Question

Consider a computer technical support center where personnel take calls and provide service. The time between calls ranges from 1 to 4 minutes. There are two technical support people – Ayşe and Burak. Ayşe is more experienced and can provide service faster than Burak. System works as follows:

• Ayşe takes the call if both technical support are idle.
• If Ayşe is busy and Burak is idle, Burak takes the call.
• If both Ayşe and Burak are busy, call waits in a queue until one of them gets idle.

Simulate the system for 8 calls. Interarrival distribution of calls and service time distributions for Ayşe and Burak are provided below.

Time between arrivals	Probability
1	0.25
2	0.40
3	0.20
4	0.15

Service time distribution for Burak	Probability
3	0.35
4	0.25
5	0.20
6	0.20

Service time distribution for Ayşe	Probability
2	0.30
3	0.28
4	0.25
5	0.17

Given that the first arrival occurs at time t = 0, create a record of hand simulation (on the empty table given below) and compute the following performance measures:

1. Average waiting time per customer
2. Average time in system
3. Average service time
4. Average interarrival time
5. Utilization of Ayşe
6. Utilization of Burak
7. Average number of customers in queue
8. Average number of customers in system

Caller Number	Random Number (Interarrival)	Interarrival Time	Arrival Time	When Ayşe becomes Available	When Burak becomes Available	Server Chosen (Ayşe/Burak)	Random Number (Service Time)	Service Time	Service Begins	Service Ends	Time in Queue	Time in System
1			0	0	0	Ayşe	57
2	61						95
3	80						33
4	58						50
5	35						69
6	25						80
7	62						49
8	43						55

In the “when Ayşe becomes available” and “when Burak becomes available” columns you can write the time when Ayşe/Burak becomes available in order to make the simulation easier for you. In the first call, since both servers are idle, Ayşe takes the call.

Answer 1

First we figure the aggregate probabilities and the interim of arbitrary numbers required for recreation.

Time between arrival:

Time b/w Arrival	Probability	Cumulative probability	Interval of random number
1	0.25	0.25	0 to less than 0.25
2	0.4	0.65	0.25 to less than 0.65
3	0.2	0.85	0.65 to less than 0.85
4	0.15	1	0.85 to less than 1

For instance the main arbitrary number for Arrival time is 26. We deviced this by 100 and get 0.26. Since 0.26 is in go 0.25 to under 0.65, the time between appearance from the above table is 3.

So also the tables for the administration times of Ayse and Burak

Ayse Service time	Probability	Cumulative probability	Interval of random number	Burak Service time	Probability	Cumulative probability	Interval of random number
2	0.3	0.3	0 to less than 0.30	3	0.35	0.35	0 to less than 0.35
3	0.28	0.58	0.30 to less than 0.58	4	0.25	0.6	0.35 to less than 0.60
4	0.25	0.83	0.58 to less than 0.83	5	0.2	0.8	0.60 to less than 0.80
5	0.17	1	0.83 to less than 1	6	0.2	1	0.80 to less than 1

(a)-(d)

Let us accept that the clock begins at 0, when the principal client shows up. We will likewise accept that since Ayse is progressively encountered, the client will go to Ayse on the off chance that he is free.

Client 1:

Shows up at 0

Since both the help are free, the administration begins at 0

Capable backings. The arbitrary number is 95. Utilizing the table for Ayse, the administration time is 5 (0.95 is in the range 0.83 to 1)

Administration finishes at 5

Capable is free at 5

Client 2:

Arbitrary number for appearance is 26, the time between appearance is 2.

Client 2 shows up at framework time of 2

Since Burak is free, the administration time start at 2.

Administration time arbitrary number is 21. The administration time is 3

Administration finishes at 5. Burak is free at 5, Ayse is free at 5

Client 3:

Arbitrary number for appearance is 98, the time between appearance is 4

Client 3 shows up at framework time of 6

Since Ayse/Burak is free, the administration time start at 6

Administration time arbitrary number is 51. The administration time is 3 (Ayse serves)

Administration finishes at 9. Capable is free at 9, Burak is free at 5

Client 4:

Arbitrary number for appearance is 90, the time between appearance is 4

Client 4 shows up at framework time of 10

Since Ayse/Burak is free, the administration time start at 10

Administration time arbitrary number is 92. The administration time is 5 (Ayse serves)

Administration finishes at 15. Capable is free at 15, Burak is free at 5

Client 5:

Arbitrary number for appearance is 26, the time between appearance is 2

Client 5 shows up at framework time of 12

Since Burak is free, the administration time start at 12

Administration time arbitrary number is 89. The administration time is 6 (Burak serves)

Administration finishes at 18. Capable is free at 15, Burak is free at 18

The accompanying table condenses the abovementioned.

Time b/w Arrival	Probability	Cumulative Probability	Interval of random number
1	0.25	0.25	0 to less than 0.25
2	0.4	0.65	0.25 to less than 0.25
3	0.2	0.85	0.65 to less than 0.25
4	0.15	1	0.85 to less than 1

Time b/w Arrival	Probability	Cumulative Probability	Interval of random number
2	0.30	0.30	0 to less than 0.30
3	0.28	0.58	0.30 to less than 0.58
4	0.25	0.83	0.58 to less than 0.83
5	0.17	1	0.83 to less than 1

Time b/w Arrival	Probability	Cumulative Probability	Interval of random number
3	0.35	0.35	0 to less than 0.35
4	0.25	0.6	0.35 to less than 0.6
5	0.2	0.8	0.6 to less than 0.8
6	0.2	1	0.80 to less than 1

Simulation:

Customer	Time b/w Arrival	Arrival Time	Service start time	Wait time	sevice Time	Completion time	Time in system	Ayse	Burak
1	0	0	0	0	5	5	5	5	0
2	2	2	2	0	3	5	3	5	5
3	4	6	6	0	3	9	3	9	5
4	4	10	10	0	5	15	5	15	5
5	2	12	12	0	6	18	6	15	18