(20 points) Demand during lead time for one brand ofTV is normally distributed w
ID: 338994 • Letter: #
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(20 points) Demand during lead time for one brand ofTV is normally distributed with a mean of 36 TVs and a standard deviation of 15 TVs. What safety stock should be carried for a90% service level? What is the appropriate reorder point? (Use the normal table shown below) 4. a. What safety stock should be carried for a 90% service level? (round your response to the nearest whole number). b. What is the appropriate reorder point? (round your response to the nearest The table below shows the total area under the normal curve for a point that is Z standard deviations to the right of the mean. 0.5199 0.5239 0.527 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 0.9032 0.9049 0.9066 0.9082 0 1.3 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9430 0.9441 9505 0.9515 0.9525 0.9535 0.9545 09474 0.9485 0.9495 0. 2.0 0.9773 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 2.8 0.9974 0.9975 0.9976 0.99770.9977 0.9978 0.9979 0.9980 0.9980 0.9981 2.9 0.9981 0.9982 0.9983 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986Explanation / Answer
It is to be noted that Z value of service level of 95%
= NORMSINV ( 0.95 )
= 1.6448
Standard deviation of demand during lead time = Sd = 15 TVs
Therefore ,
Required safety stock
= Z value of service level 95 % x Standard deviation of demand during lead time
= 1.6448 x 15
= 24.67 ( 25 rounded to nearest whole number )
Reorder point = Mean demand during lead time + safety stock = 36 + 25 = 61 TVs
SAFETY STOCK = 25 TVs
REORDER POINT = 61 TVs
SAFETY STOCK = 25 TVs
REORDER POINT = 61 TVs
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