Little’s Law Firm has just one lawyer. Customers arrive randomly at an average rate of 6 per 8 hour workday. Service times have a mean of 50 minutes and a standard deviation of 20 minutes. How long does a customer spend at Little’s Law Firm on average?
a. Approximately one hour
b. Approximately 98 minutes
c. Approximately 75%
d. Approximately 48 minutes

Respuesta :

Answer:

option (b) Approximately 98 minutes

Explanation:

Given:

Average arrival rate, λ = 6 per 8 hour or [tex]\frac{\textup{6}}{\textup{8}\times60}[/tex]  = 0.0125 / minute

Service Rate, μ = [tex]\frac{\textup{1}}{\textup{50}}[/tex]  = 0.02/min

Standard Deviation = 20 minutes

Now,.

Utilization Rate, ρ = [tex]\frac{\lambda}{\mu}[/tex]

or

= [tex]\frac{\textup{0.0125}}{0.02}[/tex]

= 0.625

and,

Number of people in Queue = [tex]\frac{\lambda^2\times\sigma^2+\rho^2}{2\times(1-\rho)}[/tex]

or

= [tex]\frac{0.0125^2\times20^2+0.625^2}{2\times(1-0.625)}[/tex]  

= 0.6042

and,

Waiting in the Queue = [tex]\frac{\textup{Number of people in Queue}}{\lambda}[/tex]

=  [tex]\frac{\textup{0.6042}}{0.0125}[/tex]

= 48.33  minutes

Thus,

Waiting Time in Office = Wait in the Queue + [tex]\frac{\textup{1}}{\textup{Service rate}}[/tex]

= 48.33  minutes + [tex]\frac{\textup{1}}{\textup{0.02}}[/tex]

= 48.33 + 50

= 98.33 minutes

hence, the answer is option (b) Approximately 98 minutes

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