em 13-21 Demand for walnut fudge ice cream at the Sweet Cream Dairy can be appro
ID: 351008 • Letter: E
Question
em 13-21
Demand for walnut fudge ice cream at the Sweet Cream Dairy can be approximated by a normal distribution with a mean of 22 gallons per week and a standard deviation of 3.1 gallons per week. The new manager desires a service level of 90 percent. Lead time is two days, and the dairy is open seven days a week. (Hint: Work in terms of weeks.)
Use Table B and Table B1.
a-1. If an ROP model is used, what ROP would be consistent with the desired service level? (Do not round intermediate calculations. Round your final answer to 2 decimal places.)
ROP gallons
a-2. How many days of supply are on hand at the ROP, assuming average demand? (Do not round intermediate calculations. Round your final answer to 2 decimal places.)
Days
b-1. If a fixed-interval model is used instead of an ROP model, what order size would be needed for the 90 percent service level with an order interval of 6 days and a supply of 8 gallons on hand at the order time? (Do not round intermediate calculations. Round your final answer to the nearest whole number.)
Order size gallons
b-2. What is the probability of experiencing a stockout before this order arrives? (Do not round intermediate calculations. Round your final answer to the nearest whole percent. Omit the "%" sign in your response.)
Probability %
c. Suppose the manager is using the ROP model described in part a. One day after placing an order with the supplier, the manager receives a call from the supplier that the order will be delayed because of problems at the supplier’s plant. The supplier promises to have the order there in two days. After hanging up, the manager checks the supply of walnut fudge ice cream and finds that 2 gallons have been sold since the order was placed. Assuming the supplier’s promise is valid, what is the probability that the dairy will run out of this flavor before the shipment arrives? (Do not round intermediate calculations. Round your final answer to the nearest whole percent. Omit the "%" sign in your response.)
Risk probability %
Explanation / Answer
Lead time , L = 2/7 weeks
Lead time demand = Weekly demand * Lead time = 22*2/7 = 44/7 = 6.286
Standard deviation of lead time demand = 3.1*(2/7) = 1.657
z value corresponding to 90% CSL = NORMSIN(0.9) = 1.28
a-1. ROP consistent with 90% CSL = Lead time time demand + z*Std dev of lead time demand = 6.286 + 1.28*1.657 = 8.41 gallons
a-2. Average daily demand = 22/7 = 3.143
Days of supply on hand at ROP = 8.41/3.143 = 2.67 days
b-1. Order upto level = Weekly demand * (Lead time in weeks + Review period in weeks) + z * Std deviation of weekly demand * ( Lead time + Review period)
= 22*(2/7 + 6/7) + 1.28*3.1(2/7 + 6/7) = 29.38 gallons
Order size needed when supply on hand is 8 gallons = Order upto level - Inventory position = 29.38 - 8 = 21.38 gallons
Order size = 21 gallons
b-2. Average demand during lead time (2/7 weeks) = 22*2/7 = 6.286
Std deviation of demand during lead time (2/7 weeks) 3.1*(2/7) = 1.657
For current stock level of 8 gallons, z = (8-6.286)/1.657 = 1.034
Probability of demand being less than 8 gallons during the lead time = NORMSDIST(1.034) = 0.8495
Probability of experiencing a stockout before the order arrives = 1 - 0.8495 = 0.1505 = 15 % (rounded off to whole percent )
c. Current inventory level = 8-2 = 6 gallons
Current lead time = 2 days
z value = (6-6.286)/1.657 = -0.1726
Probability that demand during the 2 days will be less than the available stock of 6 gallons = 0.4315
Risk probability = 1-0.4315 = 0.5685 ~ 57 % (rounded off to the nearest whole percent)
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