1. Queuing Model = mms, Arrival Rate `lambda=30` per 1 hr, Service Rate `mu=20` per 1 hr, Number of servers `s=2`

Solution:
Arrival Rate `lambda=30` per 1 hr and Service Rate `mu=20` per 1 hr (given)

Queuing Model : M/M/s

Arrival rate `lambda=30,` Service rate `mu=20,` Number of servers `s=2` (given)


1. Traffic Intensity
`rho=lambda/mu`

`=(30)/(20)`

`=1.5`


2. Probability of no customers in the system
`P_0=[sum_{n=0}^(s-1) (rho^n)/(n!) + (rho^s)/(s!)*(s mu)/(s mu-lambda)]^(-1)`

`=[1+(1.5)^1/(1!) + (1.5)^2/(2!)*(2*20)/(2*20-30)]^(-1)`

`=[1+(1.5)/(1) + (2.25)/(2)*(40)/(10)]^(-1)`

`=[1+1.5 + 4.5]^(-1)`

`=[7]^(-1)`

`=0.14285714` or `0.14285714xx100=14.285714%`


3. Average number of customers in the queue
`L_q=((rho^s)/((s-1)!)*(lambda mu)/(s mu-lambda)^2)*P_0`

`=((1.5)^2/(1!)*(30*20)/(2*20-30)^2)*0.14285714`

`=((2.25)/1*(600)/(10)^2)*0.14285714`

`=(13.5)*0.14285714`

`=1.92857143`


4. Average Time spent in the queue
`W_q=L_q/lambda`

`=1.92857143/30`

`=0.06428571` hr or `0.06428571xx60=3.85714286` min

Or
`W_q=((rho^s)/((s-1)!)*(mu)/(s mu-lambda)^2) * P_0`

`=((1.5)^2/(1!)*(20)/(2*20-30)^2)*0.14285714`

`=((2.25)/1*(20)/(10)^2)*0.14285714`

`=(0.45)*0.14285714`

`=0.06428571` hr or `0.06428571xx60=3.85714286` min


5. Average number of customers in the system
`L_s=L_q+rho`

`=1.92857143+1.5`

`=3.42857143`


6. Average Time spent in the queue
`W_s=L_s/lambda`

`=3.42857143/30`

`=0.11428571` hr or `0.11428571xx60=6.85714286` min

Or
`W_s=W_q+1/mu`

`=0.06428571+1/20`

`=0.06428571+0.05`

`=0.11428571` hr or `0.11428571xx60=6.85714286` min


7. Utilization factor
`U=rho/s=(lambda)/(s mu)`

`=1.5/2`

`=0.75` or `0.75xx100=75%`

Or
`U=(L_s-L_q)/s`

`=(3.42857143-1.92857143)/(2)`

`=0.75` or `0.75xx100=75%`


8. Probability that there are n customers in the system
`P_n={((rho^n)/(n!)*P_0, "for "0<=n< s),((rho^n)/(s!*s^(n-s))*P_0, "for "n>=2):}`

`P_n={(((1.5)^n)/(n!)*P_0, "for "0<=n<2),(((1.5)^n)/(2!*2^(n-2))*P_0, "for "n>=2):}`

`P_1=((1.5)^1)/(1!)*P_0=1.5/1*0.14285714=0.21428571`

`P_2=((1.5)^2)/(2!*2^(2-2))*P_0=2.25/(2*2^(0))*0.14285714=0.16071429`

`P_3=((1.5)^3)/(2!*2^(3-2))*P_0=3.375/(2*2^(1))*0.14285714=0.12053571`

`P_4=((1.5)^4)/(2!*2^(4-2))*P_0=5.0625/(2*2^(2))*0.14285714=0.09040179`

`P_5=((1.5)^5)/(2!*2^(5-2))*P_0=7.59375/(2*2^(3))*0.14285714=0.06780134`

`P_6=((1.5)^6)/(2!*2^(6-2))*P_0=11.390625/(2*2^(4))*0.14285714=0.050851`

`P_7=((1.5)^7)/(2!*2^(7-2))*P_0=17.0859375/(2*2^(5))*0.14285714=0.03813825`

`P_8=((1.5)^8)/(2!*2^(8-2))*P_0=25.62890625/(2*2^(6))*0.14285714=0.02860369`

`P_9=((1.5)^9)/(2!*2^(9-2))*P_0=38.44335938/(2*2^(7))*0.14285714=0.02145277`

`P_10=((1.5)^10)/(2!*2^(10-2))*P_0=57.66503906/(2*2^(8))*0.14285714=0.01608958`