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 when I solve for this one, $\delta = 0$ is the solution with the smallest absolute value (unreasonable!)
For the 2nd one, we would have the system $\ D/E_2/1$, but I could not find the formula for computing $W_q$ of such system derived in the literature, so
My question: Could someone please help with this difficult problem? I got stuck on this problem for several hours, so any help would greatly be appreciated.
To my best knowledge, no-closed form formula exists for $W_q$ in the $D$/$E_2$/$1$ queue. To find the numerical value of $W_q$, you have to solve the equilibrium equations of a discrete-time Markov chain. The state of the Markov chain is the number of uncompleted service phases just before the arrival of a new customer (think of the Erlang-2 service time as the sum of two independent exponential phases). A simple approximation formula for $W_q$ is given on p. 425 in the book of Henk Tijms, A First Course to Stochastic Models.