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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 |  
        |  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` |  
        | 
 
 
 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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