Question
Assign the tasks shown in the accompanying precedence diagram (times are in minutes) to workstations using the following rules:
a. In order of most following tasks. Tiebreaker: greatest positional weight.
b.In order of greatest positional weight.
Explanation / Answer
Review Problems and Solutions for Chapter 6: Process Selection and Facility Layout For the following three problems (1, 2, 7), we assume that parallel workstations are not allowed. 1. An assembly line with 17 tasks is to be balanced. The longest task is 2.4 minutes, and the total time for all tasks is 18 minutes. The line will operate for 450 minutes per day. a. What are the minimum and maximum cycle times? b. What range of output is theoretically possible for the line? c. What is the minimum number of workstations needed if the maximum output rate is to be sought? d. What cycle time will provide an output rate of 125 units per day? e. What output potential will result if the cycle time is (1) 9 minutes? (2) 15 minutes? Solution: OT = 450 minutes a. Minimum cycle time = length of longest task, which is 2.4 minutes. Maximum cycle time = S task times = 18 minutes. b. Range of output: 25 units 18 @18min. : 450 187.5 units 2.4 @2.4min.: 450 = = c. 7.5, which rounds to 8 450 187.5(18) OT N Dx t = = S = d. 3.6minutes per cycle 125 Solving for CT, CT 450 CT Output = OT = = e. Potential output: (1) 50 units 9 450 CT CT = 9min.: OT = = (2) 30 units 15 CT =15min.: 450 = 2. A manager wants to assign tasks to workstations as efficiently as possible, and achieve an hourly output of 33? units. Assume the shop works a 60-minute hour. Assign the tasks shown in the accompanying precedence diagram (times are in minutes) to workstations using the following rules: a. In order of most following tasks. Tiebreaker: greatest positional weight. b. In order of greatest positional weight. c. What is the efficiency? Solution: Desired output = 33.33 units per hour Operating time = 60 minutes per hour 1.80minutes per unit 33.33 units per hour 60minutes per hour Desired output CT = Operating time = = a. Task Number of following tasks Positional Weight A 7 6 B 6 4.6 C 2 1.6 D 2 2.2 E 2 2.3 F 1 1.0 G 1 1.5 H 0 0.5 Assembly Line Balancing Table (CT = 1.8) Work Station Task Task Time Time Remaining Feasible tasks Remaining I A 1.4 0.4 – II B 0.5 1.3 C, D, E E 0.8 0.5 – D 0.7 1.1 C C 0.6 0.5 F III F 0.5 0 – IV G 1.0 0.8 H H 0.5 0.3 – b. Assembly Line Balancing Table (CT = 1.8) Work Station Task Task Time Time Remaining Feasible tasks Remaining I A 1.4 0.4 – II B 0.5 1.3 C, D, E E 0.8 0.5 – D 0.7 1.1 C C 0.6 0.5 F III F 0.5 0 – IV G 1.0 0.8 H H 0.5 0.3 – c. 83.3% 7.2 6.0 CTx no.of stations Efficiency = Total time = = 7. For the set of tasks given below, do the following: a. Develop the precedence diagram. b. Determine the minimum and maximum cycle times in seconds for a desired output of 500 units in a 7-hour day. Why might a manager use a cycle time of 50 seconds? c. Determine the minimum number of workstations for output of 500 units per day. d. Balance the line using the largest positional weight heuristic. Break ties with the most following tasks heuristic. Use a cycle time of 50 seconds. e. Calculate the percentage idle time for the line. Solution: a. b. = = = 500 7(60) D CT OT .84 minutes = 50.4 seconds (maximum cycle time) Minimum cycle time = maximum task time = 45 seconds (results in 560 units of production) c. 3.83 or 4 stations 50.4 193 CT N t = = S = d. Task Number of followers *PW A 6 106 B 5 61 C 4 50 D 4 106 E 3 56 F 2 30 G 2 31 H 2 29 I 1 19 J 0 10 *Positional weight CT = 50 seconds Work Station Task Task Time Time Remaining Feasible tasks Remaining I A 45 5 – III D 50 – – III B 11 39 C, E a b c d e h g f i j E 26 13 C, F C 9 4 – G 12 38 H, F F 11 27 H H 10 17 I IV I 9 8 – V J 10 40 – e. 22.8% (50)(5) I =1- 193 = Optional Problem (Parallel workstations are allowed). Consider the following precedence diagram: Suppose that OT (operating time per day) = 480 minutes, and our target output rate D is 960 for each day. Balance the line. Solution: CT = OT/D = 0.5 minutes. Suppose that we use the number of followers to break ties, then Work Station Task Task Time Time Remaining Feasible tasks Remaining I a 0.1 0.4 – II 0.7 min (task time for c) >0.5 min, new workstations are needed II & III c 0.7 0.3 - IV 1.0 min (task time for b) >0.5 min, new workstations are needed IV & V b 1.0 0 - VI d 0.5 0 - VII e 0.2 0.3 - Notice that we can use smaller number of workstations. In fact, we can use two workstations (instead of three) on tasks a and c, i.e., each of these two work station works on tasks a and c. In other words, we can treat tasks a and c as one task, with a task cycle time = 0.8 min < (0.5 min x 2 = 1.0 min). Then it is clear that two workstations are enough for the task (a and c together).