Question 1193728: The are two postal workers serving clients at the same time at the post office. Service time per client for the first postal worker is X1∼ E(λ1=0.94), second one - X2∼ E(λ2=1.0). When two new clients A and B walk into the postal office both workers are busy with a client already. Both clients want to be served by different workers. How much time on average would both clients A and B need to wait to be served, if A should be served first?
Answer by proyaop(69)   (Show Source):
You can put this solution on YOUR website! Certainly, let's calculate the average waiting time for clients A and B.
**1. Waiting Time for Client A**
* Client A waits for the first worker to finish serving the current client.
* Since the service time of the first worker follows an exponential distribution with rate λ1 = 0.94, the waiting time for client A is also exponentially distributed with the same rate.
* **Average waiting time for Client A:**
E[Waiting Time for A] = 1 / λ1 = 1 / 0.94 ≈ 1.0638
**2. Waiting Time for Client B**
* Client B waits for two events to occur:
* The first worker finishes serving client A.
* The second worker finishes serving the current client.
* The waiting time for Client B follows a minimum distribution of two independent exponential random variables.
* **Minimum of two exponential distributions:**
If X1 ~ Exp(λ1) and X2 ~ Exp(λ2), then the minimum of X1 and X2 follows an exponential distribution with rate (λ1 + λ2).
* **Average waiting time for Client B:**
E[Waiting Time for B] = 1 / (λ1 + λ2) = 1 / (0.94 + 1.0) ≈ 0.5263
**Therefore:**
* **Average waiting time for Client A:** Approximately 1.0638 units of time
* **Average waiting time for Client B:** Approximately 0.5263 units of time
**Note:**
* This analysis assumes that the service times for both workers are independent and identically distributed.
* The units of time will depend on the units used for the service rates (e.g., minutes, seconds).
Let me know if you have any other questions or would like to explore different scenarios!
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