Advanced Systems Lab Exercise Sheet 2

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Queuing Exercises Sheet 2

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Exercise 1

For M/M/1/B the book gives this formula: on page 539.

Variant 1

Here is my attempt at a solution:

Taking from the book that for M/M/1/B (I sure hope they give us this equation on the exam :))

Calculate E[n] like this:


Variant 2

Here an other solution if you haven't got any formulas:

Considering geometric rows formula:

See also: http://de.wikipedia.org/wiki/Geometrische_Reihe

Solved after :

So:


Exercise 2

We have to show that the average response times for M/M/1 scenario are greater than the M/M/5 scenario. I used the examples 31.2 and 31.4 to figure this out.

If we choose to model the system as M/M/5, in order to get E[r] we need to use the formulas on pages 528 and 529 (everything is known).


arrival_rate = 10 j/s

service_rate = 10 j/s

system_load = 0.2


1. Calculate probability of zero jobs in system (=0.3678)

2. Calculate probability of queueing (=0.00383125)

3. Calculate E[r] for our M/M/m queue (=0.1sec)


If we choose to model the system as five separate M/M/1-s we have the following data


arrival_rate = 2 j/s (we split it 5-ways)

service_rate = 10 j/s

system_load = 0.2


Using the M/M/1 forumla for E[r] we get that the mean response time is 0.125sec. Since 0.1sec for M/M/5 is less than the 0.125sec for the M/M/1 we conclude that it's better to implement it as a M/M/5.

Exercise 3

For M/M/m the traffic intensity is calculated with:

For M/M/1 the traffic intensity is calculated with: but the arrival rate needs to be split between the servers:

The traffic intensity of both systems is then:

Exercise 4

Using Operational Law U=X*S we have X = 400/s and S = 0.002s

Then U = 0.8 (80%)

Is it really that trivial? I would say! A different approach: The service rate (mu) is observed requests per second. 2000 requests divided by 5 seconds = 400 requests/second. The you compute the utilization as percentage: observed service rate over max. service rate * 100: 400/500*100 = 80 %

Exercise 5

We also use U=X*S, where U=0.5 all the time. We have two different values for S (0.005s, 0.00125s) which gives the two different Throughput results:

X1 = 100/s and X2 = 400/s

Or even simpler: We have given the full throughput, if utilization is only .5 we can also half the throughput. That gives us a throughput range from 100 (=200/2) to 400 (=800/2) requests per second.