Question

Problem 2. 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. Find the reorder point necessary to provide a 98 percent service probability. Suppose the production manager is asked to reduce the safety stock of this time by 50 percent. If she does so, what will the new service probability be?

Answer 1

Annual demand = 15600 units

Weekly demand = 300 units

Standard deviations =90 units

Ordering cost = $31.20

Holding cost = $0.10 per unit

Lead time = 4 weeks

a. Service level =98%, Z = 2.05

Reorder point = lead time*demand + Z*σ *√lead time*demand

= 4*300 + 2.05*90*√(4*300)

RP= 7591.27 units

b. Safety stock @ 98% service level = ROP – LT demand = 7591.27 - 4*300 = 6391.27 units

Now the safety stock is made half i.e. 6391.27/2 = 3195.64

New ROP = Safety stock + LT demand = 3195.64 + 4*300 = 4395.64 units

Thus new value of Z is:

  

z =  

                                 z = 1.02

Thus from the Z table, service level is = 84.61%

a. EOQ Calculation:EOQ = √ ((2∗annual demand ∗ordering cost)/holding cost per unit per year)

EOQ = √ ((2*15600*31.20)/0.10)

EOQ = 3120

E0Q =3120

Current safety stock = 6931.27 units

Reduced safety stock= 3195.64 units

Reorder point = 7591.27 units

Service level = 84.61%