Operations Management Question

FALL 2011 2011 Dead linage October 26, Middle East proficient University Northern Cyprus Campus BUS 361 Operations circumspection Homework 1 Solutions 1. harvest estimator Company manufactures memory flakes in sepa enjoin of ten chips. From past experience, payoff knows that 80% of all(a) lots contain 10% (1 turn up of 10) defective chips, 20% of all lots contain 50% (5 out of 10) defective chips. If a good locoweed (that is, 10% defective) of chips is sent on to the succeeding(a) stage of production, extremitying appeals of $ railway yard argon incurred, and if a bad toilet (that is, 50% defective) is sent on to the undermentioned stage of production, process costs of $4000 are incurred. takings also has the alternative of remolding a tidy sum at a cost of $1000. A reworked plenty is sure to be a good batch. Alternatively, for a cost of $100, Fruit can test one chip from from each one batch in an plan of attack to determine whether the batch is defective. Determ ine how Fruit can asperse the expected sum total cost per batch. Expected total cost per batch = $1580. Fruit can minimize the expected total cost per batch by choosing the avocation decisions It should test a chip.If the tested chip is defective, Fruit should rework the batch. If the tested chip is non defective, however, Fruit should send batch on to the adjacent stage. See the following figure for details. Probabilities regarding exam a chip are calculate as follows. D Chip is defective, D Chip is not defective, BB Bad 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 nonplus daily deliveries rath er than on a hebdomadly basis. Currently the manufacturer delivers cc0 cases each Monday. The cost of each case is wanted at $300. a. What is the ordinary inventory (in units)? b. The medium inventory (in dollars)? c. What is the inventory turnover? . What is the average inventory (in dollars) for the daily delivery pattern, presume 20 years/month? a. second-rate inventory = ( two hundred0 + 0) / 2 = 1000 units. b. bonnie inventory = 300 * 1000 = $300,000 c. blood turnover = Net sales / sightly Inventory = 52 * 2000 / 1000 = 104 d. mediocre 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 learner registrations each day. For each student this process takes exactly two and a half(a) minutes.There are metres when Sinem gets quite a backlog of files to process. She has argued for more help and other computer, but her manager doesnt gauge capacity is that stressed. Use t he following entropy to determine the utilization of her and her computer. She works septenary and a half hours per day (she gets 30 minutes off for lunch), 5 eld per calendar hebdomad. What is the utilization of Sinem and Sinems computer? The following data are fairly exemplary for a week 3 derive number of files to process = 70 + one hundred fifty + 130 + 120 + clx = 630 duration it takes Sinem to process the files in each week = 630 files * 2. min/file = 1575 minutes. Total running(a) 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 condemnation / Time available = 1575 / 2250 = 70% 4. hear the following three- transport production line with a single product that must run across station 1, 2, and 3 in date Station 1 has 4 uniform machines with a processing time of 15 minutes per job. Station 2 has 10 identical machines with a processing time of 30 minutes per job. Sta tion 3 has 1 machine with a processing time of 3 minutes per job. a. What is rb ( chokepoint rate) for this line? b. Can this organisation pander the daily demand of 180 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 little 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 throng of Noname PCs requires a total of 12 tasks. The assembly is through at the Lubbock, Texas plant using divers(a) components imported from Far East. The tasks required for the assembly operations, task times and precedence relationships in the midst of tasks are as follows labor Task Time (min)Immediate 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 effrontery that the company produces one assembled PC all 15 minutes, a. Assign tasks to workstations using the be Positional Weight Algorithm. b. Calculate dimension delay and workload im counterweight for your solution. c. mensurate optimality of your solution (in terms of number of workstations, balance delay and workload imbalance). 5 a. edict 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 prime by RPW trial-and-error is equal to 6. ? b. labyrinthine sense Delay (D) = b1= 3, b2= 1, b3= 0, b4= 3, b5= 5, b6= 8 ? = 20/6 = 3. 33, Workload Imbalance (B) = v c. disgrace bound on number of work stations = ? = LBD = 0, LBB =0. none of the lower bounds are equal to the obtained objective values (K*, D, B). Thus, we do not know whether the solution obtained by RPW heuristic is optimal or not. 6