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# NU OPNS 430 - Chapter 6 - Inventory Analysis

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Chapter 6: Inventory Analysis6.3 Solutions to the Problem SetProblem 6.1Problem 6.2Problem 6.3c. Inventory turns = R/I, where average inventory I = Q/2 with cycle stock only.Problem 6.4Problem 6.5Problem 6.6Chapter 6 CHAPTER 6: INVENTORY ANALYSIS6.3 Solutions to the Problem SetProblem 6.1 The data in the question is: flow rate R = 50,000 parts/yr, fixed setup cost S = \$800, purchasing cost C = \$4/part, and cost of capital r = 20%/yr. Thus, the annual unit holding cost is H = rC = \$0.8/yr. The economic order quantity tells us to purchase each timea) Q = 8.0800000,5022 HRS = 10,000 units.b) Order R/Q = 5 times per year. Problem 6.2BIM Computers: Assume 8 working hours per day. - We know Q = 4 wks supply = 1,600 units, R = 400units/wk = 20,000 units/yr, purchase cost per unit C = \$1250-20% = \$1,000. Thus, holding cost H = rC = 20%/year * \$1,000 = \$200/yr. Switch over or setup cost S = \$2,000 + 1/2hr*\$1,500/day*1day/8hr = \$2,093.75. Thus, # of setups per year = R/Q = 20,000 units/yr / 1600 units/setup = 12.5 setups/yr. Thus, - Annual setup cost = (R/Q) S = 12.5setups/yr * \$2,093.75/setup = \$26,172/yr. - Annual Purchasing Cost = RC = 20,000 units/yr * \$1,000/unit = \$ 20 M/yr. - Annual Holding Cost = (Q/2)*H = 800*\$200/yr = \$160,000/yr. - Thus, total annual production and inventory cost = \$20,186,172. -- EOQ = 2 2 20000 2093.75200RSH� �= = 647 units. - #setups = R/Q = 20,000 /647 = 30.91. Thus, annual setup cost = 30.91setups/yr * \$2,093.75/setup = \$64,711/yr. - annual holding cost = (Q/2) * H = 323.5 * \$200/yr = \$64,711/yr (notice that at optimal EOQ annual holding cost equal setup costs) - annual purchasing cost remains \$20M/yr - the resulting annual savings equals \$186,172 - \$129,422 = \$56,750. Problem 6.3Victor's data: flow unit = one dress, flow rate R = 30 units/wk, purchase cost C = \$150/unit, order lead time L = 2 weeks, fixed order cost S = \$225, cost of capital r = 20%/yr. Victor currently orders ten weeks supply at a time, hence Q = 10wks * 30 units/wk = 300 units.a. Costs for Victor's current inventory management:- Annual variable ordering (purchasing) cost = RC = \$150/unit * 30 units/wk * 52 wks/yr = \$234,000/yr.Chapter 6- Annual fixed ordering (setups) cost = (# of orders/yr) S = (R/Q) S = (30*52/yr/300) \$225 = \$1,170/yr.- Annual holding cost = H (Q/2) = (rC) (Q/2) = \$30/yr * 150 = \$4,500/yr.- Total annual costs = \$239,670.b. To minimize costs, Victor should order in batches of Q* = EOQ = 30225523022 HRS = 153 units.- Thus, he should place an order for 153 units two weeks before he expects to run out. That is,whenever current inventory drops to RL = 30 units/wk * 2 wks = 60 units, which is the re-order point.- His annual cost will be RC + 30225523022 RSH+ \$234,000 = \$4,589 + \$234,000 = \$238,589.c. Inventory turns = R/I, where average inventory I = Q/2 with cycle stock only.- Current policy: turns = R/(Q/2)=2*R/Q = 2*30units/wk / 300units = 2/10week = 52*2/10 per year = 10.4 times per year.- Proposed policy: Q is roughly halved, so turns roughly double to 20.4 times per year.Problem 6.4The retailer: Current Fixed Costs S1 = \$1000. Current optimal lot size Q1 = 400. New, desired lot size Q2 = 50. We must find the fixed cost S2 at which Q2 is optimal. Since Q1 is optimal for S1, we haveQ1 = 400 = HRHRS 1000221. So, R/H = 160000/2000 = 80.Now, Q2 = 50 = HRS22, or S2 = 502 /(2*80) = 15.625. So ABC should try to reduce the fixed costs to \$15.625.Problem 6.5Major Airlines: This question illustrates the basic tradeoff between fixed and variable costs in a service industry; thus the concepts of EOQ discussed in class in the context of inventory management are much more generic.The process view here is illuminating and it goes as follows: flow unit = one flight attendant (FA). The process transforms an input (= "un-trained" FA) into an output (= "quitted" FA). The sequence of activities is: undergo training for 6 weeks, go on vacation for one week, wait in a buffer of "trained, but not assigned FA" until being assigned, serve as a FA on flights, and finally quit the job. UntrainedFATraining VacationPool oftrained FAsServe on flightsQuittedFAT = 6 wks 1 wks ? wks 2 yearsRChapter 6 The question asks for the tradeoff between training costs (higher class size is preferred) versus 'holding costs' in the buffer (smaller class size -> fewer attendants waiting in buffer is preferred).- (a)- Flow rate R = 1000 every two years = 500 attendants per year = 10 per week. - Fixed costs of training involves hiring ten instructors and support personnel for 6 weeks. Thus, fixed costs of training S = 10 * (\$220+\$80) * 6 weeks = \$18,000 per training session. - Annual holding cost is the cost incurred to hold one flow unit (FA) in the buffer for one year: H = \$500 per month * 12 = \$6,000 / person / year. - Thus, Economic Class Size (EOQ) = 54.77 or 55 per class. Thus, we should run R/Q = 500 / 55 = 9.09 classes per year - Per person variable cost of training is the stipend paid for 6 weeks of training + stipend for a week of vacation = \$500/mo,person * 7 wk * 12 mo/yr / 50 wk/yr = \$840 per person. Notice that the annual variable cost is constant \$840/person * 500 person/yr = \$420,000/yr regardless of the class size.- Total Annual Cost = Fixed Costs of Training + Variable Costs of Training + Holding Costs = (\$18,000 * 9.09) + (\$840 * 500) + (55/2)(\$6000) = \$748,636.36 per year. - Time Between starting consecutive classes (say, T) = Q/R = 5.5 weeks. Thus, we will have two classes overlap for a 1/2 week (and thus we need two sets of trainers and training class rooms). The inventory-time diagram looks as follows (assuming for simplicity that we start the training process at time 0):t (weeks)0I (in training)5.5655110t (weeks)0I (on vacation)6 755t (weeks)0I (in buffer)6 755Class 1 Class 2 Class 3Class 1 Class 2 Class 3Class 1 Class 2 Class 3Chapter 6- (b): This part of the question illustrates the following: Often, in reality, people wish to adopt policies that are simple (e.g., starting training every 6 weeks is simpler than trying to track the exact days to start training when subsequent trainings start every 5.5 weeks. But what is the implication of deviation from the optimal? In this case, quite small. This is because the optimal cost structure near the (optimal) EOQ is quite flat. Thus any solution close to optimality will suffice. - If time between classes (T = Q/R ) has to be 6 weeks, then Q = T R = 6 wks * 10 attendants/wk = 60

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