A company has a choice of hiring one of two individuals to operate its single-channel facility. One man's times were approximately exponentially distributed with a mean rate of $6$/day, while the other man's times were distributed according to an Erlang-2 with a mean rate of $5$/day. Given the arrival rate of 4 per day, which man should be hired?
My thought: For the first man with exponential service time, we see that we would have a $D/M/1$ queue system. So $W_q = \frac{1}{u} \frac{\delta}{1-\delta}$ where $\delta$ is the solution with smallest absolute value to the equation $\delta = e^{-24(1-\delta)}$, and $\delta\in (0,1)$ (since $u = 6$ and $\beta= 4$). But this equation does not have any solution $\delta$.
For the 2nd one, we would have the system $\ D/E_2/1$. But I have no idea how to compute the expected wait time for such system.
My question: Could someone please help with this difficult problem??
HINT: This is a solution sketch...
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Assume jobs arrive to a"single channel facility" randomly, which we are going to model using an M/G/1 queue. The service distribution or worker 1 is exponentially distributed with rate of 6 per day, while worker 2's service time is Erlangian-2 distributed with rate of 5 per day. Arrivals come to both facilities at a rate of 4 per day.
Thus, the utilization for each of the facilities are:
Worker 1: utilization = $\rho_1$ = 4 / 6.
Worker 2: utilization = $\rho_2$ = 4 / 5.
For Worker 1, the service time is exponentially distributed. Thus to compute the mean response time using the PK formula, we need the 1st and 2nd moments.
Thus, we have
$E[S] = \frac{1}{\mu } = \frac{1}{6 }$
$E[S^2] = \frac{2}{\mu ^2} =\frac{2}{36}$
So, the mean response time $E[R]$ is given by:
$E[R] = E[S] + \frac{\lambda E[S^2]}{2(1-\rho)}$ = $\frac{1}{6} + \frac{4 \frac{2}{36}} {2(1 -\frac{4}{6})} = 1/2$
For Worker 2, the service time is $E_2$ distributed. Thus, for 1st and 2nd moments we have:
$E[S] = \frac{2}{\mu } = \frac{1}{5 }$
$E[S^2] = \frac{6}{\mu ^2} =\frac{3}{50}$
$E[R] = E[S] + \frac{\lambda E[S^2]}{2(1-\rho)} = \frac{4}{5}$
So, we should definitely hire Worker 1 since it takes him 1/2 day to complete the job while Worker 2 requires most of the day (4/5).