Operations Management Question

FALL 2011 2011 Dead statement October 26, Middle East Technical University Federal Cyprus Campus BUS 361 Operations Management Homework 1 Solutions 1. result figurer Company manufactures memory snatchs in lots of ten chips. From past experience, Fruit be intimates that 80% of all lots contain 10% (1 out of 10) sorry chips, 20% of all lots contain 50% (5 out of 10) spoilt chips. If a good troop (that is, 10% defective) of chips is sent on to the conterminous stage of production, bear on bes of $ metre are incurred, and if a mischievousness batch (that is, 50% defective) is sent on to the next stage of production, impact costs of $4000 are incurred.Fruit also has the alternative of reworking a batch at a cost of $1000. A reworked batch is sure to be a good batch. Alternatively, for a cost of $100, Fruit can streamlet superstar chip from sepa grazely batch in an attempt to determine whether the batch is defective. Determine how Fruit can lessen the expected total cost per batch. Expected total cost per batch = $1580. Fruit can minimize the expected total cost per batch by choosing the following decisions It should test a chip.If the tested chip is defective, Fruit should rework the batch. If the tested chip is not defective, however, Fruit should send batch on to the next stage. cover the following figure for details. Probabilities regarding testing a chip are careful as follows. D Chip is defective, D Chip is not defective, BB big Batch, GB Good Batch P(GB) = 0. 8, P(BB) = 0. 2, P(D GB) = 0. 1, P(D GB) = 0. 9, P(D BB) = 0. 5, P(D BB) = 0. 5, P(D) = (0. 8)(0. 1) + (0. 2)(0. 5) = 0. 18, P(D) = 1 P(D) = 0. 82P(GB D) = (P(DGB) P(GB) + P(DBB)P(BB)) / P(D) = 8/18 P(BB D) = 1 P(GB D) = 10/18 P(GB D) = (P(DGB) P(GB) + P(DBB)P(BB)) / P(D) = 72/82 P(BB D) = 1 P(GB D) = 10/82 1 2. A retailer of electronic products has asked a particular manufacturer to begin daily deliveries rather than on a weekly basis. Currently the manufacturer delive rs two hundred0 cases for each one Mon twenty-four hour period. The cost of each case is valued at $300. a. What is the average instrument (in units)? b. The average history (in dollars)? c. What is the inventory turnover? . What is the average inventory (in dollars) for the daily delivery pattern, presume 20 days/month? a. Average inventory = (2000 + 0) / 2 = 1000 units. b. Average inventory = 300 * 1000 = $300,000 c. Inventory turnover = salary sales / Average Inventory = 52 * 2000 / 1000 = 104 d. Average inventory = (2000/5 + 0) / 2 = 200 units Average inventory = 300 * 200 = $60,000 3. METU NCC Student Affairs officer, Sinem, is checking the accuracy of assimilator registrations each day. For each student this process takes exactly two and a half minutes.There are propagation when Sinem gets quite a backlog of files to process. She has argued for more help and another computer, hardly her manager doesnt think capacity is that stressed. Use the following entropy to dete rmine the utilization of her and her computer. She works seven and a half hours per day (she gets 30 minutes off for lunch), 5 days per week. What is the utilization of Sinem and Sinems computer? The following data are fairly typical for a week 3 Total number of files to process = 70 + cl + 130 + 120 + 160 = 630 Time it takes Sinem to process the files in each week = 630 files * 2. min/file = 1575 minutes. Total working hours available in a week = 7. 5 hours/day * 5 days = 7. 5 * 5 = 37. 5 hours = 37. 5 * 60 minutes = 2250 minutes / week Utilization = Actual working time / Time available = 1575 / 2250 = 70% 4. Consider the following three- home production cable television service with a single product that must visit station 1, 2, and 3 in sequence position 1 has 4 identical instruments with a processing time of 15 minutes per job. transmit 2 has 10 identical machines with a processing time of 30 minutes per job. Station 3 has 1 machine with a processing time of 3 minutes per job. a. What is rb (bottleneck rate) for this line? b. Can this system satisfy the daily demand of one hundred eighty units (assume 2 shifts in a day, and 4 hours in a shift)? c. What is T0 (raw processing time) for this line? d. What is W0 (critical WIP) for this line? Station 1 Production rate (jobs/min) Production rate (jobs/day) = 128 Station 2 Station 3 = 160 = 160 a. Station 1 is the bottleneck station, which has bottleneck rate, rb = 4/15. b.Because the bottleneck stations production rate of 128 is less than the daily demand of 180 units, this system cannot satisfy the daily demand. 4 c. T0 = 15 + 30 + 3 = 48 minutes. d. W0 = rb * T0 = 4/15 * 48 = 12. 8 13 units. 5. The final assembly of Noname PCs requires a total of 12 tasks. The assembly is through with(p) at the Lubbock, Texas plant using various components imported from Far East. The tasks take for the assembly operations, task times and precedence relationships between tasks are as follows Task Task Time (min)Immediat e Predecessors 1 2 2 2 2 3, 4 7 5 6, 9 8, 10 11 Positional Weight 70 58 31 27 20 29 25 18 18 17 13 7 Rank 1 2 3 4 5 6 7 8 9 10 11 12 12 6 6 2 2 12 7 5 1 4 6 7 1 2 3 5 7 4 6 8 9 10 11 12 Given that the company produces one assembled PC any 15 minutes, a. Assign tasks to workstations using the Ranked Positional Weight Algorithm. b. numerate balance delay and workload imbalance for your solution. c. Evaluate optimality of your solution (in footing of number of workstations, balance delay and workload imbalance). 5 a. Order of tasks 1, 2, 3, 6, 4, 7, 5, 8, 9, 10, 11, 12WS 1 1 15 3 WS 2 2, 3, 4 15 9 3 1 WS 3 6, 5, 9 15 3 1 0 WS 4 7, 8 15 8 3 WS 5 10, 11 15 11 5 WS 6 12 15 8 Thus, the number of workstations found by RPW heuristic is equal to 6. ? b. remnant Delay (D) = b1= 3, b2= 1, b3= 0, b4= 3, b5= 5, b6= 8 ? = 20/6 = 3. 33, Workload Imbalance (B) = v c. Lower bound on number of workstations = ? = LBD = 0, LBB =0. None of the lower edge are equal to the obtained objective value s (K*, D, B). Thus, we do not know whether the solution obtained by RPW heuristic is optimal or not. 6