311 2912 313 53
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Step 5: Simulating the Demand
Billy now needs to find the demand level associated with each of the random numbers generated for each week. He could do this by hand. For example, Billy could look at the first random number, 74, and compare it to the interval table. He would notice that 74 falls within the range defined by 56 to 86. Since that interval is associated with a demand level of 125 newspapers, Billy would use that demand for the first week.
But Billy knows that this is ponderous and that Excel can do this automatically by using the VLOOKUP function. Billy selects the first cell of the Demand column and clicks on fx and then on VLOOKUP. The dialog box that appears requires 4 pieces of information.
1. Lookup_value = the cell with the Random Number for this demand level
2. Table_array = the interval table (F4 this to freeze it for all subsequent cells)
3. Col_index_num = 3 (the column number in the table_array containing the demand (the 3rd column)
4. Range_lookup = TRUE
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Once Billy gets this to work in the first Demand cell, he then copies it to all 13 weeks and his template now looks as follows.
WeekRandom Number Demand
1 74 1252 46 1003 90 1504 15 755 35 1006 54 1007 86 1258 35 1009 89 15010 33 10011 29 10012 3 5013 53 100
Now Billy knows that all the hard work is done. He has successfully simulated demand for the next 13 weeks and he quickly expands his model to project his weekly and overall profit based on a fixed order quantity.
Billy produces the following model. He initially chooses an order quantity of 110, because that number is closest to the expected value of the probability distribution. He realizes that the Order Quantity cell can be changed and starts to look for a better solution through trial and error.
His dad wanders in after having unsuccessfully tried to teach the family dog, Euclid, how to randomize the number of times she needs to be let out every night by using the random number table book. Euclid, while seemingly uninterested in the detail of random number tables, did seem quite interested in biting the book and shaking it until all the pages flew out. Das looks at the screen and remarks, “I’m impressed, son. But why are you fumbling around with trial and error to optimize your purchase quantity? Why don’t you just use the Solver and maximize the value of the total profit cell by changing the order quantity cell?”
Billy looks up at his dad and says, “Solver? Have you been snooping on my work when I’m not looking? You can barely use the mouse and you know about the Solver?”
Dad kindly replies, “I guess your old man stills has some ‘mad skills’ left in him.”
Smiling to himself, Dad chose not to tell Billy that his ‘old school’ skills had created a world in which old Dad earned more money from compound interest in the 20 minutes it would take Billy to solve the model than Billy will make all quarter from sitting out in the cold selling newspapers.
But he also knew that letting Billy develop these skills on his own is the best way to put Billy in exactly the same position when he becomes “the old man.”
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WeekRandom Number Demand Units Sold
Gross Revenue Total Cost
Net Profit (Loss)
1 74 125 100 250.00$ 150.00$ 100.00$ 2 46 100 100 250.00$ 150.00$ 100.00$ 3 90 150 100 250.00$ 150.00$ 100.00$ 4 15 75 75 187.50$ 150.00$ 37.50$ 5 35 100 100 250.00$ 150.00$ 100.00$ 6 54 100 100 250.00$ 150.00$ 100.00$ 7 86 125 100 250.00$ 150.00$ 100.00$ 8 35 100 100 250.00$ 150.00$ 100.00$ 9 89 150 100 250.00$ 150.00$ 100.00$ 10 33 100 100 250.00$ 150.00$ 100.00$ 11 29 100 100 250.00$ 150.00$ 100.00$ 12 3 50 50 125.00$ 150.00$ (25.00)$ 13 53 100 100 250.00$ 150.00$ 100.00$
Average Demand = 105.77
Order Quantity 100 Total Profit 1,112.50$
Sales Price per unit 2.50$ Purchase cost per unit 1.00$ Rent cost per week 50.00$
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Problem 15-27 (Dr. Greenberg) Annotated Solution
This document provides a step-by-step walk through of the problem. First, we will develop a static solution using the random number tables provided in the text. Then, we will examine the logic and formulas necessary to build a live simulation for the problem.
Step 1 - Analyzing the problem.Dr. Greenberg needs to leave by 12:15 and would like to use simulation to help him understand whether this is likely to happen. At first glance, it appears he will be fine. His last patient, Hinkel, has an 11:45 appointment and the procedure Hinkel requires should take 15 minutes. In theory, Hinkel should be done at 12:00, leaving Dr. Greenberg 15 minutes before he must leave.
