Queuing Theory, M/M/infinity Queuing Model Example-4: lambda=4 per 1 hr, mu=1 per 10 min

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Home > Operation Research calculators > Queuing Theory M/M/Infinity Queuing Model example

7. Queuing Theory, M/M/infinity Queuing Model example ( Enter your problem )

( Enter your problem )

Algorithm and examples

Formula
Example-1: $lambda=8,mu=9$
Example-2: $lambda=6,mu=7$
Example-3: $lambda=10$ per 8 hr, $mu=1$ per 30 min
Example-4: $lambda=4$ per 1 hr, $mu=1$ per 10 min
Example-5: $lambda=30$ per 1 day, $mu=1$ per 36 min
Example-6: $lambda=96$ per 1 day, $mu=1$ per 10 min

Other related methods

4. Example-3:

$lambda=10$ per 8 hr,

$mu=1$ per 30 min
(Previous example)

6. Example-5:

$lambda=30$ per 1 day,

$mu=1$ per 36 min
(Next example)

5. Example-4: $lambda=4$ per 1 hr, $mu=1$ per 10 min

Queuing Model = mminf, Arrival Rate $lambda=4$ per 1 hr, Service Rate $mu=1$ per 10 min

Solution:
Arrival Rate

$lambda=4$ per 1 hr and Service Rate

$mu=1$ per 10 min (given)

So, Arrival Rate

$lambda=4$ per hr and Service Rate

$mu=0.1xx60=6$ per hr

Queuing Model : M/M/ $oo$

Arrival Rate

$lambda=4,$ Service Rate

$mu=6$ (given)

1. Traffic Intensity
$rho=lambda/mu$

$=(4)/(6)$

$=0.66666667$

2. Probability of no customers in the system
$P_0=e^(-rho)$

$=e^(-0.66666667)$

$=0.51341712$ or

$0.51341712xx100=51.341712%$

3. Probability that there are n customers in the system
$P_n=rho^n/(n!)*P_0$

$P_n=(0.66666667)^n/(n!)*P_0$

$P_1=((0.66666667)^1)/(1!)*P_0=0.66666667/1*0.51341712=0.34227808$

$P_2=((0.66666667)^2)/(2!)*P_0=0.44444444/2*0.51341712=0.11409269$

$P_3=((0.66666667)^3)/(3!)*P_0=0.2962963/6*0.51341712=0.02535393$

$P_4=((0.66666667)^4)/(4!)*P_0=0.19753086/24*0.51341712=0.00422566$

$P_5=((0.66666667)^5)/(5!)*P_0=0.13168724/120*0.51341712=0.00056342$

$P_6=((0.66666667)^6)/(6!)*P_0=0.0877915/720*0.51341712=0.0000626$

4. Average number of customers in the system
$L_s=rho$

$=0.66666667$

5. Average number of customers in the queue
$L_q=0$

6. Average time spent in the system
$W_s=1/mu$

$=1/(6)$

$=0.16666667$ hr or

$0.16666667xx60=10$ min

7. Average Time spent in the queue
$W_q=0$

This material is intended as a summary. Use your textbook for detail explanation.
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