Queuing Model = mm1, Arrival Rate `lambda=10` per 8 hr, Service Rate `mu=1` per 30 min
Solution:
Arrival Rate `lambda=10` per 8 hr and Service Rate `mu=1` per 30 min (given)
So, Arrival Rate `lambda=1.25` per hr and Service Rate `mu=0.03333333xx60=2` per hr
Queuing Model : M/M/1
Arrival Rate `lambda=1.25,` Service Rate `mu=2` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(1.25)/(2)`
`=0.625`
2. Probability of no customers in the system
`P_0=1-rho`
`=1-0.625`
`=0.375` or `0.375xx100=37.5%`
3. Average number of customers in the system
`L_s=lambda/(mu-lambda)`
`=(1.25)/(2-1.25)`
`=(1.25)/(0.75)`
`=1.66666667`
4. Average number of customers in the queue
`L_q=L_s-rho`
`=1.66666667-0.625`
`=1.04166667`
Or
`L_q=(lambda^2)/(mu(mu-lambda))`
`=((1.25)^2)/(2*(2-1.25))`
`=(1.5625)/(2*(0.75))`
`=(1.5625)/(1.5)`
`=1.04166667`
5. Average time spent in the system
`W_s=L_s/lambda`
`=(1.66666667)/(1.25)`
`=1.33333333` hr or `1.33333333xx60=80` min
Or
`W_s=1/(mu-lambda)`
`=1/(2-1.25)`
`=1/(0.75)`
`=1.33333333` hr or `1.33333333xx60=80` min
6. Average Time spent in the queue
`W_q=L_q/lambda`
`=(1.04166667)/(1.25)`
`=0.83333333` hr or `0.83333333xx60=50` min
Or
`W_q=(lambda)/(mu(mu-lambda))`
`=(1.25)/(2*(2-1.25))`
`=(1.25)/(2*(0.75))`
`=(1.25)/(1.5)`
`=0.83333333` hr or `0.83333333xx60=50` min
7. Utilization factor
`U=L_s-L_q`
`=1.66666667-1.04166667`
`=0.625` or `0.625xx100=62.5%`
8. Probability that there are n customers in the system
`P_n=rho^n*P_0`
`P_n=(0.625)^n*P_0`
`P_1=(0.625)^1*P_0=0.625*0.375=0.234375`
`P_2=(0.625)^2*P_0=0.390625*0.375=0.14648438`
`P_3=(0.625)^3*P_0=0.24414062*0.375=0.09155273`
`P_4=(0.625)^4*P_0=0.15258789*0.375=0.05722046`
`P_5=(0.625)^5*P_0=0.09536743*0.375=0.03576279`
`P_6=(0.625)^6*P_0=0.05960464*0.375=0.02235174`
`P_7=(0.625)^7*P_0=0.0372529*0.375=0.01396984`
`P_8=(0.625)^8*P_0=0.02328306*0.375=0.00873115`
`P_9=(0.625)^9*P_0=0.01455192*0.375=0.00545697`
`P_10=(0.625)^10*P_0=0.00909495*0.375=0.00341061`
This material is intended as a summary. Use your textbook for detail explanation.
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