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Pitt IE 1055 - HW3-17 Solution

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THIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017. IE 1055/2025 Homework 3 Due Jan. 25, 2017 1. Problem 2.25 from the text. (This problem was on the last assignment.) Assume that part A must pass through machines 1 and 2 to be completed. For example, machine 1 drills a hole and machine 2 puts a taper in the hole made at machine 1. Similarly, part B must pass through both machines 3 and 4 to be completed. Please note that on problem 2.25 that "the desired output 06" means the output of station 6 and that "input 11 to machine 1" is a typo and means the input to machine 1 (it ought to read "input to machine 1"). (20 pts) a. Determine an equation for the required input for both machines 1 and 3 (not just machine 1 as stated in the problem.) b. Determine the required input for both machines 1 and 3 if the following scrap percentages apply: d1=.10, d2=.05, d3=.05, d4=.02, d5=.04, d6=.01. c. If the desired output of parts at station 6 is 70 parts per hour determine the input rate of parts required at stations 1 and 3 for the problem variations described in parts b and c of this problem. d. How does the answer to part b change if 75% of the defects at machine 1 and 99% of the defects at machine 5 can be reworked? Give both a qualitative and quantitative answer – the quantitative answer will relate I1 to O6 for example but will not be a number such as start with 100 parts because no numeric value is specified for O6. Answer (20 pts) a) If O6 is the desired output for part C, then the required inputs I1 to machine 1 and I3 to machine 3 are )1)(1)(1)(1(652161PPPPOI )1)(1)(1)(1(654363PPPPOI Part A Part B M 1M 3M 4M 6M 5M 2Part C P1 P6 P5 P4 P3 P2THIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017. b) Given p1 = 0.10, p2= 0.05, p3 = 0.05, p4 = 0.02, p5 = 0.04, p6 = 0.01. If O6 is the desired output for part C, then the required inputs I1 to machine 1 and I3 to machine 3 are 666652161*230.1813.0)01.01)(04.01)(05.01)(10.01()1)(1)(1)(1(OOOppppOI 666654363130.1885.0)01.01)(04.01)(02.01)(05.01()1)(1)(1)(1(OOOppppOI c) If the desired output for part C is 70, then the required inputs to machine 1 and 3 are 8714.86)01.01)(04.01)(05.01)(10.01(70)1)(1)(1)(1(652161ppppOI Part A Part B M 1M 3M 4M 6M 5M 2Part C0.90 0.990.960.98 0.95 0.95THIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017. 801.79)01.01)(04.01)(02.01)(05.01(70)1)(1)(1)(1(654363ppppOI Note that we need to round up to the next greatest integer in order to insure that we meet the required throughput. d) Given q1 = 0.75, q5 = 0.99 If O6 is the desired output for part C, then the required inputs I1 to machine 1 and I3 to machine 3 are 6666521551161092.1813.0*888.01)(04.01)(05.01)(10.01(*04.01)(75.0*10.01()1)(1)(1)(1()1)(1(OOOpppppqpqOI Part A Part B M 1M 3M 4M 6M 5M 2Part C0.90 0.990.960.98 0.95 0.95 q5 q1 66665435563085.1885.0*960.0)01.01)(04.01)(02.01)(05.01()99.0*04.01()1)(1)(1)(1()1(OOOpppppqOITHIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017. 2. 2.23 except change the quantity required to 300 non-defective parts. (10 pts) Answer (10 pts) S= 15 min. H= (8hrs./shift)*(60min/hr) = 480 min/shift E=15min./20min. = 0.75 Q 3001 0.2 375units R = 7/8 = 0.875 F (15min/unit)(375 units/shift)(0.75)(480min/shift)(7/8)17.86 => 18 Machines You could also choose to use an S value of 20 and then drop E from the denominator. 3. Rework problem 2.23 (still change the quantity required to 300 non-defective parts) if one half of the defective items can be reworked. Assume that the reworked items also have an 80% probability of being non-defective and that the parts can be reworked indefinitely (thus, the reworked parts have the same probability of being good as the new parts do - these assumptions are similar to what we assumed in the class discussion.) How does your answer compare to the answer for problem 2.23 from problem 2? Discuss any implications. (10 pts) Answer (10 pts) The equipment fraction does not change because the processing rate is still 375 units. The reason is that with rework, the number of parts to start with should be pqpOI1)1( but the processing rate PR I1 qpO(1qp)1 p1 qpO1 p 375 (since the last expression is exactly what we had before). It is important to note that now there is less scrap and therefore the input rate would be lower too. 4. You work for SteelPro a fabrication company that specializes in machining specialty alloys. One of your best customers is Bombardier Transportation. Bombardier has asked you to fill an order for 20 of their part 107. Bombardier will pay you $500 for each good part 107 but will only accept exactly 20 good ones – no more, no less. The raw materials required to make a 107 cost $150. Unfortunately you have yield loss issues related to manufacturing the 107. Assume that the 107s are produced independently of each other with the probability of an individual 107 being good being equal to 0.85. Defective 107s can be scrapped for $50. There is also the technical challenge that the way youTHIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017. process the 107s is that you machine the entire batch (whatever batch size you choose) and then you heat treat and further process them and only after the heat treating and further processing do the defects become detectable. Thus, you must determine the batch size to run prior to knowing what your precise yield will be. Assume you only have time to run one batch. (20 pts) i. Given the economics stated in the problem, what is the best batch size to run for an order of 20 107s to maximize the expected profit? ii. Bombardier has indicated that they will order part 107, or a part very similar to it made by similar processes, three times per year. Suppose you could invest $1500 to improve the yield for the 107 and related parts


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Pitt IE 1055 - HW3-17 Solution

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