## 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.