|
4. Queuing Theory, M/M/s Queuing Model (M/M/c) example
( Enter your problem )
|
Algorithm and examples
- Formula
- Example-1: `lambda=30`, `mu=20`, `s=2`
- Example-2: `lambda=10`, `mu=3`, `s=4`
- Example-3: `lambda=1` per 6 min, `mu=1` per 4 min, `s=2`
- Example-4: `lambda=15` per 1 hr, `mu=1` per 5 min, `s=2`
- Example-5: `lambda=20` per 8 hr, `mu=1` per 40 min, `s=3`
- Example-6: `lambda=1` per 5 hr, `mu=1` per 1 hr, `s=4`
|
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
|
|
2. Example-1: `lambda=30`, `mu=20`, `s=2`
1. Queuing Model = mms, Arrival Rate `lambda=30` per 1 hr, Service Rate `mu=20` per 1 hr, Number of servers `s=2`
Solution:
Arrival Rate `lambda=30` per 1 hr and Service Rate `mu=20` per 1 hr (given)
Queuing Model : M/M/s
Arrival rate `lambda=30,` Service rate `mu=20,` Number of servers `s=2` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(30)/(20)`
`=1.5`
2. Probability of no customers in the system
`P_0=[sum_{n=0}^(s-1) (rho^n)/(n!) + (rho^s)/(s!)*(s mu)/(s mu-lambda)]^(-1)`
`=[1+(1.5)^1/(1!) + (1.5)^2/(2!)*(2*20)/(2*20-30)]^(-1)`
`=[1+(1.5)/(1) + (2.25)/(2)*(40)/(10)]^(-1)`
`=[1+1.5 + 4.5]^(-1)`
`=[7]^(-1)`
`=0.14285714` or `0.14285714xx100=14.285714%`
3. Average number of customers in the queue
`L_q=((rho^s)/((s-1)!)*(lambda mu)/(s mu-lambda)^2)*P_0`
`=((1.5)^2/(1!)*(30*20)/(2*20-30)^2)*0.14285714`
`=((2.25)/1*(600)/(10)^2)*0.14285714`
`=(13.5)*0.14285714`
`=1.92857143`
4. Average Time spent in the queue
`W_q=L_q/lambda`
`=1.92857143/30`
`=0.06428571` hr or `0.06428571xx60=3.85714286` min
Or
`W_q=((rho^s)/((s-1)!)*(mu)/(s mu-lambda)^2) * P_0`
`=((1.5)^2/(1!)*(20)/(2*20-30)^2)*0.14285714`
`=((2.25)/1*(20)/(10)^2)*0.14285714`
`=(0.45)*0.14285714`
`=0.06428571` hr or `0.06428571xx60=3.85714286` min
5. Average number of customers in the system
`L_s=L_q+rho`
`=1.92857143+1.5`
`=3.42857143`
6. Average Time spent in the queue
`W_s=L_s/lambda`
`=3.42857143/30`
`=0.11428571` hr or `0.11428571xx60=6.85714286` min
Or
`W_s=W_q+1/mu`
`=0.06428571+1/20`
`=0.06428571+0.05`
`=0.11428571` hr or `0.11428571xx60=6.85714286` min
7. Utilization factor
`U=rho/s=(lambda)/(s mu)`
`=1.5/2`
`=0.75` or `0.75xx100=75%`
Or
`U=(L_s-L_q)/s`
`=(3.42857143-1.92857143)/(2)`
`=0.75` or `0.75xx100=75%`
8. Probability that there are n customers in the system
`P_n={((rho^n)/(n!)*P_0, "for "0<=n< s),((rho^n)/(s!*s^(n-s))*P_0, "for "n>=2):}`
`P_n={(((1.5)^n)/(n!)*P_0, "for "0<=n<2),(((1.5)^n)/(2!*2^(n-2))*P_0, "for "n>=2):}`
`P_1=((1.5)^1)/(1!)*P_0=1.5/1*0.14285714=0.21428571`
`P_2=((1.5)^2)/(2!*2^(2-2))*P_0=2.25/(2*2^(0))*0.14285714=0.16071429`
`P_3=((1.5)^3)/(2!*2^(3-2))*P_0=3.375/(2*2^(1))*0.14285714=0.12053571`
`P_4=((1.5)^4)/(2!*2^(4-2))*P_0=5.0625/(2*2^(2))*0.14285714=0.09040179`
`P_5=((1.5)^5)/(2!*2^(5-2))*P_0=7.59375/(2*2^(3))*0.14285714=0.06780134`
`P_6=((1.5)^6)/(2!*2^(6-2))*P_0=11.390625/(2*2^(4))*0.14285714=0.050851`
`P_7=((1.5)^7)/(2!*2^(7-2))*P_0=17.0859375/(2*2^(5))*0.14285714=0.03813825`
`P_8=((1.5)^8)/(2!*2^(8-2))*P_0=25.62890625/(2*2^(6))*0.14285714=0.02860369`
`P_9=((1.5)^9)/(2!*2^(9-2))*P_0=38.44335938/(2*2^(7))*0.14285714=0.02145277`
`P_10=((1.5)^10)/(2!*2^(10-2))*P_0=57.66503906/(2*2^(8))*0.14285714=0.01608958`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
|
|
|
