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7. A shop has a demand for water is approximated by a normal distribution with a mean of 31 gallons per week and a standard d
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Answer #1

Weekly Demand d = 31 gallon

SD pf Weekly Demand = 4

Service Level = 90%

Z = 1.28

Lead Time LT = 3 days = 3/7 week

a.

ROP = d*LT + Z*Sd of Demand*(LT)^(1/2)

ROP = 31*(3/7) + 1.28*4*(3/7)^(1/2)

ROP = 16.64 gallons

ROP = 17 gallons

b.

Order Interval P = 7 days = 1 week

On Hand Inventory I = 10

Order Quantity Q = d*(LT+P) + Z*Sd of Demand*(LT+P)^(1/2) - I

LT +P = 3/7 + 1 = 10/7 weeks

Q = 31*(10/7) + 1.28*4*(10/7)^(1/2) - 10

Q = 50.41 - 10

Q = 40.41

Q = 40 gallons

c.

Stock on Hand = ROP - Used Stock = 17 - 7 = 10 galoon

Expected Receiving day = 2 days = 2/7 weeks

Now, we will treat 10 gallon as ROP with 2 days lead time and we will find the value of Z. After that we will find the service level basiz Z value and after that wwe will find the stock out probabiity bu subtractting service level from 1.

10 = 31*(2/7) + Z*4*(2/7)^(1/2)

10 = 8.8571 + Z*2.1381

Z = (10-8.8571)/2.1381

Z = 0.5345

Service Level = 0.7035

Stock out Probability = 1 - Service Level

Stock out Probabiity = 1 - 0.7035

Stock out Probability = 0.2965

Stock out Probability = 0.30

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