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# Calculating the EOQ and inventory costs

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Ross White's machine shop uses 2,500 brackets during the course of a year, and this usage is relatively constant throughout the year. These brackets are purchased from a supplier 100 miles away for \$15 each, and the lead time is 2 days. The holding cost per bracket per year is \$1.50 (or 10% of the unit cost) and the ordering cost per order is \$18.75. There are 250 working days per year.

Ross White wants to reconsider his decision of buying the brackets and is considering making the brackets in-house. He has determined that setup costs would be \$25 in machinist time and lost production time, and 50 brackets could be produced in a day once the machine has been set up. Ross estimates that the cost (including labor time and materials) of producing one bracket would be \$14.80. The holding cost would be 10% of this cost.
(a) What is the daily demand rate?
(b) What is the optimal production quantity?
(c) How long will it take to produce the optimal quantity? How much inventory is sold during this time?
(d) If Ross uses the optimal production quantity, what would be the maximum inventory level? What would be the average inventory level? What is the annual holding cost?
(e) How many production runs would there be each year? What would be the annual setup cost?
(f) Given the optimal production run size, what is the total annual inventory cost?
(g) If the lead time is one-half day, what is the ROP?

#### Solution Preview

(a) What is the daily demand rate?
Total demand=D=2500
Number of working days=250
Daily demand rate=d=2500/250=10

(b) What is the optimal production quantity?
Daily production rate=p=50
Holding cost per unit=H=14.80*10%=\$1.48 per unit
Set up cost=S=\$25
Optimal production quantity=EPR=((2DS)/(H*(1-d/p)))^0.5=((2*2500*25)/(1.48*(1-10/50)))^0.5=324.922 ...

#### Solution Summary

Solution describes the steps to calculate EOQ, holding cost, ordering cost and ROP in the given case.

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