Question

Customers arrive at Rich Dunn’s Styling Shop at a rate of 3 per hour, distributed in a Poisson fashion. Rich’s service times follow a negative exponential distribution, and Rich can complete an average of 5 haircuts per hour. a) Find the average number of customers waiting for haircuts. b) Find the average number of customers in the shop. c) Find the average time a customer waits until it is his or her turn. d) Find the average time a customer spends in the shop. e) Find the percentage of time that Rich is busy.

Answer 1

Answer:

a) 0.9,b) 1.5,c) 0.3hrs, d) 0.5hrs,e)  60%

Step-by-step explanation:

Given Data:

rate of arrival   = 3customers/hr ;

rate of service = 5 haircuts/hr    ;

a)

Average number of customers = La = λ²/[μ(μ-λ)]

                                                           = 3²/[(5(5-3)]

Average number of customers = La = 0.9

b)

Number of customers in system = Ls = λ/(μ-λ)

                                                             = 3/(5-3)

Number of customers in system = Ls = 1.5

c)

Average waiting time = Ta = λ/[μ(μ-λ)]

                                             = 3/[(5(5-3)]

Average waiting time = Ta =0.3hrs or 18mins

d)

Average time spent by customer = Ts = 1/(μ-λ)

                                                               = 1/(5-3)

Average time spent by customer = Ts = 0.5hrs or 30mins

e)

% of time  = Tr = λ/μ

                        = 3/5

% of time  = Tr = 0.6 or 60%

Answer 2

The arrival of customers follows a Poisson distribution

• The average number of customers waiting for haircut is 0.9
• The average number of customers in the shop is 1.5
• The average time of waiting for haircut is 0.3 hour
• The average time spent in the shop is 0.5 hour
• Rich is busy 60% of the time

The given parameters are:

$\mathbf{\lambda = 3}$ --- rate of arrival

$\mathbf{\mu= 5}$ ---- rate of service

(a) Average number of customers waiting

This is calculated using:

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

So, we have:

$\mathbf{L_a = \frac{3^2}{5(5 - 3)}}$

$\mathbf{L_a = \frac{9}{5 \times 2}}$

$\mathbf{L_a = \frac{9}{10}}$

$\mathbf{L_a = 0.9}$

Hence, the average number of customers waiting for haircut is 0.9

(b) Average number of customers in the shop

This is calculated using:

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

So, we have:

$\mathbf{L_s = \frac{3}{5 - 3}}$

$\mathbf{L_s = \frac{3}{2}}$

$\mathbf{L_s = 1.5}$

Hence, the average number of customers in the shop is 1.5

(c) Average time of waiting

This is calculated using:

$\mathbf{T_a = \frac{\lambda}{\mu(\mu - \lambda)}}$

So, we have:

$\mathbf{T_a = \frac{3}{5(5 - 3)}}$

$\mathbf{T_a = \frac{3}{5 \times 2}}$

$\mathbf{T_a = \frac{3}{10}}$

$\mathbf{T_a = 0.3}$

Hence, the average time of waiting for haircut is 0.3 hour

(d) Average time spent in the shop

This is calculated using:

$\mathbf{T_s = \frac{1}{\mu - \lambda}}$

So, we have:

$\mathbf{T_s = \frac{1}{5 - 3}}$

$\mathbf{T_s = \frac{1}{2}}$

$\mathbf{T_s = 0.5}$

Hence, the average time spent in the shop is 0.5 hour

(e) Percentage of time Rich is busy

This is calculated as:

$\mathbf{T = \frac{\lambda}{\mu}}$

So, we have:

$\mathbf{T = \frac{3}{5}}$

Divide

$\mathbf{T = 0.6}$

Express as percentage

$\mathbf{T = 60\%}$

Hence, Rich is busy 60% of the time

Read more about Poisson distribution at:

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