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Solution
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Solution provided by AtoZmath.com
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Queuing Theory, M/M/1/N Queuing Model (M/M/1/K) calculator
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1. Arrival Rate `lambda=8`, Service Rate `mu=9`, Capacity `N=3`
2. Arrival Rate `lambda=6`, Service Rate `mu=7`, Capacity `N=3`
3. Arrival Rate `lambda=1`, Service Rate `mu=1.2`, Capacity `N=6`
4. Arrival Rate `lambda=25`, Service Rate `mu=40`, Capacity `N=12`
5. Arrival Rate `lambda=1.5`, Service Rate `mu=2.1`, Capacity `N=10`
6. Arrival Rate `lambda=1/10`, Service Rate `mu=1/4`, Capacity `N=5`
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Example1. Queuing Model = mm1n, Arrival Rate `lambda=8` per 1 hr, Service Rate `mu=9` per 1 hr, Capacity `N=3`
Solution: Arrival Rate `lambda=8` per 1 hr and Service Rate `mu=9` per 1 hr (given)
Queuing Model : M/M/1/N
Arrival rate `lambda=8,` Service rate `mu=9,` Capacity `N=3` (given)
1. Traffic Intensity `rho=lambda/mu`
`=(8)/(9)`
`=0.88888889`
2. Probability of no customers in the system `P_0=(1-rho)/(1-rho^(N+1))`
`=(1-0.88888889)/(1-(0.88888889)^(3+1))`
`=(0.11111111)/(1-(0.88888889)^4)`
`=(0.11111111)/(0.37570492)`
`=0.29574037` or `0.29574037xx100=29.574037%`
3. Probability of N customers in the system `P_N=rho^N*P_0`
`=(0.88888889)^3*0.29574037`
`=0.70233196*0.29574037`
`=0.20770791`
4. Average number of customers in the system `L_s=rho/(1-rho) - ((N+1)*rho^(N+1))/(1-rho^(N+1))`
`=0.88888889/(1-0.88888889) - ((3+1)*(0.88888889)^(3+1))/(1-(0.88888889)^(3+1))`
`=0.88888889/0.11111111 - (4*(0.88888889)^4)/(1-(0.88888889)^4)`
`=8 - (4*(0.62429508))/(1-(0.62429508))`
`=8 - (2.49718031)/(0.375704923030026)`
`=8 - 6.64665314`
`=1.35334686`
5. Effective Arrival rate `lambda_e=lambda(1-P_N)`
`=8*(1-0.20770791)`
`=6.33833671`
6. Average number of customers in the queue `L_q=L_s-(lambda_e)/(mu)=L_s-(lambda(1-P_N))/(mu)`
`=1.35334686-6.33833671/9`
`=0.64908722`
7. Average time spent in the system `W_s=(L_s)/(lambda_e)=(L_s)/(lambda(1-P_N))`
`=(1.35334686)/(6.33833671)`
`=0.21351767` hr or `0.21351767xx60=12.81105991` min
8. Average Time spent in the queue `W_q=(L_q)/(lambda_e)=(L_q)/(lambda(1-P_N))`
`=(0.64908722)/(6.33833671)`
`=0.10240655` hr or `0.10240655xx60=6.14439324` min
9. Utilization factor `U=L_s-L_q`
`=1.35334686-0.64908722`
`=0.70425963` or `0.70425963xx100=70.425963%`
10. Probability that there are n customers in the system `P_n=rho^n*P_0`
`P_n=(0.88888889)^n*P_0`
`P_1=(0.88888889)^1*P_0=0.88888889*0.29574037=0.26288032`
`P_2=(0.88888889)^2*P_0=0.79012346*0.29574037=0.2336714`
`P_3=(0.88888889)^3*P_0=0.70233196*0.29574037=0.20770791`
`P_4=(0.88888889)^4*P_0=0.62429508*0.29574037=0.18462925`
`P_5=(0.88888889)^5*P_0=0.55492896*0.29574037=0.16411489`
`P_6=(0.88888889)^6*P_0=0.49327018*0.29574037=0.1458799`
`P_7=(0.88888889)^7*P_0=0.43846239*0.29574037=0.12967103`
`P_8=(0.88888889)^8*P_0=0.38974434*0.29574037=0.11526313`
`P_9=(0.88888889)^9*P_0=0.34643942*0.29574037=0.10245612`
`P_10=(0.88888889)^10*P_0=0.30794615*0.29574037=0.09107211`
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