|
Cars Demanded |
Probability |
|
25 |
.25 |
|
30 |
.20 |
|
35 |
.15 |
|
40 |
.20 |
|
45 |
.20 |
For the optimal order quantity, find a 95% confidence interval for the expected profit.
a)
Simulation model is following:

EXCEL FORMULAS:
| Simulation Table | |||||
| Month | Cars Demanded | Cars Sold | Shortage | Excess | Profit |
| 1 | =LOOKUP(RAND(),$C$2:$C$6,$A$2:$A$6) | =MIN(B13,$C$9) | =B13-C13 | =$C$9-C13 | =B13*$G$3+E13*$G$5-$C$9*$G$2-D13*$G$4 |
| 2 | =LOOKUP(RAND(),$C$2:$C$6,$A$2:$A$6) | =MIN(B14,$C$9) | =B14-C14 | =$C$9-C14 | =B14*$G$3+E14*$G$5-$C$9*$G$2-D14*$G$4 |
| 3 | =LOOKUP(RAND(),$C$2:$C$6,$A$2:$A$6) | =MIN(B15,$C$9) | =B15-C15 | =$C$9-C15 | =B15*$G$3+E15*$G$5-$C$9*$G$2-D15*$G$4 |
Formulas of Simulation Table are copied down to 1000 rows for 1000 simulation trials.
| Simulation Output | |
| Expected profit | =AVERAGE(F13:F1012) |
| Standard deviation of profit | =STDEV(F13:F1012) |
| 95% confidence interval (CI) for expected profit | |
| z-value for 95% CI | =NORMSINV(1-(1-0.95)/2) |
| Lower limit | =I13-I19*I15 |
| Upper limit | =I13+I19*I15 |
In order to find the optimal order quantity,
Change the number of cars ordered in cell C9 to order quantity 25, 30, 35, 40, 45 and note down expected profit for each order quantity.




Based on simulation of expected profit for different order quantities, the summary is following:
| Order quantity | Expected Profit |
| 25 | 153005 |
| 30 | 157220 |
| 35 | 155700 |
| 40 | 151935 |
| 45 | 142640 |
Highest expected profit is 157,220 for order quantity 30
Therefore, optimal order quantity = 30 cars
95% confidence interval for expected profit for each order quantity is computed in respective worksheet and is shown .
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