Queuing Model = mm1nn, Arrival Rate `lambda=1/10` per 1 hr, Service Rate `mu=1/4` per 1 hr, Limited Customer `N=5`
Solution:
Arrival Rate `lambda=1/10` per 1 hr and Service Rate `mu=1/4` per 1 hr (given)
Queuing Model : M/M/1/N/N
Arrival rate `lambda=0.1,` Service rate `mu=0.25,` Machine `N=5` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(0.1)/(0.25)`
`=0.4`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(N) (N!)/((N-n)!)*rho^n]^(-1)`
`=[sum_{n=0}^(5) (5!)/((5-n)!)*(0.4)^n]^(-1)`
`=[1+(5!)/(4!)*(0.4)^1+(5!)/(3!)*(0.4)^2+(5!)/(2!)*(0.4)^3+(5!)/(1!)*(0.4)^4+(5!)/(0!)*(0.4)^5]^(-1)`
`=[1+(5)*(0.4)+(5xx4)*(0.16)+(5xx4xx3)*(0.064)+(5xx4xx3xx2)*(0.0256)+(5xx4xx3xx2xx1)*(0.01024)]^(-1)`
`=[1+2+3.2+3.84+3.072+1.2288]^(-1)`
`=[14.3408]^(-1)`
`=0.06973112` or `0.06973112xx100=6.973112%`
3. Probability that there are n customers in the system
`P_n=(N!)/((N-n)!)*rho^n*P_0`
`P_n=(5!)/((5-n)!)*(0.4)^n*P_0`
`P_1=(5!)/((5-1)!)*(0.4)^1*0.06973112=0.13946223`
`P_2=(5!)/((5-2)!)*(0.4)^2*0.06973112=0.22313957`
`P_3=(5!)/((5-3)!)*(0.4)^3*0.06973112=0.26776749`
`P_4=(5!)/((5-4)!)*(0.4)^4*0.06973112=0.21421399`
`P_5=(5!)/((5-5)!)*(0.4)^5*0.06973112=0.0856856`
4. Average number of customers in the system
`L_s=sum_{n=0}^(N) nP_n`
`=sum_{n=0}^(5) n*P_n`
`=0*P_0+1*P_1+2*P_2+3*P_3+4*P_4+5*P_5`
`=0*0.06973112+1*0.13946223+2*0.22313957+3*0.26776749+4*0.21421399+5*0.0856856`
`=2.67432779`
Or
`L_s=N-mu/lambda(1-P_0)`
`=5-0.25/0.1(1-0.06973112)`
`=5-2.32567221`
`=2.67432779`
5. Average number of customers in the queue
`L_q=sum_{n=1}^(N) (n-1)P_n`
`=sum_{n=1}^(5) (n-1)*P_n`
`=0*P_1+1*P_2+2*P_3+3*P_4+4*P_5`
`=0*0.13946223+1*0.22313957+2*0.26776749+3*0.21421399+4*0.0856856`
`=1.74405891`
Or
`L_q=N-((lambda+mu)/lambda)(1-P_0)`
`=5-((0.1+0.25)/0.1)*(1-0.06973112)`
`=5-(3.5)*(0.93026888)`
`=5-3.25594109`
`=1.74405891`
6. Effective Arrival rate
`lambda_e=lambda(N-L_s)`
`=0.1*(5-2.67432779)`
`=0.23256722`
7. Average time spent in the system
`W_s=(L_s)/(lambda_e)=(L_s)/(lambda(N-L_s))`
`=(2.67432779)/(0.23256722)`
`=11.49916047` hr or `11.49916047xx60=689.94962821` min
8. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=(L_q)/(lambda(N-L_s))`
`=(1.74405891)/(0.23256722)`
`=7.49916047` hr or `7.49916047xx60=449.94962821` min
9. Utilization factor
`U=L_s-L_q`
`=2.67432779-1.74405891`
`=0.93026888` or `0.93026888xx100=93.026888%`
This material is intended as a summary. Use your textbook for detail explanation.
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