# k-server queueing system: find the expected number of busy server. Why does this simple solution not work?

Example: (Ross, p.338 #48(a)). Consider an $$n$$ -server parallel queueing system where customers arrive according to a Poisson process with rate $$\lambda,$$ where the service times are exponential random variables with rate $$\mu,$$ and where any arrival finding all servers busy immediately departs without receiving any service. If an arrival finds all servers busy,

Question: find the expected number of busy servers found by the next arrival.

It is easy to show that if we had a single server, and at $$t=0$$, it was busy, than the probability that the customers find the server busy is $$\frac{\lambda}{\mu + \lambda}$$. Hence $$T_1 := \text{# of busy servers} = 1 * \frac{\lambda}{\mu + \lambda} + 0*(1-\frac{\lambda}{\mu + \lambda}) = \frac{\lambda}{\mu + \lambda}.$$

Given that we have $$n$$ out of $$n$$ busy servers at $$t=0$$ which are independent from each other, the expected number of busy servers should be $$n* \frac{\lambda}{\mu + \lambda},$$ but in the lecture notes where I got this example, the author first finds $$T_1$$, then considers a $$k$$-server system with all servers busy and separates the cases according to whether first a customers arrives or at least one of servers finishes it's job. Then it tries to form a recursive relation between $$T_i's$$, and obtains quite a complex answers.

However, I cannot understand what is wrong with my answers. After all, we throw $$n$$ independent coins, each of which having probability $$p$$ to get heads, then we would expect to see $$p*n$$ heads. But, in this case, for some reason the auhors thinks that this logic does not work. Why?

I just made small simulation of the above system, with $$n = 10$$ and #of simulations = 100000. As far as numbers tell, the expected number of busy servers found by the customers is quite close to

$$\sum_{i=1}^n \frac{u_i}{u_i+\lambda}.$$ and this is the answer the author gives ## 1 Answer

Your approach and result are correct. You mention independence twice. The linearity of expectation does not require independence; your approach and result would be correct even without it. Since you didn’t give any details about the author’s more complicated solution, there’s not much more to say.

• but the simulation does not agree with my answer? – Our Mar 7 '20 at 17:28
• see the updated main post also. – Our Mar 7 '20 at 17:30
• @onurcanbektas: I did see the update, but didn't understand it, since it contains variables $u_i$ that you didn't introduce. – joriki Mar 7 '20 at 21:54
• @onurcanbektas: There's little point in repeating that without replying to my question above about the result of the simulation. In any case, what do you think is more likely – that there's a bug in your simulation (or perhaps you're misinterpreting its results), or that you, I, and Ross in the solution manual all made the same mistake? – joriki Mar 7 '20 at 22:45
• There was indeed a bug in the code; thanks a lot for the help. – Our Mar 8 '20 at 11:22