Question

# A server whose service time is uniformly distributed with an interval of (10, 20) minutes. The...

A server whose service time is uniformly distributed with an interval of (10, 20) minutes. The customer inter-arrival time is also uniformly distributed with an internal (15, 25) minutes.

Determine the expected waiting time of customers of the queue?

LET US CONSIDER

C = Average number of customers in the system C

C'I = Average number of customers in the queue
W = Average time a customer spends in the systemC

W'l = Average time a customer spends in the queueC

WE HAVE

• Service time uniform distribution = (10,20) min
• Customer arrival time uniform distribution = (15,25) min

THEN

• λ= mean arrival rate

= 1/15 m

= 0.07 min

• 1/μ = mean service time

= 1/10 min

= 0.1 min

• σ = standard deviation of service time

= 20 min

Using M/G/1 queuing system,

• average number of customers in system (C) is given as,

C = ((λσ)²+(λ/µ)²) / 2(1-( λ/µ) + (λ/µ)

C= ((0.07*20)²+(0.07*0.1)²) / 2(1-(0.07*0.1) + (0.07*0.1)

C = 0.9939

= 1

• Average number of customers in queue (C’I) can calculated using Little’s formula

C’I =C-λ/µ

C'I = 1- 0.07*0.1

C’I = 0.993

• Average time customer spends in queue, W’I is given as,

W’I =C’I/ λ

W’I = 0.993 / .07

W’I = 141.85 min.