Problem 2

xlsx

School

University of Illinois, Urbana Champaign *

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Course

MISC

Subject

Economics

Date

Jan 9, 2024

Type

xlsx

Pages

3

Uploaded by ColonelFlower12122

Report
Given, Daily demand, d = 60 Demand Standard deviation, σd = 15 Service level = 0.98 z = 2.054 Lead Time, L = 12 days Lead Time Standard deviation, σL = 2 a. Expected demand during lead time = d*L 720 b. Standard deviation of demand during lead time, σdLt =Sqrt (σd^2*L + (d^2*σL^2) 130.77 c. Average Safety stock = z*σdLt 268.5954 d. The reorder point= dL+ safety stock 988.5954 e. Reducing the lead time by 10%: Lead Time, L = 10.8 Expected demand during lead time = d*L 648 Standard deviation of demand during lead time, σdLt =Sqrt (σd^2*L + (d^2*σL^2) 129.7305 Average Safety stock = z*σdLt 266.4664 The reorder point= dL+ safety stock 914.4664 Reducing the standard deviation of lead time by 10% Lead Time Standard deviation, σL = 1.8 Standard deviation of demand during lead time, σdLt =Sqrt (σd^2*L + (d^2*σL^2) 119.8499 Average Safety stock = z*σdLt 246.1717 The reorder point= dL+ safety stock 966.1717 Reducing the service level by 10%
Service level = 0.882 1.185044 Average Safety stock = z*σdLt 154.9646 The reorder point= dL+ safety stock 874.9646 So, we can see that reducing service level by 10% lowers the reorder point by the greatest amount if the h to lower it, which is good for the hospital however the associated cons of this strategy is it might create d among the customers due to lower service level.
hospital desires dissatisfaction
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