0
1.4kviews
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-
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.

1
35views

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 $(\frac{3}{12} \times 10Q) 30 \%$