HW03_key-Sp10

# HW03_key-Sp10 - MGMT 36100 HW 03 40 Points Due March 1 2010...

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HW03_key Sp10.docx Page 1 MGMT 36100: HW 03 40 Points Due: March 1, 2010 Question 1: (12 points – 4 + 4 + 4) - Solve problem 8 as follows: Part a: See the textbook; however round off the optimal lot size to the nearest multiple of 100. D=27,000 [units/month], Q=4,000 [units], H=\$.18/unit/month, and S=\$60/order. From the EOQ formula we obtain Q 0 = 2DS/H = 4242 [units] 4200 [units]. AIC(Q 0 =4200) = (HQ 0 /2 + SD/Q 0 ) * 12 = \$9164.57/year. AIC(Q=4000) = (HQ/2 + SD/Q ) * 12 = \$9180.00/year. The difference between AIC(Q 0 =4200) and AIC(Q=4000) is the penalty incurred by the company by ordering at a quantity lower than the optimal quantity (EOQ). So, Penalty = AIC(Q 0 =4200) - AIC(Q=4000) = \$9180.00-\$9164.57= \$15.43/year or \$1.29/month. Part b: See the textbook. If they order 10 times a month, then the order quantity should be equal to Q 1 = 27,000/10=2,700 [units]. Let S 1 be the new ordering cost that enables the manager to justify ordering every other day. There are different interpretations of this problem. Here is the most logical interpretation: At what value of S will the lot size 2700 be optimal? Q 0 = 2DS 1 /H = 2700 . This will result in S 1 = \$24.3 Part c: Do not solve part C from the textbook. Instead, draw the graph of annual cost for holding, ordering and

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