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If I initially have an M/M/1/∞ queue with arrival rate of 6/hour (people arrive according to a Poisson distribution) and service rate 5/hour (average service time is exponentially distributed). But I want to reduce the arrival rate to about 4/hour and to do so I provide a limited number of seats in a waiting room, so now I have an M/M/1/N queue with arrival rate of about 4 and service rate 5/hour. In the new queueing system, when all seats are taken, new arriving customers go elsewhere.

I want to determine what N should be in order to achieve an arrival rate of about 4.

What I thought of doing is, taking different values of λ (e.g. 5,4,3) and calculating the proportion of lost customers when switching from the M/M/1/∞ to and M/M/1/N, which equals to λ times the probability that there are N customers in the system, for each value of λ. Then I would get different values of N for different λ.

Is this the right way to solve this question?

Thanks in advance!

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1 Answer 1

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Since you have an arrival rate of $6/$hour and you need an arrival rate of $4/$hour you need to compute a probability $p$ of accepting a customer to the system, such that $6p=4\implies p=2/3$. The probability $1-p=\frac13$, which is the probabitily of rejecting a customer, is equal to the probability that an arriving customer sees $N$ customers in the system (waiting seats and service), i.e. equal to $p_{N}$. Hence, you need $N$ such that $$p_{N}=1-\frac23=\frac13$$ Letting $\rho=λ/μ=6/5$ you have that $$p_{N}=\frac{ρ^{N}(1 − ρ)}{1 − ρ^{N+1}} \iff \frac13=\frac{(6/5)^{N}(-1/5)}{1-(6/5)^{N+1}}\iff N=\frac{\ln{(5/3)}}{\ln(6/5)}\approx 2.8$$ and hence $N$ must me an integer, you get $N=3$.

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  • $\begingroup$ But my logic is right, isn't it? $\endgroup$
    – annahow95
    Commented Nov 15, 2016 at 6:45
  • $\begingroup$ I think you made a mistake in your calculations, N ≈ 3. @jimmyR $\endgroup$
    – annahow95
    Commented Nov 15, 2016 at 7:13
  • $\begingroup$ @annahow95 Yes, you are right. I had a mistake in the calculations. It is indeed $N\approx 3$. Yes, your logic is right. I also edited the text of my answer to make it clearer. $\endgroup$
    – Jimmy R.
    Commented Nov 15, 2016 at 8:32

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