THIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017 IE 1055/2025 Homework 2 Due Jan. 18, 2017 1. In a just-in-time and total quality management environment, defects are often reduced or eliminated. Describe at least five strategies to reduce or eliminate defects and explain the impact of reducing defects on facilities design. (15 pts) Answer (20 pts) 1. Employee certification 2. Better process/operation documentation with emphasis on quality concerns 3. Mistake proofing (poka yokes) 4. Preventive/predictive maintenance practices 5. Autonomous maintenance (cleaning, lubrication, minor adjustments) Scrap consumes resources and requires additional space. Eliminating scrap (by reducing defects) reduces space and flow requirements and impacts activity interrelationships. 2. You are a strategic planner for Ford Motor Company. Your current assignment is to examine whether or not the hood and trunk latches should be made in-house instead of procuring them from outside vendors. The latches are basically a small metal assembly consisting of several stamped sheet metal parts and some additional components (springs, bolts, etc.) (15 pts) - Discuss some of the reasons that it would be a good idea to make the latches in-house. - Discuss some of the reasons that it would not be a good idea to make the latches in-house. - What are some of the key questions that would have to be answered before a decision could be made? Answer (15 pts) - Discuss some of the reasons that it would be a good idea to make the latches in-house. o The company has the capability to produce the latches with reasonable quality and costs. o It is cheaper to produce the latches in-house. o The company has the required equipment to produce the latches. o The latches are required in high volume and demand is stable for the long term. o Etc. - Discuss some of the reasons that it would not be a good idea to make the latches in-house. o The company is unable to produce the latches. o It is cheaper to buy the latches. o The latches are only needed over the short-term.THIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017 o Demand for the latches is low. o Latches are probably not an important strategic asset in vehicle design. o Etc. - What are some of the key questions that would have to be answered before a decision could be made? o Can the latches be purchased? o Can we produce the latches? o Is it cheaper for us to produce than to buy? o Is the capital available to allow us to produce the latches?THIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017 3. 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 you 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. - What is the minimum number of 107s you need to schedule to have a 0.90 probability that you get 20 good 107s? (20 pts) Answer (25 pts) Since the parts are produced independent of each other, and the probability of an individual casting being good is equal to 0.85, the number of good parts follows a binomial distribution with parameter 0.85. We need the probability that we get 20 good parts to be 0.90, thus it is, 0.90),85.0,,19(1)15.085.019...15.085.0115.085.00(1...15.085.02115.085.020parts) X among parts good 20least at are (there1919110021212020TRUEXBINOMDISTXXXXXpXXXXX We use the BINOMDIST function in Excel to get the following table, X BINOMDIST(19, X, 0.85, TRUE) 1-BINOMDIST(19, X, 0.85, TRUE) 20 0.961 0.039 21 0.845 0.155 22 0.662 0.338 23 0.460 0.540 24 0.287 0.713 25 0.162 0.838 26 0.083 0.917 27 0.040 0.960 28 0.018 0.982 29 0.007 0.993THIS SOLUTION MAY ONLY BE USED FOR IE 1055/2025 SPRING 2017 30 0.003 0.997 31 0.001 0.999 32 0.0004 0.9996 This function BINOMDIST(19, X, 0.85, TRUE) will let me know the probability that there are 19 successes or less in X trials (where X is the left column in the table) if the probability of success in each trial is 0.85. The idea is that I am finding the probability that 19 or fewer are good. One minus that is the probability that at least 20 are good. Therefore, to have a .90 chance of having 20 good I need 26 parts. Alternative way: Since the function BINOMDIST in excel is the probability that there are n successes or less in X trials, we can think of the bad parts as successes with probability 0.15. So we need the bad parts to be X-20 or less. Then we use BINOMDIST(X-20, X, 0.1, TRUE) to reach the same conclusion as above. X BINOMDIST(X-20, X, 0.15, TRUE) 20 0.039 21 0.155 22 0.338 23 0.540 24 0.713 25 0.838 26 0.917 27 0.960 28 0.982 29 0.993 30 0.997 31 0.999 32