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Suppose the bank has three service lines. Customers arrive to the first, second, and third lines according to an exponential distribution with rates $\lambda = 3$, $\lambda = 4$, and $\lambda = 8$.

Furthermore, sometimes the service clerks must call the supervisor to approve a request. The probabilities of needing to call the supervisor at each desk are

$$p_1 = \frac{1}{4},\quad p_2 = \frac{1}{6},\quad p_3=\frac{1}{9}$$

respectively. What is the average time, in minutes, between customers being referred to the supervisor?

If I understand correctly, this is a M/G/$3$ system. Hence, the average time between referrals would be

$$ \mathbb{E}W_{\text{M}/\text{G}/3} = \left(\frac{K^2}{2} + \frac{1}{2}\right)\mathbb{E}W_{\text{M}/M/3}, $$

where $C \approx .4$, which is the standard deviation divided by the mean of $(p_1, p_2, p_3)$.

However, there is suppose to be a relatively straight forward solution, and the above doesn't seem to get me that. Any tips?

jimCao
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1 Answers1

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This can be solved using two somewhat well known results:

  1. We note that when looking only at a single service line $i$ ($i \in \{1,2,3\}$) we can use the result that the average referral time $\tau_i$ is again exponentially distributed with parameter $p_i\lambda_i$. A proof using generating functions can be found here: Using Moment Generating Functions to solve for exponential distribution

  2. The minimum of independent exponentially distributed r.vs is again exponentially distributed with parameter $\sum \lambda_i$. For a proof see here: How to prove that minimum of two exponential random variables is another exponential random variable?

Now in our case it should be clear that what we are looking for is $\mathbb{E}(\tau)$, with $\tau= \mathrm{min}\{\tau_1,\tau_2,\tau_3\}$. Hence

$ \mathbb{E}(\tau) = \frac{1}{\sum (p_i\lambda_1)}= 1/(3/4+2/3+8/9) = 83/36 $

Jay
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