Problem 2 Packard Electric Co., a wire harness manufacturing plant in Warren, OH

Question

Problem 2 Packard Electric Co., a wire harness manufacturing plant in Warren, OH uses 124,460 plastic connectors annually. The supplier of this part is nearby, such that immediately upon receiving an order, the supplier begins delivery at the rate of 1,200 connectors per day. Packard Electric operates 254 days per year. Packard Electric Co. has estimated that the ordering cost is $55 every time they place a new order, and the cost of carrying connectors in inventory before they are processed is $0.55 per connector on an annual basis. Determine the following a) The optimal order size b) The maximum level of inventory held for this part c) The number of operating days required to receive a complete order d) The number of operating days between orders e) The cost of holding inventory per year f The total annual inventory cost associated with the optimal order quantity

Explanation / Answer

Demand = 124460

Demand rate d = (124460/254) = 490

Delivery rate p = 1200/day

Number of days operating = 254/year

Ordering cost s =55 $

Holding cost = 0.55 $/connecter

a) Optimal order size Q = SQRT(2*D*S/H*(1-d/p)) = SQRT(2*124460*55/0.55*(1-490/1200)) = 6486

b) Maximum inventory level = Q*(1-d/p) = 6486*(1-490/1200) = 3838

c) The number of operating days to receive a complete order = 6486/1200 = 5.4 days

d) The number of operating days between orders = D/Q = 124460/6486 = 19.19 days

e) Cost of holding inventory per year = H*Q*(1-d/p)/2 = 0.55*(6486/2)*(1-490/1200) = 1056 $

f) Total annual inventory cost = Ordering cost+Holding cost= (D/Q)*s+1056 = (124460/6486)*55+1056 = 2112 $

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.