Performance analysis I
Consider the process in figure 4.46.
Q 1
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.
Q 2
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.
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.
A 1
Occupation rate
For task 1
$20$ tasks arrive per hour;
$60/2 = 30$ tasks are dealt per hour;
occupation rate: $20/30= .67$.
For task 2
$20$ tasks arrive per hour;
$60/2.5 = 24$ tasks are dealt per hour;
occupation rate: $20/24= .83$.
Average WIP
$\rho$ task 1/ $(1-\rho)$ task 1 + $\rho$ task 2/ $(1-\rho)$ task 2 = 7
Average flow time
For task 1
$2/30*60+2=6$ min
For task 2
$5/24*60+2.5=15$ min
So the throughput time is $21$ min.
Average waiting time
Task 1= $4$ min ; Task 2= $12.5$ min.
A 2
Occupation rate
For task 1
$20$ tasks arrive per hour;
$60/2 = 30$ tasks are dealt per hour;
occupation rate: $20/30= .67$.
For task 2
$20$ tasks arrive per hour;
$60/6 = 10$ tasks are dealt per hour;
occupation rate: $10/20= .5$.
Average WIP
$\rho$ task 1/ $(1-\rho)$ task 1 + $\rho$ task 2/ $(1-\rho)$ task 2 = 3
Average flow time
$(60/10+6)*.25+6=9$ min
So the throughput time is $9$ min.
Average waiting time
Task 1= $4$ min ; Task 2= $6$ min.
