Queuing Theory, M/M/s/N Queuing Model (M/M/c/K) calculator

SolutionHelp

1. Arrival Rate

$lambda=30$ , Service Rate

$mu=20$ , Number of servers

$s=2$ , Capacity

$N=3$

2. Arrival Rate

$lambda=10$ , Service Rate

$mu=3$ , Number of servers

$s=2$ , Capacity

$N=3$

3. Arrival Rate

$lambda=40$ , Service Rate

$mu=1$ , Number of servers

$s=10$ , Capacity

$N=10$

4. Arrival Rate

$lambda=45$ , Service Rate

$mu=15$ , Number of servers

$s=2$ , Capacity

$N=12$

5. Arrival Rate

$lambda=1/10$ , Service Rate

$mu=1/4$ , Number of servers

$s=2$ , Capacity

$N=5$

6. Arrival Rate

$lambda=1/10$ , Service Rate

$mu=1/4$ , Number of servers

$s=2$ , Capacity

$N=5$

Example

1. Queuing Model = mmsn, Arrival Rate $lambda=30$ per 1 hr, Service Rate $mu=20$ per 1 hr, Number of servers $s=2$ , Capacity $N=3$

Solution:
Arrival Rate

$lambda=30$ per 1 hr and Service Rate

$mu=20$ per 1 hr (given)

Queuing Model : M/M/s/N

Arrival rate

$lambda=30,$ Service rate

$mu=20,$ Number of servers

$s=2,$ Capacity

$N=3$ (given)

1. Traffic Intensity
$rho=lambda/mu$

$=(30)/(20)$

$=1.5$

2. Probability of no customers in the system
$P_0=[sum_{n=0}^(s-1) (rho^n)/(n!) + sum_{n=s}^(N) (rho^n)/(s! *s^(n-s))]^(-1)$

$=[sum_{n=0}^(1) (rho^n)/(n!) + sum_{n=2}^(3) (rho^n)/(2! * 2^(n-2))]^(-1)$

$=[(1+(1.5)^1/(1!)) + (((1.5)^2)/(2! * 2^(0))+((1.5)^3)/(2! * 2^(1)))]^(-1)$

$=[(1+(1.5)^1/(1)) + (((1.5)^2)/(2*1)+((1.5)^3)/(2*2))]^(-1)$

$=[(1+1.5) + (1.125+0.84375)]^(-1)$

$=[4.46875]^(-1)$

$=0.22377622$ or

$0.22377622xx100=22.377622%$

3. Probability that there are n customers in the system
$P_n={((rho^n)/(n!)*P_0, "for "0<=n< s),((rho^n)/(s!*s^(n-s))*P_0, "for "s<=n<= N):}$

$P_n={(((1.5)^n)/(n!)*P_0, "for "0<=n<2),(((1.5)^n)/(2!*2^(n-2))*P_0, "for "2<=n<=3):}$

$P_1=((1.5)^1)/(1!)*P_0=1.5/1*0.22377622=0.33566434$

$P_2=((1.5)^2)/(2!*2^(2-2))*P_0=2.25/(2*2^(0))*0.22377622=0.25174825$

$P_3=((1.5)^3)/(2!*2^(3-2))*P_0=3.375/(2*2^(1))*0.22377622=0.18881119$

4. Average number of customers in the system
$L_s=sum_{n=0}^(N) nP_n$

$=sum_{n=0}^(3) n*P_n$

$=0*P_0+1*P_1+2*P_2+3*P_3$

$=0*0.22377622+1*0.33566434+2*0.25174825+3*0.18881119$

$=1.40559441$

5. Average number of customers in the queue
$L_q=sum_{n=s}^(N) (n-s)P_n$

$=sum_{n=2}^(3) (n-2)*P_n$

$=(2-2)*P_2+(3-2)*P_3$

$=0*0.25174825+1*0.18881119$

$=0.18881119$

6. Effective Arrival rate
$lambda_e=lambda*(1-P_N)$

$=30*(1-0.18881119)$

$=24.33566434$

7. Average Time spent in the queue
$W_q=(L_q)/(lambda_e)=L_q/(lambda*(1-P_N))$

$=(0.18881119)/(24.33566434)$

$=0.00775862$ hr or

$0.00775862xx60=0.46551724$ min

8. Average Time spent in the queue
$W_s=(L_s)/(lambda_e)=L_s/(lambda*(1-P_N))$

$=(1.40559441)/(24.33566434)$

$=0.05775862$ hr or

$0.05775862xx60=3.46551724$ min

Or
$W_s=W_q+1/mu$

$=0.00775862+1/20$

$=0.00775862+0.05$

$=0.05775862$ hr or

$0.05775862xx60=3.46551724$ min

9. Utilization factor
$U=(L_s-L_q)/s$

$=(1.40559441-0.18881119)/(2)$

$=0.60839161$ or

$0.60839161xx100=60.839161%$