However, we are told that both the arrival times and service times can vary. Patients are sometimes early or late for their appointments and the procedures can sometimes take more or less time than planned. This variability creates a classic queuing problem in which we must simulate both arrivals and service times in order to understand what may happen during the morning.
In addition, we are told that Dr. Greenberg will be ready for work at 9:30, so an early arrival by the first patient will not result in an earlier start time. Finally, Dr. Greenberg takes patients according to the order of their scheduled exam. If patient 3 arrives before patient 2, patient 2 is still seen first.
The development of the simulation begins with the development of probability interval tables for arrival and service times. Based on the probabilities provided, we develop the following tables:
Arrival Factor P(x) C P(x) Lower Upper
Arrival Factor
-20 0.20 0.20 1 20 -20
-10 0.10 0.30 21 30 -10
0 0.40 0.70 31 70 0
10 0.25 0.95 71 95 10
20 0.05 1.00 96 100 20
Service Factor P(x) C P(x) Lower Upper
Service Factor
-20% 0.15 0.15 1 15 -20%
0% 0.50 0.65 16 65 0%
20% 0.25 0.90 66 90 20%
40% 0.10 1.00 91 100 40%
The tables distribute the random numbers in accordance with the relative probability of each arrival or service event. Note that we are including a second column for arrival factor and service factor, respectively, in each table to accommodate the requirements of the =VLOOKUP function.
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The next step is to simulate the arrivals. This requires choosing a set of random numbers, relating the random numbers back to an arrival event and then calculating the simulated arrival time for each patient. For simplicity, we will use the first 8 random numbers in the first column of the random number table 15.5 in your text.
The following table is produced:
PatientRandom Number
Arrival Factor
Scheduled Arrival Time
Simulated Arrival Time
Adams 52 0 9:30 9:30
Brown 37 0 9:45 9:45
Crawford 82 10 10:15 10:25
Dannon 69 0 10:30 10:30
Erving 98 20 10:45 11:05
Fink 96 20 11:15 11:35
Graham 33 0 11:30 11:30
Hinkel 50 0 11:45 11:45
Random Number: Chosen from table 15.5
Arrival Factor: Selected from the table above based on the random number.
Ex 1: random number 52 fits in the range 96 to 100, identified with an arrival factor of +20 minutes. This patient is simulated to arrive 20 minutes late..
Ex 2: random number 98 fits in the range 16 to 65, identified with an arrival factor of 0. This patient is simulated to arrive on time.
Scheduled Arrival Time: Given in the problem
Simulated Arrival Time: Scheduled arrival time plus or minus the simulated arrival factor.
Ex 1: Erving is scheduled to arrive at 10:45, but based on the random number of 98, is simulated to be 20 minutes late. Erving's simulated arrival time is 11:05.
When this table is completed, we have a full set of simulated arrival times based on the random numbers chosen.
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The next step is to simulate the service times. This requires choosing a second set of random numbers, relating the random numbers back to an arrival event and then calculating the simulated arrival time for each patient. For simplicity, we will use the first 8 random numbers in the second column of the random number table 15.5 in your text.
The following table is produced:
PatientRandom Number
Service Factor
Scheduled Service Time
Simulated Service Time
Adams 6 -20% 15 12
Brown 63 0% 20 20
Crawford 57 0% 15 15
Dannon 2 -20% 10 8
Erving 94 40% 30 42
Fink 52 0% 15 15
Graham 69 20% 20 24
Hinkel 33 0% 15 15
Random Number: Chosen from table 15.5
Service Factor: Selected from the table above based on the random number.
Ex 1: random number 6 fits in the range 1 to 15, identified with an service factor of -20%. This service is simulated to require 20% less time than planned.
Ex 2: random number 94 fits in the range 91 to 100, identified with an service factor of +40%. This service is simulated to require 40% more time than planned.
Scheduled Service Time: Given in the problemSimulated Service Time: Scheduled arrival time plus or minus the simulated arrival factor.
