|
6. Queuing Theory, M/M/s/N/N Queuing Model (M/M/c/K/K) example
( Enter your problem )
|
Algorithm and examples
- Formula
- Example-1: `lambda=30`, `mu=20`, `s=2`, `N=3`
- Example-2: `lambda=30`, `mu=20`, `s=2`, `N=3`
- Example-3: `lambda=40`, `mu=1`, `s=10`, `N=10`
- Example-4: `lambda=45`, `mu=15`, `s=2`, `N=12`
- Example-5: `lambda=1.5`, `mu=2.1`, `s=3`, `N=10`
- Example-6: `lambda=1/10`, `mu=1/4`, `s=2`, `N=5`
|
Other related methods
- M/M/1 Model
- M/M/1/N Model (M/M/1/K Model)
- M/M/1/N/N Model (M/M/1/K/K Model)
- M/M/s Model (M/M/c Model)
- M/M/s/N Model (M/M/c/K Model)
- M/M/s/N/N Model (M/M/c/K/K Model)
- M/M/Infinity Model
|
|
6. Example-5: `lambda=1.5`, `mu=2.1`, `s=3`, `N=10`
Queuing Model = mmsnn, Arrival Rate `lambda=1.5` per 1 hr, Service Rate `mu=2.1` per 1 hr, Number of servers `s=3`, Limited Customer `N=10`
Solution:
Arrival Rate `lambda=1.5` per 1 hr and Service Rate `mu=2.1` per 1 hr (given)
Queuing Model : M/M/s/N/N
Arrival rate `lambda=1.5,` Service rate `mu=2.1,` Number of servers `s=3,` Machine `N=10` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(1.5)/(2.1)`
`=0.71428571`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(s-1) (N!)/((N-n)!*n!)*rho^n + sum_{n=s}^(N) (N!)/((N-n)!*s!*s^(n-s))*rho^n]^(-1)`
`=[sum_{n=0}^(2) (10!)/((10-n)!*n!)*(0.71428571)^n + sum_{n=3}^(10) (10!)/((10-n)!*3!*3^(n-3))*(0.71428571)^n]^(-1)`
`=[(1+(10!)/(9!*1!)*(0.71428571)^1+(10!)/(8!*2!)*(0.71428571)^2) + ((10!)/(7!*3!*3^(0))*(0.71428571)^3+(10!)/(6!*3!*3^(1))*(0.71428571)^4+(10!)/(5!*3!*3^(2))*(0.71428571)^5+(10!)/(4!*3!*3^(3))*(0.71428571)^6+(10!)/(3!*3!*3^(4))*(0.71428571)^7+(10!)/(2!*3!*3^(5))*(0.71428571)^8+(10!)/(1!*3!*3^(6))*(0.71428571)^9+(10!)/(0!*3!*3^(7))*(0.71428571)^10)]^(-1)`
`=[(1+(10)/(1)*(0.71428571)^1+(10xx9)/(2)*(0.71428571)^2) + ((10xx9xx8)/(6*1)*(0.71428571)^3+(10xx9xx8xx7)/(6*3)*(0.71428571)^4+(10xx9xx8xx7xx6)/(6*9)*(0.71428571)^5+(10xx9xx8xx7xx6xx5)/(6*27)*(0.71428571)^6+(10xx9xx8xx7xx6xx5xx4)/(6*81)*(0.71428571)^7+(10xx9xx8xx7xx6xx5xx4xx3)/(6*243)*(0.71428571)^8+(10xx9xx8xx7xx6xx5xx4xx3xx2)/(6*729)*(0.71428571)^9+(10xx9xx8xx7xx6xx5xx4xx3xx2xx1)/(6*2187)*(0.71428571)^10)]^(-1)`
`=[(1+7.14285714+22.95918367) + (43.73177843+72.88629738+104.12328197+123.95628805+118.05360767+84.32400548+40.15428832+9.56054484)]^(-1)`
`=[627.89213295]^(-1)`
`=0.00159263` or `0.00159263xx100=0.159263%`
3. Probability that there are n customers in the system
`P_n={((N!)/((N-n)!*n!)*rho^n*P_0, "for "0<=n< s),((N!)/((N-n)!*s!* s^(n-s))*rho^n*P_0, "for "s<=n<= N):}`
`P_n={((10!)/((10-n)!*n!)*(0.71428571)^n*P_0, "for "0<=n<3),((10!)/((10-n)!*3!*3^(n-3))*(0.71428571)^n*P_0, "for "3<=n<=10):}`
`P_1=0.01137593`
`P_2=0.03656549`
`P_3=0.06964855`
`P_4=0.11608092`
`P_5=0.16582989`
`P_6=0.19741653`
`P_7=0.18801575`
`P_8=0.13429696`
`P_9=0.06395093`
`P_10=0.01522641`
4. Average number of customers in the system
`L_s=sum_{n=0}^(N) nP_n`
`=sum_{n=0}^(10) n*P_n`
`=0*P_0+1*P_1+2*P_2+3*P_3+4*P_4+5*P_5+6*P_6+7*P_7+8*P_8+9*P_9+10*P_10`
`=0*0.00159263+1*0.01137593+2*0.03656549+3*0.06964855+4*0.11608092+5*0.16582989+6*0.19741653+7*0.18801575+8*0.13429696+9*0.06395093+10*0.01522641`
`=5.88973334`
5. Average number of customers in the queue
`L_q=sum_{n=s+1}^(N) (n-s)P_n`
`=sum_{n=4}^(10) (n-3)*P_n`
`=1*P_4+2*P_5+3*P_6+4*P_7+5*P_8+6*P_9+7*P_10`
`=1*0.11608092+2*0.16582989+3*0.19741653+4*0.18801575+5*0.13429696+6*0.06395093+7*0.01522641`
`=2.95382858`
6. Effective Arrival rate
`lambda_e=lambda(N-L_s)`
`=1.5*(10-5.88973334)`
`=6.16539999`
7. Average time spent in the system
`W_s=(L_s)/(lambda_e)=(L_s)/(lambda(N-L_s))`
`=(5.88973334)/(6.16539999)`
`=0.95528812` hr or `0.95528812xx60=57.31728691` min
8. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=(L_q)/(lambda(N-L_s))`
`=(2.95382858)/(6.16539999)`
`=0.47909764` hr or `0.47909764xx60=28.74585834` min
9. Utilization factor
`U=(L_s-L_q)/s`
`=(5.88973334-2.95382858)/(3)`
`=0.97863492` or `0.97863492xx100=97.863492%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
|
|
|
