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## Q. 12.EX.8

A PRODUCTION ORDER QUANTITY MODEL. Nathan Manufacturing, Inc., makes and sells specialty hubcaps for the retail automobile aftermarket. Nathan’s forecast for its wire-wheel hubcap is 1,000 units next year, with an average daily demand of 4 units. However, the production process is most efficient at 8 units per day. So the company produces 8 per day but uses only 4 per day. The company wants to solve for the optimum number of units per order. ( Note: This plant schedules production of this hubcap only as needed, during the 250 days per year the shop operates.)
APPROACH $\blacktriangleright$ Gather the cost data and apply Equation (12-7):
Annual demand = D = 1,000 units
Setup costs = S = $10 Holding cost = H =$0.50 per unit per year
Daily production rate = p = 8 units daily
Daily demand rate = d = 4 units daily

$Q^*_p = \sqrt{\frac{2DS}{H[1 - (d/p)]}}$             (12-7)

## Verified Solution

SOLUTION $\blacktriangleright$

$Q^*_p = \sqrt{\frac{2DS}{H[1 – (d/p)]}}$

$Q^*_p = \sqrt{\frac{2(1,000)(10)}{0.50[1 – (4/8)]}} = \sqrt{\frac{20,000}{0.50(1/2)}} = \sqrt{80,000} = 282.8$ hubcaps, or 283 hubcaps

INSIGHT $\blacktriangleright$ The difference between the production order quantity model and the basic EOQ model is that the effective annual holding cost per unit is reduced in the production order quantity model because the entire order does not arrive at once.

LEARNING EXERCISE $\blacktriangleright$ If Nathan can increase its daily production rate from 8 to 10, how does $Q^*_p$ change? [Answer: $Q^*_p$ = 258.]
RELATED PROBLEMS $\blacktriangleright$ 12.18, 12.19, 12.20, 12.30 (12.37 is available in MyOMLab)
EXCEL OM Data File Ch12Ex8.xls can be found in MyOMLab.
ACTIVE MODEL 12.2 This example is further illustrated in Active Model 12.2 in MyOMLab.