Jacksons Networks/Queing Theory - Utilisation > 1!

Webmonkey

Hey

This is a long shot but....


Right I've to solve waiting times and number of customers in each queue for a jackson network.

I've a problem that the utilisation of inner queues comes out over 1! - I know that this can happen but I thought the whole point of the jackson's theory is that all queues must be utilised under 1.

Take Question 2 on the attached exam paper for example.

I've worked out the lamda values for each of the queues, 1,2 and 3 but 1 of them come out greater than their departure rates.

The equations I get are:

(lamda1) = 1 + 0.6(lamda3)
(lamda2) = 2 + 0.4(lamda3)
(lamda3) = 0 + 0.5(lamda1) + (lamda2)

Solving these using regular simultaneous equation solving methods I get:

(lamda1) = 6.0
(lamda2) = 5.3
(lamda3) = 8.3

The problem is the depature rates of all queues are 7,7,7.

So That results in utilisation3 = 8.3/7 = 1.186!

I mean I can work away and work out waiting times and amount of customers in queue using littles theorem and so forth but surely this is all wrong considering utilisation is greater than 1.

I've also experienced it in several other exam papers, am i doing it incorrectly or have past students been doing it wrong!

Probably nobody out there but might as well try just incase!

Cheers

Edit: Change to simulataneous equation, should r2 = 2. Answers worked out on this though.

Tau

I know this theory in a completely different notation, but I ran the numbers anyway and got the same final answers as you (although you've switched a 1 for a 2 in one of your simultaneous equations above).

So the third queue is infinite (or overflowing it buffer). You can still work out utilisations for the other queues, which might be what they're looking for.

Either that or both of us are wrong - if so, I'd REALLY like to see the right solution!

Webmonkey

Tau wrote: »

I know this theory in a completely different notation, but I ran the numbers anyway and got the same final answers as you (although you've switched a 1 for a 2 in one of your simultaneous equations above).

So the third queue is infinite (or overflowing it buffer). You can still work out utilisations for the other queues, which might be what they're looking for.

Either that or both of us are wrong - if so, I'd REALLY like to see the right solution!

Cheers for that, glad to see someone out there!

Strange one alright, did you end up with exact same lamda values? - Thought the simultanous equations right up there.

Anyways, I seem to be getting similar results for other exam papers too, i mean about 2 more questions, the utilisation ends up > 1, even 3 queues in 1 of them ended up with utilisation > 1. I don't know if all these were done up wrong down through the years.

We got a new lecturer this year and he knows what he's doing and he came across a similar problem with a past exam paper in class before. He didn't do too many with us so I dont know where to get the answers!

Thanks


Edit: See what you mean now, I wrote R2 as 1, actually worked answers on paper with R2 = 2 however. Hmm...

Tau

I got 6, 5 1/3, 8 1/3 which is what I assumed you had got, you just rounded off. (but I called them alphas) (traffic constants?)

I figured you used the right equations, just posted them wrong.

You can still get an equilibrium measure for the whole network, which gives an equilibrium distribution for queues 1 and 2, but I can't see any way around queue 3 overflowing.

Webmonkey

The lamda's are mean's of a poisson distribution to the queue.

The number of customers in each queue is essentially (utilisation/(1-utilisation)). I can work these out away but they will be all wrong and negative since the utilisation is greater than 1. I can get the total waiting time as (1/(depature-lamda)) but this is also negative!

The whole thing ends up negative since its not a stable system. Surely the exam papers can't be wrong, they'd never get through external examiners..... Its really confusing and doesn't help study!

Thanks for your help

Tau

That calculation of number of people in the queue still applies for queues 1 and 2. For queue 3, it doesn't apply - the queue size is infinite, not negative.

Which is what I'd put down in an exam if I really had to put something down.

Good luck!

Webmonkey

Thanks, yeah first 2 queues ok. I've other questions where the utilisation comes out over 1 for almost all queues which is very bad indeed. On plus side, the ones that lecturer done with us all work out just these past exam papers dodgy.

Thanks for your help once again.

Webmonkey

Here I am again, This is just another example Q3 from 2007 paper in attached paper:

(lamda1) = r1 = 2
(lamda2) = r2 + P32(lamda3) = 3 + 0.4(lamda3)
(lamda3) = P13(lamda1) + P23(lamda2) = 1.4 + (lamda2)



I get results of:

(lamda1) = 2
(lamda2) = 5.9333
(lamda3) = 7.3333

: : :


utilisation1 = 2/3 = 0.67
utilisation2 = 1.1866
utilisation3 = 1.4666

Clearly the simultaneous equations are worked out correctly and the utilisations are as that quite simple. It must be the way in which the simultaneous equations set up?

It seems most of them end up like this now so surely I'm doing it wrong, even though Tau, you are arriving at similar conclusions.

The Exam paper is one from another year so surely if something was wrong on the paper it would be noticed and not happen a year later again!

This is driving me insane!

Webmonkey

By the way, Lecturer agreed papers were wrong. How the externs didn't pick up I don't know - disgrace tbh.

Anyways all is well, exam done and his ones were proper!

Cheers

Tau

Glad to hear it!

I figured we couldn't both be wrong, but then I started worrying... (I have an exam in something similar in June)