Question

Customers can be served by any of three servers, where the service times of server \(i\) are exponentially distributed with rate \(\mu_{i}, i=1,2,3 .\) Whenever a server becomes free, the customer who has been waiting the longest begins service with that server. (a) If you arrive to find all three servers busy and no one waiting, find the expected time until you depart the system. (b) If you arrive to find all three servers busy and one person waiting, find the expected time until you depart the system.

Short answer

(a) The expected time until a customer departs the system when all three servers are busy and no one is waiting: \(E(T_A) = \frac{1}{\mu_1+\mu_2+\mu_3}\) (b) The expected time until a customer departs the system when all three servers are busy and there is one person waiting: \(E(T_B) = \frac{2}{\mu_1+\mu_2+\mu_3}\)

Step-by-step solution

Define the Variables

Let \(X_1, X_2, X_3\) be independent exponential random variables representing the remaining service time for each of the three servers with rates \(\mu_1, \mu_2,\) and \(\mu_3\), respectively. Let \(T_A\) and \(T_B\) represent the expected time until a customer departs the system in case (a) and case (b), respectively.

Conditional Expectations for Case (a)

In this case, a customer leaves the system as soon as one of the servers become free. The customer's waiting time \(T_A\) is equal to the minimum of the remaining service times for all three servers: \(T_A = \min(X_1, X_2, X_3)\). We will use the properties of exponential random variables and conditional expectations to find the expected time until departure \(E(T_A)\). Using the conditional expectation formula, we can write: \(E(T_A) = E[T_A | X_i = \min(X_1, X_2, X_3)] P(X_i = \min(X_1, X_2, X_3)) + E[T_A | X_j = \min(X_1, X_2, X_3)] P(X_j = \min(X_1, X_2, X_3)) + E[T_A | X_k = \min(X_1, X_2, X_3)] P(X_k = \min(X_1, X_2, X_3)),\) where \(\{i, j, k\} = \{1, 2, 3\}\).

Calculate the Probabilities and Expectations

We will now calculate the probabilities and expectations for case (a): 1. The probability that server \(i\) finishes first: \(P(X_i = \min(X_1, X_2, X_3)) = \frac{\mu_i}{\mu_1 + \mu_2 + \mu_3}\) for \(i = 1, 2, 3\). 2. The expectation of waiting time, given that server \(i\) finishes first: \(E[T_A | X_i = \min(X_1, X_2, X_3)] = \frac{1}{\mu_i}\) for \(i = 1, 2, 3\). Now, substitute these probabilities and expectations into the conditional expectation formula: \(E(T_A) = (\frac{1}{\mu_1}\times\frac{\mu_1}{\mu_1+\mu_2+\mu_3}) + (\frac{1}{\mu_2}\times\frac{\mu_2}{\mu_1+\mu_2+\mu_3}) + (\frac{1}{\mu_3}\times\frac{\mu_3}{\mu_1+\mu_2+\mu_3})\) Simplifying the expression, we get: \(E(T_A) = \frac{1}{\mu_1+\mu_2+\mu_3}\)

Calculate Expected Time for Case (b)

For part (b), there is one customer waiting when the new customer joins the line. The new customer has to wait for one other customer to finish before they can leave. We can find the expected time for this case by adding the expected service time of the first customer \(E(T_A)\) (calculated in Step 3) to the expected service time of the new customer joining the line, which is also \(E(T_A)\): \(T_B = T_A + T_A\) Therefore, \(E(T_B) = 2E(T_A) = \frac{2}{\mu_1+\mu_2+\mu_3}\)

Final Answer

(a) The expected time until a customer departs the system when all three servers are busy and no one is waiting: \(E(T_A) = \frac{1}{\mu_1+\mu_2+\mu_3}\) (b) The expected time until a customer departs the system when all three servers are busy and there is one person waiting: \(E(T_B) = \frac{2}{\mu_1+\mu_2+\mu_3}\)