Problem Set 4
OPIM Assignment 4 1. Cu = 24-11 = $13
Co = 11-7 = $4
Critical ratio = 13/(13+4) = 0.7647 μ = 30,000 σ = 10,000
Using normal distribution function (=norminv(0.7647,30000,10000)), the optimum order quantity is 37,216 jerseys to maximize profit.
2. Quantity = 32,000
First, we normalize the order quantity to find the z-statistic z=Q-μσ=32,000-25,00010,000=0.7 We then look up the standard normal loss function. The expected lost sale is given by.
Lz=0.1429
Therefore, the expected lost sales = 10,000 * 0.1429 ≈ 1,429
Expected sales = 30,000 – 1,429 = 30,571 jerseys
3. First, we normalize the order quantity to find its z-statistic z=Q-μσ=28,000-20,00010,000=0.8 In-stock probability = normdist(0.8, 0, 1, 1) = 0.788145
Therefore, the probability of filling all demand is 0.788145
4. Quantity = 8,000
First, we normalize the order quantity to find the z-statistic z=Q-μσ=8,000-15,00010,000=-0.7 We then look up the standard normal loss function. The expected lost sale is given by.
Lz=0.8429
Therefore, the expected lost sales = 10,000 * 0.8429 = 8,429
Expected sales = 15,000 – 8,429 = 6,571 jerseys
Expected leftover = 8000 – 6,571 = 1429 jerseys
Therefore, Nike has to sell 1,429 on discount
5. Cu = 16-11 = $5
Co = 9-7 = $2
Critical ratio = 5/(5+2) = 0.7143 μ = 40,000 σ = 10,000
Using normal distribution (=norminv(0.7143,40000,10000)), the optimum order quantity is 45660 jerseys in the first order to maximize profit.
6. Cu = 0.02-0.01 = $0.01
Co = 0.01-0 = $0.01
Critical ratio = 0.01/(0.01+0.01) = 0.5 μ = 300 σ = 75
Using normal distribution (=norminv(0.5,300,75)), the optimum order quantity is 300 million minutes from Vmail to minimize the expected capacity expense.
7. Q = 375
First, we normalize the order quantity to find the z-statistic z=Q-μσ=375-30075=1 We then look up the standard normal loss function. The expected lost sale is given by.
Lz=0.0833
Therefore, the expected lost sales = 75 * 0.0833 =
Co = 11-7 = $4
Critical ratio = 13/(13+4) = 0.7647 μ = 30,000 σ = 10,000
Using normal distribution function (=norminv(0.7647,30000,10000)), the optimum order quantity is 37,216 jerseys to maximize profit.
2. Quantity = 32,000
First, we normalize the order quantity to find the z-statistic z=Q-μσ=32,000-25,00010,000=0.7 We then look up the standard normal loss function. The expected lost sale is given by.
Lz=0.1429
Therefore, the expected lost sales = 10,000 * 0.1429 ≈ 1,429
Expected sales = 30,000 – 1,429 = 30,571 jerseys
3. First, we normalize the order quantity to find its z-statistic z=Q-μσ=28,000-20,00010,000=0.8 In-stock probability = normdist(0.8, 0, 1, 1) = 0.788145
Therefore, the probability of filling all demand is 0.788145
4. Quantity = 8,000
First, we normalize the order quantity to find the z-statistic z=Q-μσ=8,000-15,00010,000=-0.7 We then look up the standard normal loss function. The expected lost sale is given by.
Lz=0.8429
Therefore, the expected lost sales = 10,000 * 0.8429 = 8,429
Expected sales = 15,000 – 8,429 = 6,571 jerseys
Expected leftover = 8000 – 6,571 = 1429 jerseys
Therefore, Nike has to sell 1,429 on discount
5. Cu = 16-11 = $5
Co = 9-7 = $2
Critical ratio = 5/(5+2) = 0.7143 μ = 40,000 σ = 10,000
Using normal distribution (=norminv(0.7143,40000,10000)), the optimum order quantity is 45660 jerseys in the first order to maximize profit.
6. Cu = 0.02-0.01 = $0.01
Co = 0.01-0 = $0.01
Critical ratio = 0.01/(0.01+0.01) = 0.5 μ = 300 σ = 75
Using normal distribution (=norminv(0.5,300,75)), the optimum order quantity is 300 million minutes from Vmail to minimize the expected capacity expense.
7. Q = 375
First, we normalize the order quantity to find the z-statistic z=Q-μσ=375-30075=1 We then look up the standard normal loss function. The expected lost sale is given by.
Lz=0.0833
Therefore, the expected lost sales = 75 * 0.0833 =