Please explain how the idle time is 29 seconds? And also how the efficiency for
ID: 366356 • Letter: P
Question
Please explain how the idle time is 29 seconds? And also how the efficiency for workstation 5 & 6 is 80% & 69.05%?
The Action Toy Company has decided to manufacture a new train set, the production of which is broken into six steps. The demand for the train is 4,600 units per 40-hour workweek Performance Time (seconds) 20 28 Task Predecessors Fig. 3 12 27 B, C D, E Cycle Time-Time Available per Demand Period/Units Demanded per Period Given the demand, the cycle time for the production of the new train set = 31.30' seconds (round your response to two decimal places) Minimum Number of Workstations = Sum of the Task Times / Cycle Time The theoretical minimum number of workstations 4 (round your response to the immediate higher whole number) Assignment of tasks to workstations can be done by using an assignment heuristic like longest task time and following the guidelines outlined below 1. Task time across all jobs assigned to a workstation should be s Cycle Time 2. Task's precedence requirements should be satisfied 3. (Cycle Time - Workstations Operational Time) should be minimized 4. New workstation, assign longest time activity available first whose precedence requirements are satisfied 5. A task can only be assigned to one workstation The assignment of tasks to workstations should be Workstation # Station 1 Station 2 Station 3 Station 3 Station 4 Station 5 Were you able to assign all the tasks to the theoretical minimum number of workstations? No Total Idle Time per Cycle- (Largest Assigned Task Time - Task Time for Station i) The total idle time per cycle for the process29 seconds (enter your response as a whole number) Task times Efficiency (Actual number of workstations) x (Largest assigned task time) The efficiency of the assembly line with 5 workstations-80.00% (enter your response as a percentage rounded to two decimal places) If one used 6 workstations instead of 5, the efficiency of the assembly line would be = 69.05 % (enter your response as a percentage rounded to two decimal places)Explanation / Answer
Station with maximum cycle time is station III with CT = 29 sec
Idle time of Station I = 29 - 20 = 9 sec, Station II = 29 - 28 = 1 sec,......, Station V = 29 - 27 = 2 sec
Total idle time = 9+1+0+17+2 = 29 sec
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Total task time = 116 sec
Actual no. of stations = 5
Largest assigned task time (that of station III) = 29 sec
So, efficiency = 116 / (5 x 29) = 80%
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Had it been a 6 station layout, all activities had a separate station and the sation with max cycle time would have been Station II with CT = 28 sec
Efficiency = 116 / (6 x 28) = 69.05%
Task Seconds Pred LPT layout Duration Station Cycle time Idle time A 20 - A 20 I 20 9 B 28 A B 28 II 28 1 C 14 A D 15 III 29 0 D 15 A C 14 E 12 B, C E 12 IV 12 17 F 27 D, E F 27 V 27 2 Totals 116 29