|
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
|
|
5. Example-4: `lambda=45`, `mu=15`, `s=2`, `N=12`
Queuing Model = mmsnn, Arrival Rate `lambda=45` per 1 hr, Service Rate `mu=15` per 1 hr, Number of servers `s=2`, Limited Customer `N=12`
Solution:
Arrival Rate `lambda=45` per 1 hr and Service Rate `mu=15` per 1 hr (given)
Queuing Model : M/M/s/N/N
Arrival rate `lambda=45,` Service rate `mu=15,` Number of servers `s=2,` Machine `N=12` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(45)/(15)`
`=3`
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}^(1) (12!)/((12-n)!*n!)*(3)^n + sum_{n=2}^(12) (12!)/((12-n)!*2!*2^(n-2))*(3)^n]^(-1)`
`=[(1+(12!)/(11!*1!)*(3)^1) + ((12!)/(10!*2!*2^(0))*(3)^2+(12!)/(9!*2!*2^(1))*(3)^3+(12!)/(8!*2!*2^(2))*(3)^4+(12!)/(7!*2!*2^(3))*(3)^5+(12!)/(6!*2!*2^(4))*(3)^6+(12!)/(5!*2!*2^(5))*(3)^7+(12!)/(4!*2!*2^(6))*(3)^8+(12!)/(3!*2!*2^(7))*(3)^9+(12!)/(2!*2!*2^(8))*(3)^10+(12!)/(1!*2!*2^(9))*(3)^11+(12!)/(0!*2!*2^(10))*(3)^12)]^(-1)`
`=[(1+(12)/(1)*(3)^1) + ((12xx11)/(2*1)*(3)^2+(12xx11xx10)/(2*2)*(3)^3+(12xx11xx10xx9)/(2*4)*(3)^4+(12xx11xx10xx9xx8)/(2*8)*(3)^5+(12xx11xx10xx9xx8xx7)/(2*16)*(3)^6+(12xx11xx10xx9xx8xx7xx6)/(2*32)*(3)^7+(12xx11xx10xx9xx8xx7xx6xx5)/(2*64)*(3)^8+(12xx11xx10xx9xx8xx7xx6xx5xx4)/(2*128)*(3)^9+(12xx11xx10xx9xx8xx7xx6xx5xx4xx3)/(2*256)*(3)^10+(12xx11xx10xx9xx8xx7xx6xx5xx4xx3xx2)/(2*512)*(3)^11+(12xx11xx10xx9xx8xx7xx6xx5xx4xx3xx2xx1)/(2*1024)*(3)^12)]^(-1)`
`=[(1+36) + (594+8910+120285+1443420+15155910+136403190+1023023925+6138143550+27621645975+82864937925+124297406887.5)]^(-1)`
`=[242098290608.5]^(-1)`
`=0` or `0xx100=0%`
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={((12!)/((12-n)!*n!)*(3)^n*P_0, "for "0<=n<2),((12!)/((12-n)!*2!*2^(n-2))*(3)^n*P_0, "for "2<=n<=12):}`
`P_1=0`
`P_2=0`
`P_3=0.00000004`
`P_4=0.0000005`
`P_5=0.00000596`
`P_6=0.0000626`
`P_7=0.00056342`
`P_8=0.00422566`
`P_9=0.02535393`
`P_10=0.11409269`
`P_11=0.34227808`
`P_12=0.51341712`
4. Average number of customers in the system
`L_s=sum_{n=0}^(N) nP_n`
`=sum_{n=0}^(12) 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+11*P_11+12*P_12`
`=0*0+1*0+2*0+3*0.00000004+4*0.0000005+5*0.00000596+6*0.0000626+7*0.00056342+8*0.00422566+9*0.02535393+10*0.11409269+11*0.34227808+12*0.51341712`
`=11.33333333`
5. Average number of customers in the queue
`L_q=sum_{n=s+1}^(N) (n-s)P_n`
`=sum_{n=3}^(12) (n-2)*P_n`
`=1*P_3+2*P_4+3*P_5+4*P_6+5*P_7+6*P_8+7*P_9+8*P_10+9*P_11+10*P_12`
`=1*0.00000004+2*0.0000005+3*0.00000596+4*0.0000626+5*0.00056342+6*0.00422566+7*0.02535393+8*0.11409269+9*0.34227808+10*0.51341712`
`=9.33333333`
6. Effective Arrival rate
`lambda_e=lambda(N-L_s)`
`=45*(12-11.33333333)`
`=30`
7. Average time spent in the system
`W_s=(L_s)/(lambda_e)=(L_s)/(lambda(N-L_s))`
`=(11.33333333)/(30)`
`=0.37777778` hr or `0.37777778xx60=22.66666667` min
8. Average Time spent in the queue
`W_q=(L_q)/(lambda_e)=(L_q)/(lambda(N-L_s))`
`=(9.33333333)/(30)`
`=0.31111111` hr or `0.31111111xx60=18.66666667` min
9. Utilization factor
`U=(L_s-L_q)/s`
`=(11.33333333-9.33333333)/(2)`
`=1` or `1xx100=100%`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
|
|
|
