工作流结构性能分析

问题: Performance analysis I

Consider the process in figure 4.46.

(a) Determine the following performance indicators:

• Occupation rate (utilization) for each resource,

• Average WIP (work in progress),

• Average flow time (throughput time), and

• Average waiting time for each task.

Task 2 is a check task. The management thinks about a selective execution

of this task where only 25% of the cases are checked. The average

service time of this new task is 6 minutes.

(b) Determine the performance indicators again:

• Occupation rate (utilization) for each resource,
• Average WIP (work in progress),
• Average flow time (throughput time), and
• Average waiting time for each task.

解答

（a）答案如下：

其中：

• λ1 = 20，λ2 = 20

• μ1 = 60/2 = 30， μ2 = 60/2.5 = 24

• ρ1 = 20/30 = 0.67， ρ2 = 20/24 = 0.83

• L1 = 0.67/（1-0.67） = 2， L2 = 0.83/（1-0.83） = 5

• W1 = 0.67/（30-20）= 0.067hours = 4mins

  W2 = 0.83/（30-24）= 0.208hours = 12.5mins

• S1 = 1 / （30-20） = 0.1hours = 6mins

  S2 = 1 / （30-24） = 0.25hours = 15mins

（b）答案如下：

其中：

• λ1 = 20，λ2 = 20*0.25 = 5

• μ1 = 60/2 = 30， μ2 = 60/6 = 10

• ρ1 = 20/30 = 0.67， ρ2 = 5/10 = 0.50

• L1 = 0.67/（1-0.67） = 2， L2 = 0.50/（1-0.50） = 1

• W1 = 0.67/（30-20）= 0.067hours = 4mins

  W2 = 0.50/（10-5）= 0.1hours = 6mins

  Wtotal = 4 + 6*0.25 = 5.5mins

• S1 = 1 / （30-20） = 0.1hours = 6mins

  S2 = 1 / （10-5） = 0.2hours = 12mins

  Stotal = 6 + 12*0.25 = 9