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?

On average, customer spends approximately  98 min at Little’s Law Firm

Step-by-step explanation:

Given the data in the question;

Arrival rate λ = 6 per 8 hours =  6 / ( 8 × 60 )min = 6 / 480 = 0.0125 per minute

Service rate δ = 1 / 50 min = 0.02 per minute

Standard deviation σ = 20 min

Now,

Utilization rate U = Arrival rate / Service rate

U = 0.0125 / 0.02

Utilization rate = 0.625

Number of people in Queue will be;

⇒ ( (λ² × σ²) + U² ) / ( 2 × ( 1 - U )

we substitute

⇒ ( (0.0125² × 20²) + 0.625² ) / ( 2 × ( 1 - 0.625 )

⇒ ( 0.0625 + 0.390625 ) / ( 2 × 0.375 )

⇒ 0.453125 / 0.75

Number of people in Queue = 0.6042

Now

Wait in the Queue = Number of people in Queue / λ

= 0.6042 / 0.0125 = 48.336

Wait Time in Office = Wait in the Queue + ( 1 / δ )

= 48.336 + ( 1 / 0.02 )

= 48.336 + 50

Wait Time in Office = 98.336 ≈ 98 min

Therefore, On average, customer spends approximately  98 min at Little’s Law Firm

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