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4. Queuing Theory, M/M/s Queuing Model (M/M/c) example
( Enter your problem )
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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`
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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
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7. Example-6: `lambda=1` per 5 hr, `mu=1` per 1 hr, `s=4`
Queuing Model = mms, Arrival Rate `lambda=1` per 5 hr, Service Rate `mu=1` per 1 hr, Number of servers `s=4`
Solution:
Arrival Rate `lambda=1` per 5 hr and Service Rate `mu=1` per 1 hr (given)
Queuing Model : M/M/s
Arrival rate `lambda=0.2,` Service rate `mu=1,` Number of servers `s=4` (given)
1. Traffic Intensity
`rho=lambda/mu`
`=(0.2)/(1)`
`=0.2`
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+(0.2)^1/(1!)+(0.2)^2/(2!)+(0.2)^3/(3!) + (0.2)^4/(4!)*(4*1)/(4*1-0.2)]^(-1)`
`=[1+(0.2)/(1)+(0.04)/(2)+(0.008)/(6) + (0.0016)/(24)*(4)/(3.8)]^(-1)`
`=[1+0.2+0.02+0.00133333 + 0.00007018]^(-1)`
`=[1.22140351]^(-1)`
`=0.81873025` or `0.81873025xx100=81.873025%`
3. Average number of customers in the queue
`L_q=((rho^s)/((s-1)!)*(lambda mu)/(s mu-lambda)^2)*P_0`
`=((0.2)^4/(3!)*(0.2*1)/(4*1-0.2)^2)*0.81873025`
`=((0.0016)/6*(0.2)/(3.8)^2)*0.81873025`
`=(0.00000369)*0.81873025`
`=0.00000302`
4. Average Time spent in the queue
`W_q=L_q/lambda`
`=0.00000302/0.2`
`=0.00001512` hr or `0.00001512xx60=0.00090718` min
Or
`W_q=((rho^s)/((s-1)!)*(mu)/(s mu-lambda)^2) * P_0`
`=((0.2)^4/(3!)*(1)/(4*1-0.2)^2)*0.81873025`
`=((0.0016)/6*(1)/(3.8)^2)*0.81873025`
`=(0.00001847)*0.81873025`
`=0.00001512` hr or `0.00001512xx60=0.00090718` min
5. Average number of customers in the system
`L_s=L_q+rho`
`=0.00000302+0.2`
`=0.20000302`
6. Average Time spent in the queue
`W_s=L_s/lambda`
`=0.20000302/0.2`
`=1.00001512` hr or `1.00001512xx60=60.00090718` min
Or
`W_s=W_q+1/mu`
`=0.00001512+1/1`
`=0.00001512+1`
`=1.00001512` hr or `1.00001512xx60=60.00090718` min
7. Utilization factor
`U=rho/s=(lambda)/(s mu)`
`=0.2/4`
`=0.05` or `0.05xx100=5%`
Or
`U=(L_s-L_q)/s`
`=(0.20000302-0.00000302)/(4)`
`=0.05` or `0.05xx100=5%`
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>=4):}`
`P_n={(((0.2)^n)/(n!)*P_0, "for "0<=n<4),(((0.2)^n)/(4!*4^(n-4))*P_0, "for "n>=4):}`
`P_1=((0.2)^1)/(1!)*P_0=0.2/1*0.81873025=0.16374605`
`P_2=((0.2)^2)/(2!)*P_0=0.04/2*0.81873025=0.0163746`
`P_3=((0.2)^3)/(3!)*P_0=0.008/6*0.81873025=0.00109164`
`P_4=((0.2)^4)/(4!*4^(4-4))*P_0=0.0016/(24*4^(0))*0.81873025=0.00005458`
`P_5=((0.2)^5)/(4!*4^(5-4))*P_0=0.00032/(24*4^(1))*0.81873025=0.00000273`
`P_6=((0.2)^6)/(4!*4^(6-4))*P_0=0.000064/(24*4^(2))*0.81873025=0.00000014`
`P_7=((0.2)^7)/(4!*4^(7-4))*P_0=0.0000128/(24*4^(3))*0.81873025=0.00000001`
`P_8=((0.2)^8)/(4!*4^(8-4))*P_0=0.00000256/(24*4^(4))*0.81873025=0`
`P_9=((0.2)^9)/(4!*4^(9-4))*P_0=0.00000051/(24*4^(5))*0.81873025=0`
`P_10=((0.2)^10)/(4!*4^(10-4))*P_0=0.0000001/(24*4^(6))*0.81873025=0`
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
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