The annual demand for a product is 15,600 units. The weekly demand is 300 units
ID: 403242 • Letter: T
Question
The annual demand for a product is 15,600 units. The weekly demand is 300 units with a standard deviation of 90 units. The cost to place an order is $31.20, and the time from ordering to receipt is four weeks. The annual inventory carrying cost is $0.10 per unit.
a. Find the reorder point necessary to provide a 98 percent service probability.
b. Suppose the production manager is asked to reduce the safety stock of this item by 50 percent. If she does so, what will the new service probability be?
.
The annual demand for a product is 15,600 units. The weekly demand is 300 units with a standard deviation of 90 units. The cost to place an order is $31.20, and the time from ordering to receipt is four weeks. The annual inventory carrying cost is $0.10 per unit.
a. Find the reorder point necessary to provide a 98 percent service probability.
b. Suppose the production manager is asked to reduce the safety stock of this item by 50 percent. If she does so, what will the new service probability be?
.
Explanation / Answer
a. Find the reorder point necessary to provide a 98 percent service probability.
D = Annual demand (units)=15600
S = Cost per order ($) =$31.20
H = Holding cost ($) = $0.10
LT = Lead Time = 4 weeks
Std Dev = Sigma = 90 unit
So EOQ = Sqrt(2*D*S/H)
= Sqrt(2*15600*31.20/0.10)
= 3120 units
Standard deviation of demand during lead time
= Sigma(dLT) = 90
2% stockout policy (service level = 98%)
Using Z Table, for an area under the curve of 98%, the Z = 2.06
Safety stock = Z*Sigma(d*LT) = 2.06*(90*Sqrt(4)) =371
Reorder point(ROP) = expected demand during lead time + safety stock
ie ROP = Lead Time*weekly demand+safety stock
ie ROP = 4*300 + 371
=1571 units
b. Suppose the production manager is asked to reduce the safety stock of this item by 50 percent. If she does so, what will the new service probability be?
New Safety stock = 371/2 = 186 units
So Safety stock = Z*Sigma(d*LT) = Z*(90*Sqrt(4)) =186
ie Z*90*2 = 186
ie Z = 186/180 = 1.0333
From Z Table, This corresponds to 0.8485 or 84.85% Service probaility
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