9. Consider the Jackson system below; the rates are given in [entities/minute] T

Question

9. Consider the Jackson system below; the rates are given in [entities/minute] The goal of this problem is to calculate various performance metrics. Populate results in the form of a table like the one shown below Server 2 .1 Hz-20 6 85 Exit 1 10 05 Server 1 H1 20 Exit 2 4 Server 3 1 ,-15 Figure 1: Jackson network Server Throughput UtilizationWait-Time Queue Length 2 (a) What is the average throughput of each queue? (b) What is the utilization of each queue? (c) What is the average wait-time of each queue? (d) What is the average queue length of each queue?

Explanation / Answer

Solution

a) Average throughput of server 1 = 10

Let arrival rates of Server 2 and 3 be x and y respectively

So, 0.85x + 0.9y = 10 --(1)

x = 0.1x + .1y + 10*0.7 --(2)

and y = 0.05x + 0.3*10 --(3)

Substituting value of y from equation (3) into equation (2), we get,

x = 0.1x + 0.1*(.05x + 3) + 7

Solving it for x, we get, x = 8.156

Substituting this value of x in either equation (1) or (3), we get, y = 3.4078

ANSWER:

throughput of server 1, 1 = 10

throughput of server 2, 2 = 8.156*0.85 = 6.9326

throughput of server 3, 3 = 3.4078*.9 = 3.0670

b)

Utilization of server 1 = 1/1 = 10/20 = 50%

Utilization of server 2 = 2/2 = 8.156/20 = 40.78%

Utilization of server 3 = 3/3 = 3.4078/15 = 22.718%

c) Average waiting time = 1/((1-1)) - 1/1

The average waiting time in queue of server 1 = 1/((1-1)) - 1/1 = 1/(20-10) - 1/20 = 0.05

The average waiting time in queue of server 2 = 0.026

The average waiting time in queue of server 3 = 0.0.178

d) Queue length = 2//(-)

Queue length of server 1 = 102/20/(20-10) = 0.5

Queue length of server 2 = 6.93262/20/(20-6.9326) = 0.183

Queue length of server 3 = 3.06702/15/(15-6.81564) = 0.076

Server Throughout Utilization wait-time Queue Length 1 10 50% 0.05 0.5 2 6.9326 40.78% 0.026 0.183 3 3.0670 22.718% 0.0178 0.076

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