Consider an $M/D/1$ queue with three types of customers. The customers arrive according to a Poisson process. Each hour, 1 type 1 customer, 2 type 2 customers and 2 type 3 customers arrive on average.
The type 1 customers have preemptive resume priority over type 2 customers. Type 2 customers have preemptive resume priority over type 3 customers. I am asked to calculate the mean sojourn time for each of the three types of customers. The service time $B_{i}$ of each customer is 10 minutes.
To this end, I would like to use the following formula: $$ E(S_{i}) = \frac{\sum_{j=1}^{i} \rho_{j} E(R_{j})}{1-(\rho_{1} + \rho_{2} + \dots + \rho_{i})(1-(\rho_{1} + \dots + \rho_{i-1}))} + \frac{E(B_{i})}{1-(\rho_{1} + \dots + \rho_{i-1} )} .$$
First, I want to calculate $E(S_{1})$. I figured that $\rho_{1} = \frac{\lambda_{1}}{\mu_{1}} = \frac{1/60}{1/10} = \frac{1}{6} $, and $E(B_{1}) = 10$.
The only thing I still need to figure out is the residual service time $E(R_{i})$ for $i = 1, 2, 3$. According to the solution, the residual service time of each customer is 5 minutes. Could you explain why this is the case?
\begin{equation} \mathbb{E}[R_i] = \frac{\mathbb{E}[B_i^2]}{2\mathbb{E}[B_i]} = \frac{100}{2 \cdot 10} = 5. \end{equation}
Each service time is exactly $10$ minutes. You arrive at any random moment in time and therefore the average time that the customer still needs to be in service is $5$ minutes.
Think of it like this: let's say you want to take a bus and you know that the bus leaves every $10$ minutes. You arrive at a random time to the bus stop. What is the expected time that you need to wait before the next bus leaves from your bus stop? Clearly this is $5$ minutes.