Question

In the previous problem, the manufacturer performs additional market research. Based on this research, they determine that they can increase the price to $150 and are able to reduce the standard deviation of the forecast to s- 30. At the same time, they have made an arrangement with an outlet store that will purchase unsold equipment for $60 each (hint: the disposal value is $60). Other parameters are the same as those in the previous problem. Specifically, each unit costs $75 to manufacture and it costs $10 to hold a unit of unsold item in inventory for the entire season. Demand is normally distributed with a mean of μ-100. Question 5 (2 points) In the previous problem, the manufacturer performs additional market research. Based on this research, they determine that they can increase the price to $150 and are able to reduce the standard deviation of the forecast to s- 30. At the same time, they have made an arrangement with an outlet store that will purchase unsold equipment for $60 each (hint: the disposal value is $60). Other parameters are the same as those in the previous problem. Specifically, each unit costs $75 to manufacture and it costs $10 to hold a unit of unsold item in inventory for the entire season. Demand is normally distributed with a mean of μ-100. What is the expected profit? 4212 6547 1294 2385 8610

Answer 1

Cs = Cost of shortage = 150-75 =75

Ce = Cost of excess = 75-60 =15

Service level = Cs / Ce+Cs = 75/90 = 0.833

which corresponds to a z value of 0.97

Optimum value of quantity = 100+30x0.97 =129

Expected lost sales with this quantity = L(0.97) = 0.089

Expected lost sales = SD x L(z) = 30x0.088 =2.64 =3

Expected sales = Mean - Expected lost sales = 100-3 =97

Expected leftover inventory = Order quantity - expected sales = 129-97 = 32

Profit = Cs x Expected sales- Ce x Expected leftover inventory

= 75 x97 - 15 x32 = 7275-480 = 6795

cost of holding unsold inventory = 32x 10 =320

Net profit = 6795-320 = 6475

Option 2 is the closest answer.