Question

Calls arrive at Lynn Ann​ Fish's hotel switchboard at a rate of 2.0 per minute. The average time to handle each is 10 seconds. There is only one switchboard operator at the current time. The Poisson and negative exponential distributions appear to be relevant in this situation.1. The probability that the operator is busy = ____________ (round your response to two decimal places). 2. The average time that a caller must wait before reaching the operator = ____________ minutes (round your response to two decimal places). 3. The average number of calls waling to be answered = ___________calls (round your response to two decimal places).

Answer 1

Answer:

a) $P(Busy) =\frac{Arrival rate}{Service Rate}= \frac{1.5/min}{6/min}=0.25$

b) $Avg. Wating time= \frac{1.5}{6(6-1.5)}=0.056 minutes$

And that means 0.06 minutes for the average waiting time.

c) $Avg. Call. Wating = \frac{1.5^2}{6(6 -1.5)}=0.0833$

And that means 0.08 minutes between each call.

Step-by-step explanation:

Previous concepts

Let X the random variable of interesr. We know that $X \sim Poisson(\lambda)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

Single Channel Queuing Model. "That division of service channels happen in regards to number of servers that are present at each of the queues that are formed. Poisson distribution determines the number of arrivals on a per unit time basis, where mean arrival rate is denoted by λ".

Data given

Arrival rate = 1.5 per minute

Service Rate = 10 seconds = 6/min

Part a

For this case we want the probability that the operator is being busy. And we can find this dividing the arrival rate by the service rate like this:

$P(Busy) =\frac{Arrival rate}{Service Rate}= \frac{1.5/min}{6/min}=0.25$

Part b

In order to find the average waiting time we can use the following formula:

$Avg. Wating time = \frac{W_q}{W_r( W_r -W_q)}$

Where $W_q$ represent the arrival rate and $W_r$ the service rate. If we replace we have:

$Avg. Wating time= \frac{1.5}{6(6-1.5)}=0.056 minutes$

And that means 0.06 minutes for the average waiting time.

Part c

In order to find the average number of call on waiting we can use the following formula:

$Avg. Call. Wating = \frac{W^2_q}{W_r( W_r -W_q)}$

And replacing we got:

$Avg. Call. Wating = \frac{1.5^2}{6(6 -1.5)}=0.0833$

And that means 0.08 minutes between each call.