 # Steps in Simulation Model Building

Essay by   •  March 25, 2016  •  Exam  •  475 Words (2 Pages)  •  990 Views

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Question 4:

Steps in simulation model building:

1. N_weeks_same_price is derived based on number of weeks (prior to the given week), where the price has been set at p. and demand is assigned a normal distribution with mean (200 – p – 0.5*N_weeks_same_price) and standard deviation equal to mean/3.
2. Adjusted demand is calculated by subtracting quantity sold till the current week from the inventory of 2000 units as restocking is not permitted.
3. Revenue is populated based on adjusted demand and current price.
4. Total revenue is calculated by adding revenues of individual weeks and revenue of unsold items at the end of 15 weeks period if any.

Results:

1. Below output is of pricing policy 1.
1. Mean revenue : \$107,818
2. SD: \$2733.
3. 95% CI : \$100,582 – \$111,121

[pic 1]

1. Results of Pricing policy2
1. Mean revenue : \$ 102,370
2. SD: \$3,460
3. 95% CI : \$94,938 – \$108,707

[pic 2]

1. So, it clearly turns out that pricing policy 1 is effective than policy 2 with high revenues and less dispersion.

Question 5:

1. Each tree gives a profit of \$30 and a loss of -\$45 if unsold, so in order to break even manager should sell 60% of the inventory and profits starts for sales above 60%. This is the tradeoff manager faces while deciding how much inventory to purchase.

1. Simulation model was built as follows:
1. Demand was assigned discrete distribution.
2. Total profit was calculated as adjusted demand (demand greater than inventory is negated)  X profit + unsold X loss.
3. Results are as follows:
1. Mean revenue : \$ 3717
2. SD: \$2403
3. 78% probability of getting profit
4. Approximately 60% probability of selling the entire inventory.

[pic 3]

1. It was clearly seen that there is 21.4% probability of getting loss if we assume that demand follows a discrete distribution.

1. If we assume normal distribution for demand then the results are as follows:
1. 90% probability of getting profit.
2. Only 10% chances of loss.

[pic 4]

Question 6:

1. Results of simulation model is as follows:
1. Expected value of profit: \$1.6 million
2. Standard deviation: \$0.2 million

[pic 5]

1. If the company accepts marketing firm’s advertising campaign then the results are as follows:
1. Expected value of profit: \$1.64 million
2. Standard deviation: \$1.4 million

[pic 6]

1. It is reasonably good choice to accept marketing firm’s advertising campaign as there is a high expected value and a large right skewed dispersion of profit.

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