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Applies to 
Microsoft Office Excel 2003 

This article was adapted from Microsoft Excel Data Analysis and Business Modeling by Wayne L. Winston. Visit Microsoft Learning to learn more about this book.
This classroomstyle book was developed from a series of presentations by Wayne Winston, a well known statistician and business professor who specializes in creative, practical applications of Excel. So be prepared — you may need to put your thinking cap on.

In this article
Who uses Monte Carlo simulation?
What happens when I enter =RAND() in a cell?
How can I simulate values of a discrete random variable?
How can I simulate values of a normal random variable?
How should a greeting card company determine how many cards to produce?
The impact of risk on our decision
Confidence interval for mean profit
Problems
Sample files You can download the sample files that relate to excerpts from Microsoft Excel Data Analysis and Business Modeling from Microsoft Office Online. This article uses the files RandDemo.xls, Discretesim.xls, NormalSim.xls, and Valentine.xls.
We would like to be able to accurately estimate the probabilities of uncertain events. For example, what is the probability that a new product’s cash flows will have a positive net present value (NPV)? What is the riskiness of our investment portfolio? Monte Carlo simulation enables us to model situations that present uncertainty and play them out thousands of times on a computer.
Note The name Monte Carlo simulation comes from the fact that during the 1930s and 1940s, many computer simulations were performed to estimate the probability that the chain reaction needed for the atom bomb would work successfully. The physicists involved in this work were big fans of gambling, so they gave the simulations the code name Monte Carlo.
Who uses Monte Carlo simulation?
Many companies use Monte Carlo simulation as an important tool for decisionmaking. Here are some examples.
 General Motors, Procter and Gamble, and Eli Lilly use simulation to estimate both the average return and the riskiness of new products. At GM, this information is used by CEO Rick Waggoner to determine the products that come to market.
 GM uses simulation for activities such as forecasting net income for the corporation, predicting structural costs and purchasing costs, determining its susceptibility to different kinds of risk (such as interest rate changes and exchange rate fluctuations).
 Lilly uses simulation to determine the optimal plant capacity that should be built for each drug.
 Wall Street firms use simulation to price complex financial derivatives and determine the Value at RISK (VAR) of their investment portfolios.
 Procter and Gamble uses simulation to model and optimally hedge foreign exchange risk.
 Sears uses simulation to determine how many units of each product line should be ordered from suppliers — for example, how many pairs of Dockers should be ordered this year.
 Simulation can be used to value "real options," such as the value of an option to expand, contract, or postpone a project.
 Financial planners use Monte Carlo simulation to determine optimal investment strategies for their clients’ retirement.
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What happens when I enter =RAND() in a cell?
When you enter the formula =RAND() in a cell, you get a number that is equally likely to assume any value between 0 and 1. Thus, around 25 percent of the time, you should get a number less than or equal to .25; around 10 percent of the time you should get a number that is at least .90, and so on. To see how the RAND function works, take a look at the file RandDemo.xls, shown in the following figure.
Note When you open the file RandDemo.xls, you will not see the same random numbers shown in the previous figure. The RAND function always recalculates the numbers it generates when a spreadsheet is opened or new information is entered in the spreadsheet.
I copied from cell C3 to C4:C402 the formula =RAND(). I named the range C3:C402 Data. Then, in column F, I tracked the average of the 400 random numbers (cell F2) and used the COUNTIF function to determine the fractions that are between 0 and .25, .25 and .50, .50 and .75 and .75 and 1. When you press the F9 key, the random numbers are recalculated. Notice that the average of the 400 numbers is always near .5 and that around 25 percent of the results are in each interval of .25. These results are consistent with the definition of a random number. Also note that the values generated by RAND in different cells are independent. For example, if the random number generated in cell C3 is a large number (say .99), this tells us nothing about the values of the other random numbers generated.
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How can I simulate values of a discrete random variable?
Suppose the demand for a calendar is governed by the following discrete random variable.
Demand 
Probability 
10,000 
.10 
20,000 
.35 
40,000 
.3 
60,000 
.25 
How can we have Excel play out, or simulate, this demand for calendars many times? The trick is to associate each possible value of the RAND function with a possible demand for calendars. The following assignment ensures that a demand of 10,000 will occur 10 percent of the time, and so on.