Ex 1: Erving's procedure is scheduled to require 30 minutes, but based on the random number of 94, is simulated to require 40% more time than planned. 40% of 30 minutes is an additional 12 minutes. Erving's simulated service time is 42 minutes.
When this table is completed, we have a full set of simulated service times based on the random numbers chosen.
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The last step in the static simulation is to integrate the simulated arrivals and services into a simple business model that tells Dr. Greenberg what will happen to his day based on this SINGLE SET OF RANDOM NUMBERS.
The following table is produced:
PatientSimulated
Arrival TimeSimulated Start Time
Simulated Service
Time
Simulated Completion
Time
Adams 9:30 9:30 12 9:42
Brown 9:45 9:45 20 10:05
Crawford 10:25 10:25 15 10:40
Dannon 10:30 10:40 8 10:48
Erving 11:05 11:05 42 11:47
Fink 11:35 11:47 15 12:02
Graham 11:30 12:02 24 12:26
Hinkel 11:45 12:26 15 12:41
Simulated Arrival Time: From Arrival simulation table above.
Simulated Start Time: Developed from the following logic:
The first patient starts at the later of 9:30 or when they arrive. In this case, Adams is simulated to be on time, so the service begins at 9:30.All other patients begin at the later of when they arrive or when Dr. Greenberg completes the prior patient's service.
Ex1: Brown arrive at 9:45. Dr. Greenberg finishes Adams at 9:42. Dr. Greenberg cannot begin working on Brown until Brown arrives. Accordingly, Dr. Greenberg is idle for 3 minutes before beginning to work on Brown.
Ex2: Fink arrives at 11:35, but Dr. Greenberg is unable to complete Erving until 11:47. Fink must wait for 12 minutes before being served.
Simulated Service Time: From Service simulation table above.
Simulated Completion Time: Calculated by adding Simulated Service Time to Simulated Start Time.
Ex 1: Dannon's service is simulated to begin at 10:40 and is simulated to require 8 minutes. Dannon is done, and Dr. Greenberg is ready for his next patient, at 10:48.
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Interpretation: Based on this simulation, Dr. Greenberg will not be done until 12:41 and will miss his flight unless he cancels a his last two patients.
Strengths: A simulation is a more realistic version of what happens every day. Patients don't always arrive on time and procedures don't always take exactly the planned amount of time. Also, Dr. Greenberg can simulate his day before it happens and in compressed time.
Weaknesses: This is only one version of what might happen, based on a single experiment with a sample size of 8 patients. While it is very unlikely that Dr. Greenberg's day will unfold exactly as planned, it is also unlikely that his day will match this simulation.
Improvement: Since we cannot increase the sample size beyond the scheduled 8 patients, we can improve the usefulness of the model by building it so that it can be easily repeated with different random numbers, allowing patterns to emerge and letting Dr. Greenberg see the likelihood that he will be able to leave on time. This leads us to the development of a "live" simulation in which everything changes as we change random numbers.
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Development of a live simulation
NOTE: Excel does not handle "time" very well. When we use Excel for this purpose, it is easier to use "fractional hours" to express time. You'll see this throughout the live simulation. Here's a quick table explaining the conversion from minute to fractional hours.
Minutes Fractional Hour Time Fractional50 0.833 9:00 9.00045 0.750 9:05 9.08330 0.500 9:10 9.16725 0.417 9:15 9.25020 0.333 9:20 9.33315 0.250 9:25 9.41710 0.167 9:30 9.5005 0.083 9:45 9.750
Part 1 - Probability Interval tables
We'll cover the Excel detail of the arrival table. The service table works exactly the same way. As seen before, the table looks as follows:
Arrival Factor: Given in the problem
P(x): The probability of each arrival event. This is given in the problem.
C P(x): The cumulative probability that X <= x. Cell C5 = B5Cell C6 = C5 + B6, C7-C9 follow this format
Lower: The lower bound of the random number interval for that arrival event. Cell D5 = 1Cell D6 = 1 + E5, D7 - D9 follow this format
Upper: The upper bound of the random number interval for that arrival event.
Cell E5 = C5 * 100 (all cells in column E follow this formula; E6 = C6 * 100, etc.)
Arrival Factor (for =VLOOKUP): F5 = A5, F6 = A6, etc.
