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The task_2 shown in this figure is the check task. The management now thinks about a selective execution of this task here only 25% of cases are checked. The average service time for this new task is 6 minutes. Determine the following performance ...

(a) Occupation rate (utilization) for each resource:

  • Resource 1: (3 x 0.75 + 1 x 0.25) / 4 = 0.6875 or 68.75%
  • Resource 2: (2 x 0.75 + 1 x 0.25) / 3 = 0.75 or 75%
  • Resource 3: (4 x 0.75 + 1 x 0.25) / 5 = 0.8 or 80%

(b) Average WIP (work in progress):

  • WIP = (arrival rate x flow time)
  • Arrival rate = 240 / 480 = 0.5 customers per minute
  • Flow time = Sum of service times + Sum of waiting times
  • Flow time for Task 1 = 3 + 1 + 0 = 4 minutes
  • Flow time for Task 2 = 5 + 0.75 = 5.75 minutes
  • Flow time for Task 3 = 7 + 2 + 0.6 = 9.6 minutes
  • Flow time for Task 4 = 6 + 1.5 + 0.45 = 7.95 minutes
  • Average WIP = (0.5 x (4 + 5.75 + 9.6 + 7.95)) / 4 = 4.425

© Average flow time (throughput time):

  • Flow time for the new task (selective execution of Task 2) = (0.25 x 6) + (0.75 x 5.75) = 5.3125 minutes
  • Average flow time = (4 + 5.3125 + 9.6 + 7.95) / 4 = 6.2156 minutes

(d) Average waiting time for each task:

  • Waiting time = Flow time - Service time
  • Waiting time for Task 1 = 4 - 3 = 1 minute
  • Waiting time for Task 2 = 5.75 - 5 = 0.75 minutes
  • Waiting time for Task 3 = 9.6 - 7 = 2.6 minutes
  • Waiting time for Task 4 = 7.95 - 6 = 1.95 minutes

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