The sketch below shows an open queuing system with three stations. External arri

Question

The sketch below shows an open queuing system with three stations. External arrivals to stations one to three occur at the Poisson rates of 5/hr, 7/hr and 10/hr respectively. 70% of the arrivals to workstation 1 leave the system while 30% proceed to work station 2. 10% of the workstation 2 arrivals are recycled back to workstation 1 while 90% proceed to work station 3. 85% complete service at workstation 3 and leave the system while 15% cycle back to workstation 2. The characteristics and service rates for each queue are given in the table below. Compute the number of customers waiting at each work station and the overall average waiting time in the system.

Explanation / Answer

Let ,

External arrival rate for queue 1 = R1 = 5 /hr

External arrival rate for queue 2 = R2 = 7 /hr

External arrival rate for queue 3 = R3 = 10 /hr

Effective arrival rate for queue 1 = A1

Effective arrival rate for queue 2 = A2

Effective arrival rate for queue 3 = A3

Now we can form the equation

A1 = R1 + 0.1A2

or, A1 = 5 + 0.1A2 ........[Equ1]

A2 = R2 + 0.3A1 + 0.15A3

or, A2 = 7 + 0.3A1 + 0.15A3 ........[Equ2]

A3 = R3 + 0.9A2

or, A3 = 10 + 0.9A2 ........[Equ3]

A1 = 5 + 0.1A2    ........[Equ1]

A2 = 7 + 0.3A1 + 0.15A3       ........[Equ2]

A3 = 10 + 0.9A2                      ........[Equ3]

Substituting the Value of A3 from Equ3 in Equ2

A2 = 7 + 0.3A1 + 0.15( 10 + 0.9A2) = 7 + 0.3A1 + 1.5 + 0.135A2

or , 0.865A2 - 0.3A1 = 8.5 ........[Equ2]

A1 = 5 + 0.1A2    ........[Equ1]

0.3Equ1 + Equ2

0.865A2 = 8.5 + 1.5 + 0.03A2

0.865A2 = 8.5 + 1.5 + 0.03A2

A2 = 11.976

A1 = 5 + 0.1A2 = 5 + 0.1*11.976 = 6.1976

A3 = 10 + 0.9A2 = 20.7784

Hence the no. of costumers waiting at queue 1 = A1 / (S1 - A1) = 6.1976 / (10 - 6.1976 ) = 1.63 costumers

Hence the no. of costumers waiting at queue 2 = A2 / (S2 - A2) = 11.976 / ( 7 -11.976) . The value is negative hence no costumer is waiting in queue 2

Hence the no. of costumers waiting at queue 3 = A3 / (S3 - A3) = 20.7784 / (9-20.7784). The value is negative hence no costumer is waiting in queue 3

Waiting time for queue 2 and queue 3 = 0

Hence the overall waiting time of the system is the waiting time for queue 1 = 1 / (10 - 6.1976 )

= 0.263 hour = 15.78 minutes

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