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Unformatted text preview: OPIM 101 Assignment 1 Spring 2012 Instructions: Submit your answers using the Take the Quiz link at the bottom of this page. You must submit your assignment by 9:30am, Tuesday, January 31, 2012. Late assignments receive zero credit. You must submit your own assignment. Your assignment is submitted only once you click the "submit" button! Do NOT forget to click submit. Once you click submit, you will be allowed to submit a different set of answers, if you submit before the due date. However, ALL of your previous responses will be deleted ‐ if you click RESUBMIT, then you will have to type in all of your answers again. To promote learning (and NOT the mere copying of answers), Spring 2012 OPIM 101 students are permitted to consult with one another while working on this assignment. Make sure your answers are in the correct units. Unless otherwise noted, report your answers to the 2nd decimal point (e.g., 4.04 rather than 4.1 or 4). There are 10 questions in this assignment, each worth 1 point. There is no partial credit. Q1. Butternut is a ski resort in Massachusetts. One of their triple chair lifts unloads 1296 skiers per hour at the top of the slope. (A triple chair lift can carry three passengers per chair.) The ride from the bottom to the top takes 5 minutes. How many skiers are riding on the lift at any one time? Solution: Use Little’s Law. 1296skier/hour*5/60=108 skiers Q2. Home Depot’s annual turns are 4.7, its Cost of Goods Sold (COGS) is $44.7 Billion, and its gross margin is 33%. Recall, gross margin = (Revenue – COGS) / Revenue. What is the average inventory it holds in $ Billion? Solution: $44.7 billion / 4.7 =$9.51 billion. Note that average inventory in $s is measured by the cost of goods sold, thus the gross margin does not play a role in the calculation. Q3. A company’s holding cost is 16% per year. Its annual inventory turns are 9.5. The company buys an item for $50. What is the average cost ($s), to hold this item in inventory? Solution: the item will be turned 9.5 times a year. Thus, for each turn it stays in inventory, the holding cost is 16%/9.5 of the cost of the item. Thus, the average cost to hold this item in inventory is $50 * (16%/9.5)=$0.84 Q4. Philadelphia Airport has 5 de‐icing stations. Each plane uses a single station and each station takes 11.5 minutes to de‐ice a plane. How many planes per hour can be de‐iced at the Philadelphia Airport? Solution: each station de‐ices 60min/11.5min planes per hour. 5 stations can de‐ice 60/11.5*5=26.09 planes. OPIM 101 Assignment 1 Spring 2012 (Q5‐7) Furniture Face Lift refinishes old wood furniture. Their process for refinishing chairs has 8 workers and 4 stations. Each chair starts at the Stripping station, then goes to Priming, then to Painting and finally to Inspection. Where there are multiple workers within a station, each worker works independently on his/her own chair. Assume inventory buffers are allowed between each station. Station Staffing Processing time (hours per chair per worker) Stripping 3 2.5 Priming 2 1.5 Painting 2 1.75 Inspection 1 0.8 Q5. What is the maximum number of chairs per hour that can be produced? Assume they start the day with inventory at each station to work on. Solution: the average time needed to finishone chair at each station are as follows, Capacity=staffing/activity time: Station Staffing Activity time (hours)
Capacity=staffing/activity time Stripping 3 2.5
1.2 Priming 2 1.5
1.333333333 Painting 2 1.75
1.142857143 Inspection 1 0.8
1.25 Painting is the bottle neck since it has the lowest capacity. The process capacity is the capacity of the bottleneck. Thus, the process capacity is 1.14 chairs/hour. Q6. Suppose at the start of the day there is no inventory of chairs in the shop. That is, there are no chairs within any of the stations or between them in any buffer. A truck loaded with 15 chairs arrives. How many hours will it take them to complete these 15 chairs? Solution: it takes 2.5+1.5+1.75+0.8=6.55 hours for the first chair to be produced. It takes 1/1.14 hours for each of the subsequent chairs. In total, it takes 6.55+1/1.14*14=18.8 hours to complete 15 chairs. Q7. Suppose now that each worker is trained to do all tasks and each worker works on a chair from start to finish, i.e., each worker does Stripping, Priming, Painting and Inspection. What is the maximum capacity of the process in chairs per hour? Solution: in this system, there will be no bottle neck, i.e., every worker is working at their full capacity. It takes each worker 2.5+1.5+1.75+0.8=6.55 hours to finish one chair. 3+2+2+1=8 works can complete 8/6.55=1.22chairs/hour. OPIM 101 Assignment 1 Spring 2012 (Q8‐10) The White Tooth Device Company is a manufacturer of high‐end electric toothbrushes. For each toothbrush, there is a sequence of assembly steps performed by five workers. Each worker does two tasks. Inventory buffers are allowed between workers. Worker Task Time (seconds) A T1 40 A T2 25 B T3 20 B T4 15 C T5 10 C T6 15 D T7 10 D T8 20 E T9 25 E T10
35 Q8. What is the capacity of this process (toothbrushes per minute)? Solution: the process time at each worker are: A: 65sec, B: 35sec, C: 25sec, D: 30sec, E: 60sec. Worker A is the bottle‐neck. The process capacity is 60sec/65sec=0.92 toothbrushes/min. Q9. Suppose two workers could be hired, F and G, and they take the same time to complete tasks as the current five workers. F and G can be assigned to work on the same pair of tasks as one of the current workers. For example, F could be assigned tasks T1 and T2 (just like worker A) while G is assigned T5 and T6 (just like worker C). They cannot be assigned tasks that are currently assigned to two workers. For example, F cannot be assigned to tasks T2 and T3 (because they are currently being done by workers A and B). What is the capacity of this process with workers F and G included (toothbrushes per minute)? Solution: assign the extra one by one to the bottleneck in the process. First assign worker F to the current bottleneck A. Thus, the time to complete T1 and T2 for one toothbrush at A and F is 65sec/2=32.5 sec. Now the bottleneck is E. Assign the other worker G to be working with E. The time to complete T9 and T10 for one toothbrush is 60sec/2=30sec. The current bottleneck is B with 35sec activity time. Thus, the process capacity is 60sec/35sec=1.71 toothbrushes/min. Q10. Return to the case of 5 workers. Suppose the assignment of tasks to workers can change but the sequence of tasks must remain the same, workers must be assigned to consecutive tasks and each task can be assigned to only one worker. For example, worker A could do tasks T1‐T3 (because they are consecutive) but cannot be assign T1,T2 and T4. If worker A is assigned to tasks T1‐T3, then worker B’s first task must be T4 (worker B cannot also be OPIM 101 Assignment 1 Spring 2012 assigned to task T3). Because a U‐shaped line is used, tasks T1 and T10 can actually be considered consecutive tasks – worker A could be assigned tasks T10, T1 and T2. What would the maximum capacity be after possibly reassigning tasks (toothbrushes per minute)? Solution: The total process time for one toothbrush is (40+25+20+15+10+15+10+20+25+25+35)= 215 sec. If we have 5 workers, the system can do no better than is 215/5=43sec/worker. We try to assign tasks such the process time at the slowest worker is as close as to 43sec/worker. The best we can do is the following: A: T1(40sec), B: T2 and T3(45sec), C: T4‐T7(50sec), D: T8‐T9 (45sec), T10(35sec). The bottleneck is worker C. The process capacity is 60sec/50sec=1.2 toothbrushes/hour. ...
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