Demand 
Random Number Assigned 
10,000 
Less than .10 
20,000 
Greater than or equal to .10, and less than .45 
40,000 
Greater than or equal to .45, and less than .75. 
60,000 
Greater than or equal to .75. 
To see a simulation of demand, look at the file Discretesim.xls, shown in the following figure.
The key to our simulation is to use a random number to key a lookup from the table range F2:G5 (named lookup). Random numbers greater than or equal to 0 and less than .10 will yield a demand of 10,000; random numbers greater than or equal to .10 and less than .45 will yield a demand of 20,000; random numbers greater than or equal to .45 and less than .75 will yield a demand of 40,000; and random numbers greater than or equal to .75 will yield a demand of 60,000. I generated 400 random numbers by copying from C3 to C4:C402 the formula RAND(). Then I generated 400 trials or iterations of calendar demand by copying from B3 to B4:B402 the formula VLOOKUP(C3,lookup,2). This formula ensures that any random number less than .10 generates a demand of 10,000; any random number between .10 and .45 generates a demand of 20,000, and so on. In the cell range F8:F11, I used the COUNTIF function to determine the fraction of our 400 iterations yielding each demand. Note that whenever you press F9 to recalculate the random numbers, the simulated probabilities are close to our assumed demand probabilities.
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How can I simulate values of a normal random variable?
If you enter into any cell the formula NORMINV(rand(), mu , sigma), you will generate a simulated value of a normal random variable having a mean mu and standard deviation sigma. I’ve illustrated this procedure in the file NormalSim.xls, shown in the following figure.
Let’s suppose we want to simulate 400 trials or iterations for a normal random variable with a mean of 40,000 and a standard deviation of 10,000. (I entered these values in cells E1 and E2 and named these cells mean and sigma, respectively.) Copying the formula =RAND() from C4 to C5:C403 generates 400 different random numbers. Copying from B4 to B5:B403 the formula NORMINV(C4,mean,sigma) generates 400 different trial values from a normal random variable with a mean of 40,000 and a standard deviation of 10,000. When we press the F9 key to recalculate the random numbers, the mean remains close to 40,000 and the standard deviation close to 10,000.
Essentially, for a random number x, the formula NORMINV(p, mu , sigma) generates the p^{th} percentile of a normal random variable with a mean mu and a standard deviation sigma. For example, the random number .8466 in cell C13 generates in cell B13 approximately the 85th percentile of a normal random variable with a mean of 40,000 and a standard deviation of 10,000.
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How should a greeting card company determine how many cards to produce?
In this section, I’ll show how Monte Carlo simulation can be used as a tool to help businesses make better decisions. Suppose that the demand for a Valentine’s Day card is governed by the following discrete random variable:
Demand 
Probability 
10,000 
.10 
20,000 
.35 
40,000 
.3 
60,000 
.25 
The greeting card sells for $4.00, and the variable cost of producing each card is $1.50. Leftover cards must be disposed of at a cost of $0.20 per card. How many cards should be printed?
Basically, we simulate each possible production quantity (10,000, 20,000, 40,000 or 60,000) many times (say, 1,000 iterations). Then we determine which order quantity yields the maximum average profit over the 1,000 iterations. You can find the work for this section in the file Valentine.xls, shown in the following figure. I’ve assigned the range names in cells B1:B11 to cells C1:C11. I’ve assigned the cell range G3:H6 the name lookup. Our sales price and cost parameters are entered in cells C4:C6.
I then enter a trial production quantity (40,000 in this example) in cell C1. Next I create a random number in cell C2 with the formula =RAND(). As previously described, I simulate demand for the card in cell C3 with the formula VLOOKUP(rand,lookup,2). (In the VLOOKUP formula, rand is the cell name assigned to cell C3, not the RAND function.)
The number of units sold is the smaller of our production quantity and demand. In cell C8, I compute our revenue with the formula MIN(produced,demand)*unit_price. In cell C9, I compute total production cost with the formula produced*unit_prod_cost.
If we produce more cards than are demanded, the number of units left over equals production minus demand; otherwise no units are left over. We compute our disposal cost in cell C10 with the formula unit_disp_cost*IF(produced>demand,produceddemand,0). Finally, in cell C11, we compute our profit as revenuetotal_var_costtotal_disposing_cost.
