Question

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?

Answer #1

**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

= mean arrival rate**λ**

= 1/15 m

*= 0.07
min*

= mean service time**1/μ**

= 1/10 min

*= 0.1
min*

= standard deviation of service time**σ**

**= 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.*

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