Question

A new shopping mall is considering setting up an information desk manned by one employee. Based upon information obtained from similar information desks, it is believed that people will arrive at the desk at a rate of 20 per hour. It takes an average of 2 minutes to answer a question. It is assumed that the arrivals follow a Poisson distribution and answer times are exponentially distributed. (a) find the probability that the employee is idle. (b) Find the proportion of the time that the employee is busy. (c) Find the average number of people receiving and waiting to receive some information. (d) Find the average number of people waiting in line to get some information. (e) Find the average time a person seeking information spends in the system. (f) Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

Answer 1

Answer:

a) $P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

b) $p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

c) $L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

d) $L_q =\frac{20^2}{30(30-20)}=1.333 people$

e) $W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

f) $W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

Step-by-step explanation:

Notation

P represent the probability that the employee is idle

$p_x$ represent the probability that the employee is busy

$L_s$ represent the average number of people receiving and waiting to receive some information

$L_q$ represent the average number of people waiting in line to get some information

$W_s$ represent the average time a person seeking information spends in the system

$W_q$ represent the expected time a person spends just waiting in line to have a question answered

This an special case of Single channel model

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".

Part a

Find the probability that the employee is idle

The probability on this case is given by:

In order to find the mean we can do this:

$\mu = \frac{1question}{2minutes}\frac{60minutes}{1hr}=\frac{30 question}{hr}$

And in order to find the probability we can do this:

$P=1-\frac{\lambda}{\mu}=1-\frac{20}{30}=0.33$ and that represent the 33%

Part b

Find the proportion of the time that the employee is busy

This proportion is given by:

$p_x =\frac{\lambda}{\mu}=\frac{20}{30}=0.66$

Part c

Find the average number of people receiving and waiting to receive some information

In order to find this average we can use this formula:

$L_s= \frac{\lambda}{\lambda -\mu}$

And replacing we got:

$L_s =\frac{20}{30-20}=\frac{20}{10}=2 people$

Part d

Find the average number of people waiting in line to get some information.

For the number of people wiating we can us ethe following formula"

$L_q =\frac{\lambda^2}{\mu(\mu-\lambda)}$

And replacing we got this:

$L_q =\frac{20^2}{30(30-20)}=1.333 people$

Part e

Find the average time a person seeking information spends in the system

For this average we can use the following formula:

$W_s =\frac{1}{\lambda -\mu}=\frac{1}{30-20}=0.1hours$

Part f

Find the expected time a person spends just waiting in line to have a question answered (time in the queue).

For this case the waiting time to answer a question we can use this formula:

$W_q =\frac{\lambda}{\mu(\mu -\lambda)}=\frac{20}{30(30-20)}=0.0667 hours$

Answer 2

This is a single queue with poisson arrival and exponential serving situation. Hence,

a -

Probability that the employee is idle = Probability of 0 customers in the system = 1 - \lambda /\mu

Where,

\lambda = Arrival rate at poisson distribution = 20 per hour

and \mu = Serving rate at exponential distribution = 2 minutes per customer = 30 customers per hour

Hence, Probability = 1 - 20/30 = 1/3 = 0.33

b -

Proportion of the time the employee is busy = 1- Proportion of time when employee is idle = 1 - 0.33 = 0.67

c -

Average number of people recieving and waiting to recieve some information = Average number of customers in the system = - )

= 20/30-20 = 2

d -

Average number of people waiting in line = (x-1)/zK

= 400 / 30 ( 30 - 20) = 4/3 = 1.33

e -

Average time a person spends in the system = 1/(μ - Α) = 1/10 = 0.1 hour = 6 minutes

f -

Expected time a person spends waiting in line = λ/μίμ - Α) = 20 / 300 = 1/15 hours = 4 minutes