Question

Each entering customer must be served first by server 1 , then by server 2 , and finally by server \(3 .\) The amount of time it takes to be served by server \(i\) is an exponential random variable with rate \(\mu_{i}, i=1,2,3 .\) Suppose you enter the system when it contains a single customer who is being served by server \(3 .\) (a) Find the probability that server 3 will still be busy when you move over to server 2 . (b) Find the probability that server 3 will still be busy when you move over to server 3 . (c) Find the expected amount of time that you spend in the system. (Whenever you encounter a busy server, you must wait for the service in progress to end before you can enter service.) (d) Suppose that you enter the system when it contains a single customer who is being served by server \(2 .\) Find the expected amount of time that you spend in the system.

Short answer

The solutions to the problem are as follows: (a) The probability that server 3 will still be busy when you move to server 2 is: \(\frac{\mu_3-\mu_2}{\mu_1}\). (b) The probability that server 3 will still be busy when you move to server 3 is: \(\frac{\mu_1\mu_3 - \mu_2\mu_1 - \mu_1\mu_2}{\mu_1(\mu_1+\mu_2)}\). (c) The expected amount of time that you spend in the system is: \(E(T_1) + E(T_2) + E(T_3) + W_1 + W_2 + W_3 = \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3} + 0 + \frac{\mu_3-\mu_2}{\mu_1\mu_3} + \frac{\mu_1\mu_3 - \mu_2\mu_1 - \mu_1\mu_2}{\mu_1(\mu_1+\mu_2)\mu_3}\). (d) The expected amount of time that you spend in the system when the first customer is being served by server 2 is: \(E(T_1) + E(T_2) + E(T_3) + W_1 + W_2 + W_3 = \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3} + \frac{1}{\mu_2}-\frac{1}{\mu_1} + \frac{\mu_2-\mu_1}{\mu_1\mu_3} + 0\).

Step-by-step solution

(a) Probability server 3 will still be busy when moving to server 2

To find the probability that server 3 will still be busy when you move to server 2, we need to find the probability that the time taken to be served by server 1 and server 2 together is less than the time taken by the first customer to be served by server 3. Let \(T_1\) and \(T_2\) be time taken by server \(1\) and server \(2\) respectively to serve you, and let \(T_3\) be the time taken by server \(3\) to serve the first customer. These times are exponentially distributed with rates \(\mu_1, \mu_2\), and \(\mu_3\). Thus, we want to find the probability: \(P(T_1 + T_2 < T_3)\). Given that \(T_1, T_2,\) and \(T_3\) are independent, and the sum of two exponential random variables follows a Gamma distribution, we have: \[P(T_1 + T_2 < T_3) = P\left(\frac{T_1}{T_1+T_2} > \frac{T_1+T_2-T_3}{T_1+T_2}\right)\] Using memoryless property of exponential random variables, we get: \[P(T_1 + T_2 < T_3) = P\left(T_1 > T_1 + T_2 - T_3 \right) = P\left(\frac{\mu_1}{\mu_1+\mu_2} > \frac{\mu_1+\mu_2-\mu_3}{\mu_1+\mu_2}\right)\] Therefore, the required probability is: \[P(T_1 + T_2 < T_3) = \frac{\mu_3-\mu_2}{\mu_1}\]

(b) Probability server 3 will still be busy when moving to server 3

To find the probability that server 3 will still be busy when you move to server 3, we need to find the probability that the time taken to be served by server 1, server 2, and server 3 together is less than the time taken by the first customer to be served by server 3 and server 2. Let \(T'_1, T'_2, T'_3\) be the remaining time taken by server \(1, 2\) and \(3\) respectively to serve you after server \(3\) finishes its service to the first customer. The total time taken for you to reach server \(3\) is: \(T'_1+T'_2+T_2'\). We want to find the probability: \(P(T'_1 + T'_2 + T_2'< T_3)\). To simplify and condition on the remaining time at server \(1\), we have: \[P(T'_1 + T'_2 + T_2' < T_3) = P\left(\frac{T'_1}{T_1+T_2} > \frac{T_1+T_2-T_3-T'_2-T_2'}{T_1+T_2}\right)\] Using the memoryless property of exponential random variables, we get: \[P(T'_1 + T'_2 + T_2' < T_3) = P\left(T_1 > T_1 + T_2 - T_3 - T'_2 - T_2' \right) = P\left(\frac{\mu_1}{\mu_1+\mu_2} > \frac{\mu_1+\mu_2-\mu_3-T'_2-T_2'}{\mu_1+\mu_2}\right)\] Therefore, the required probability is: \[P(T'_1 + T'_2 + T_2' < T_3) = \frac{\mu_1\mu_3 - \mu_2\mu_1 - \mu_1\mu_2}{\mu_1(\mu_1+\mu_2)}\]

(c) Expected time spent in the system

According to question(c) , you need to find the expected amount of time that you spend in the system which includes waiting time at server 1, server 2, and server 3. Denote the expected waiting times at each server as \(W_1, W_2, W_3\), and let \(E(T_i)\) be the expected service time for server \(i\). We need to find \(E(T_1) + E(T_2) + E(T_3) + W_1 + W_2 + W_3\). Since the servers are exponentially distributed, we know \[E(T_i) = \frac{1}{\mu_i}\] We can find the expected waiting time from the probabilities calculated earlier. For instance, server 2 waits for server 3 with probability \(P(T_1+T_2\,<\,T_3)\). Thus, \[ W_1 = 0,\; W_2 = \frac{\mu_3-\mu_2}{\mu_1} \cdot \frac{1}{\mu_3},\; W_3 = \frac{\mu_1\mu_3 - \mu_2\mu_1 - \mu_1\mu_2}{\mu_1(\mu_1+\mu_2)} \cdot \frac{1}{\mu_3}\] So the expected time that you spend in the system is: \[ E(T_1) + E(T_2) + E(T_3) + W_1 + W_2 + W_3 = \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3} + 0 + \frac{\mu_3-\mu_2}{\mu_1\mu_3} + \frac{\mu_1\mu_3 - \mu_2\mu_1 - \mu_1\mu_2}{\mu_1(\mu_1+\mu_2)\mu_3}\]

(d) Expected time spent in the system when first customer is at server 2

According to question(d) now you need to find the expected amount of time that you spend in the system, given that the first customer is served by server 2. Similar to question (c), you need to find the waiting times \(W_1, W_2, W_3\). This time, \(W_1\) is determined by the waiting time for Server \(1\) to be free. \[W_1 = \frac{1}{\mu_2}-\frac{1}{\mu_1}\] Since the customer ahead of you now only has to complete the service with server 2, \begin{align*} W_2 &= P(T_1<T_2')\cdot \frac{1}{\mu_3} \\ &= \frac{\mu_2-\mu_1}{\mu_1}\cdot \frac{1}{\mu_3} \end{align*} And \(W_3\) will be zero because there's no customer ahead of you when you arrive at server 3. Thus, we get the expected time that you spend in the system if the first customer is being served at server 2: \[ E(T_1) + E(T_2) + E(T_3) + W_1 + W_2 + W_3 = \frac{1}{\mu_1} + \frac{1}{\mu_2} + \frac{1}{\mu_3} + \frac{1}{\mu_2}-\frac{1}{\mu_1} + \frac{\mu_2-\mu_1}{\mu_1\mu_3} + 0\]