Consider a single server system. Let the arrival distribution be uniformly distributed between 1 and 10 minutes and the service time distribution is as follows-

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written 5.6 years ago by

teamques10 ★ 68k

• modified 4.6 years ago

Service Time(min)	1	2	3	4	5	6
Probability	0.04	0.20	0.10	0.26	0.35	0.05

Develop the simulation table and analyze the system by simulating the arrival and service of 10 customers. Random digits for interarrival time and service time are as follows.

Customer	1	2	3	4	5	6	7	8	9	10
R.D for Interarrival Time	-	853	340	205	99	669	742	301	888	444
R.D for Service Time	71	59	12	88	97	66	81	35	29	91

Also calculate server utilization and maximum queue length.

computer simulation and modelling

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1 Answer

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written 5.6 years ago by

teamques10 ★ 68k

• modified 5.6 years ago

Probability of occurrence of each interarrival time = $\frac{1}{10} = 0.1$

Random Digit Assignment for Interarrival time

Interarrival time	Probability	Cummulative Prob	R.D.
1	0.100	0.100	001 - 100
2	0.100	0.200	102 - 200
3	0.100	0.300	201 - 300
4	0.100	0.400	301 - 400
5	0.100	0.500	401 - 500
6	0.100	0.600	501 - 600
7	0.100	0.700	607 - 700
8	0.100	0.800	701 - 800
9	0.100	0.900	801 - 900
10	0.100	1.000	901 - 000

Random Digit Assignment for Service Time

Service Time	Prob	Cumulative Prob	R.D.
1	0.04	0.04	01 - 04
2	0.20	0.24	05 - 24
3	0.10	0.34	25 - 34
4	0.26	0.6	35 - 60
5	0.35	0.95	61 - 95
6	0.05	1	96 - 00

Random Digit for Interarrival and Service Time

Customer No.	R.D. for Interarrival	R.D. for Service time
1	--	71
2	853	59
3	340	12
4	205	88
5	99	97
6	669	66
7	742	81
8	301	35
9	888	29
10	444	91

Simulation Table

Cust No.	R.D. Arrival	Time b/w arri	Arri Time	RD Service	Serv Time	Time Serv Begins	Time Cust in queue	Time serv ends	Time cust in s/m	Idle Time of server
1	-	-	0	71	5	0	0	5	5	0
2	853	9	9	59	4	9	0	13	4	4
3	340	4	13	12	2	13	0	15	2	0
4	205	3	16	88	5	16	0	21	5	2
5	99	1	17	97	6	21	4	27	10	0
6	669	7	24	66	5	27	3	32	8	0
7	742	8	32	81	5	32	0	37	5	0
8	301	4	36	35	4	37	1	41	5	0
9	888	9	45	29	3	45	0	48	3	4
10	444	5	50	91	5 ---- 244	50	0	55	5	2

Over a period of 55 minutes, the server was busy for 44 minutes i.e. $\frac{44}{55} \times 100 = 80 \% $ utilization.

Out of 10 customers only 3 had to wait $(\frac{3}{12} \times 10Q) 30 \%$

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