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Part 2 - Simulation table for arrivals
After setting up the random number interval table, we can simulate arrivals and produce the following:
Random Number: Cells B12 through B19 all use =RANDBETWEEN(1,100). When you use this function the random numbers will change every time anything is done to the spreadsheet. While confusing at first, this is what we want to happen, each change creates a new simulation.
Arrival Factor: The arrival factors change automatically with the changes in random numbers when the =VLOOKUP function is used to "lookup" the random number in the random number interval table (see prior page) and the "deliver" the correct arrival factor back to this cell.
Example: Cell C15 contains =VLOOKUP(B15,$D$5:$F$9,3)
Scheduled Arrival Time: Given in the problem, entered as fractional hours.
Simulated Arrival Time: = Scheduled Arrival Time + Arrival Factor. Cell E17 = C17 + D17
Watch this table as you hit "F9" and you can see that the random numbers, arrival factor and simulated arrival time all change.
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Part 3 - Simulation table for service
The construction of the service simulation table is almost identical to the arrival simulation table. Again, it relies upon the random number interval table for service.
Here are the tables:
Random Number: Cells I12 through I19 all use =RANDBETWEEN(1,100). When you use this function the random numbers will change every time anything is done to the spreadsheet. While confusing at first, this is what we want to happen, each change creates a new simulation.
Service Factor: The arrival factors change automatically with the changes in random numbers when the =VLOOKUP function is used to "lookup" the random number in the random number interval table (see prior page) and the "deliver" the correct arrival factor back to this cell.
Example: Cell J15 contains =VLOOKUP(I15,$K$5:$M$8,3). The easiest way to do this is to use the dialog box for this function as shown on the prior page.
Scheduled Service Time: Given in the problem, entered as fractional hours.
Simulated Service Time: = Scheduled Service Time + (Service Factor * Scheduled Service Time). Cell L13 =K13+(J13*K13)
Watch this table as you hit "F9" and you can see that the random numbers, arrival factor and simulated arrival time all change.
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Part 4 - The business modell
Simulated Arrival Time: From the arrival simulation table Cell B22 = E12
Simulated Start Time: The logic is exactly the same as in the static model. The first patient starts at the later of 9:30 or when they arrive. In this case, Adams is simulated to be on time, so their service begins at 9:30. Cell C22 =MAX(B22, 9.5)All other patients begin at the later of when they arrive or when Dr. Greenberg completes the prior patient's service. Cell B23 = MAX(B23, E22)
Simulated Service Time: From the service simulation table Cell D22 = L12
Simulated Completion Time: = Simulated Start Time + Simulated Service Time. Cell E22 = C22 + D22
Will he be late? Cell E31 uses an IF function to indicate whether Dr. Greenberg will finish on time or be late. The formula in cell E31 is =IF(E29>12.25, "late", "on time"), but it is easier to use the dialog box for the function as shown:
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Conclusion
We now have a fully live simulation of Dr. Greenberg's day.
Hit "F9" 20 times and keep track of the number of time Dr. Greenberg will be on time versus late. It will vary a bit for each experiment, but my results indicate that the model predicts he will be late about 70% to 80% of the time.
If Dr. Greenberg now understands that there is a 70% to 80% chance that he will be late, how could he use this information to make sure he is on time for his flight AND respectful of his patient's time?
This document walks through a single queuing problem, but most of the work is applicable to all of the problems we have studied.
There will always be at least 1 and usually 2 random number interval tables needed.
Anything with a random number interval table will need to be simulated via a Monte Carlo / random number process.
The business model will vary for each type of problem and may even vary for different problems of the same type (2 different queuing problems) but the development of the business model is the part that needs YOUR sharp thinking and clear understanding of the business being studied.
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Activating the add-ins in Excel 2007.
The course requires the use of 2 Excel add-ins. Both are provided as part of every copy of Excel. To activate them, please do the following.
1. Select the big button and then click on EXCEL OPTIONS.
2. Select Add-Ins.
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3. Verify that Excel Add-in is in the “Manage” box and select “Go”.
4. Left click check marks in the boxes for Analysis ToolPak and Solver Add-in. Click on “OK”.
5. Verify that the Add-Ins are installed by clicking on Data.
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