Normal Distribution and Egress

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The first step in building our simulation model was to confirm whether or not a normal distribution was in fact reasonable for the unknown demand of Egress’ ski jackets in the coming year. To do this we input Egress’ estimated demands into Best Fit. After analyzing the output (Exhibit 1 and 2) we see that the p-value for Chi Squared is .5637, this indicates that the assumption of a normal distribution of demand is indeed reasonable. Based on this output we determined that the mean of demand was 12,000 and the standard deviation was 3,497 (rounded up from 3496.8). Our next step was to build a simulation model based on the data provided to us by Egress using @RISK. Based upon our findings in Best Fit we used @RISK’s normal distribution function to generate random inputs of demand and the RISKSIMTABLE function to run the simulation multiple times using different production quantities. For each simulation run we did 1,000 iterations, and used the @RISK output function to analyze profit. At Egress’ request we ran our first simulation for a production quantity of 7,800 and production quantity at the mean demand of 12,000. From the output (Exhibit 3 and 4) we see that expected profit at Q = 7800 is $42,249 and the standard deviation is $52,026. At Q = 12,000 the expected profit is $42,296 with a standard deviation of $143,181. While the mean demand quantity of 12,000 yields a slightly better expected profit, it is a much riskier option. First, the standard deviation of Q = 12,000 is more than three times larger than that of Q = 7,800. Second, we can see in Exhibit 6 that there is a 16.4% chance that at Q = 12,000 Egress will lose $100,000 or more. In comparison when Q = 7,800 there is only a 3.3% chance that Egress will lose more than $100,000 (Exhibit 5). Finally, while at Q = 12,000 the company’s maximum profit is considerable larger than it is at Q = 7,800, the company stands to loose much more at Q = 12,000 as is reflected in the minimum profit...
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