We would like an efficient way to press F9 many (say 1,000 times) for each production quantity and tally up our expected profit for each production quantity. This situation is one in which a twoway data table comes to our rescue. The data table I used in this example is shown in the following figure.
In the cell range A16:A1015, I entered the numbers 11000 (corresponding to our 1,000 trials). One easy way to create these values is to enter a 1 in cell A16, select the cell, and then, on the Edit menu, click Fill Series. In the Series dialog box, shown in the following figure, enter a step value of 1 and a stop value of 1000. Under Series in, click Columns, and then click OK. The numbers 1 through 1000 will be entered automatically in column A, starting in cell A16.
Next we enter our possible production quantities (10,000, 20,000, 40,000, 60,000) in cells B15:E15. We want to calculate profit for each trial number (1 through 1,000) and each production quantity. We refer to the formula for profit (calculated in cell C11) in the upper left cell of our data table (A15) by entering =C11.
We are now ready to trick Excel into simulating 1,000 iterations of demand for each production quantity. Select the table range (A15:E1014), and then click Table on the Data menu. To set up a twoway data table, we select any blank cell (we chose cell I14) as our column input cell and choose our production quantity (cell C1) as the row input cell. When you click OK, Excel simulates 1,000 demand values for each order quantity.
To illustrate why this works, consider the values placed by the data table in the cell range C16:C1015. For each of these cells, Excel will use a value of 20,000 in cell C1. In C16, the column input cell value of 1 is placed in a blank cell and the random number in cell C2 recalculates. The corresponding profit is then recorded in cell C16. Then the column cell input value of 2 is placed in a blank cell, and the random number in C2 again recalculates. The corresponding profit is entered in cell C17.
By copying from cell B13 to C13:E13 the formula AVERAGE(B16:B1015), we compute average simulated profit for each production quantity. By copying the formula STDEV (B16:B1015) from cell B14 to C14:E14, we compute the standard deviation of our simulated profits for each order quantity. Each time we press F9, 1,000 iterations of demand are simulated for each order quantity. Producing 40,000 cards always yields the largest expected profit. Therefore, it appears as if producing 40,000 cards is the proper decision.
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The impact of risk on our decision
If we produce 20,000 cards instead of 40,000 cards, our expected profit drops approximately 22 percent, but our risk (as measured by the standard deviation of profit) drops almost 73 percent. Therefore, if we are extremely risk averse, producing 20,000 cards might be the right decision. By the way, producing 10,000 cards always has a standard deviation of zero cards because if we produce 10,000 cards, we will always sell all of them and have none left over.
Note In this worksheet, I set the Calculation option to Automatic Except For Tables. (See the Calculation tab on the Options dialog box.) This setting ensures that our data table will not recalculate unless we press F9, which is a good idea because a large data table will slow down your work if it recalculates every time you type something into your spreadsheet. Note that in this example, whenever you press F9, the mean profit will change. This happens because each time you press F9 a different sequence of 1,000 random numbers is used to generate demands for each order quantity.
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Confidence interval for mean profit
A natural question to ask in this situation is "Into what interval are we 95 percent sure the true mean profit will fall?" This interval is called the 95 percent confidence interval for mean profit. A 95 percent confidence interval for the mean of any simulation output is computed by the following formula.
In cell J11, I computed the lower limit for the 95 percent confidence interval on mean profit when 40,000 calendars are produced with the formula D131.96*D14/SQRT(1000). In cell J12, I computed the upper limit for our 95 percent confidence interval with the formula D13+1.96*D14/SQRT(1000). These calculations are shown in the following figure.
We are 95 percent sure that our mean profit when 40,000 calendars are ordered is between $53,860 and $59,934.
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Problems
 A GMC dealer believes that demand for 2005 Envoys will be normally distributed with a mean of 200 and standard deviation of 30. His cost of receiving an Envoy is $25,000, and he sells an Envoy for $40,000. Half of all leftover Envoys can be sold for $30,000. He is considering ordering 200, 220, 240, 260, 280, or 300 Envoys. How many should he order?
 A small supermarket is trying to determine how many copies of People magazine they should order each week. They believe their demand for People is governed by the following discrete random variable.
Demand 
Probability 
15 
.10 
20 
.20 
25 
.30 
30 
.25 
35 
.15 
 The supermarket pays $1.00 for each copy of People and sells each copy for $1.95. They can return each unsold copy of People for $0.50. How many copies of People should the store order?
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