Question #89722
a) For Rail booking there are two reservation counters for customers, who arrive in Poisson fashion at an average rate of 10 per hour. The service time for booking clerks at both the counters are exponentially distributed with mean time 5 minutes. These counters remain open for 12 hours per day.
i) Find the hours of the day for which all the clerks are busy.
ii) Find the probability that both the clerks are idle, one is idle.
iii) Find the expected proportion of idle time for clerks.
iv) Find the expected waiting time of customers in the
1
Expert's answer
2019-05-15T08:35:20-0400

i) Find the hours of the day for which all the clerks are busy. 

Two counters can serve:


2 customers5 min=10 customers25 min{2\ customers \over 5\ min}={10\ customers \over 25\ min}

So clerks are busy 25 minutes in one hour.

Then clerks are busy per day:


2512=300min=5 h25\cdot 12=300 \min=5\ h

The clerks are busy 5 hours per day.


ii) Find the probability that both the clerks are idle, one is idle. 

When n=0, the system is idle.

Pn = probability of exactly n customers in the system.


ρ=λ2μ\rho={\lambda \over 2\mu}λ=10,μ=12,ρ=102(12)=512\lambda=10, \mu=12, \rho={10 \over 2(12)}={5 \over 12}


P0=1ρ1+ρ=15121+512=7170.411765P_0={1-\rho \over 1+\rho}={1-{5 \over 12} \over 1+{5 \over 12}}={7 \over 17}\approx0.411765




P1=λμP0=2ρ(1ρ)1+ρP_1={\lambda \over \mu}P_0={2\rho(1-\rho) \over 1+\rho}

P1=1012717=351020.343137P_1={10 \over 12}\cdot{7 \over 17}={35 \over 102}\approx0.343137

The probability that both the clerks are idle is 7170.411765.{7 \over 17}\approx0.411765.


The probability that one is idle is 351020.343137.{35 \over 102}\approx0.343137.



iii) Find the expected proportion of idle time for clerks. 


1ρ=1λ2μ1-\rho=1-{\lambda \over 2\mu}

1ρ=1102(12)=7120.5833331-\rho=1-{10\over 2(12)}={7 \over 12}\approx0.583333

iv) Find the expected waiting time of customers in the system WW and expected waiting time of customers in queue WqW_q

For the M/M/2 queue,


ρ=λ2μ\rho={\lambda \over 2\mu}




P0=1ρ1+ρP_0={1-\rho \over 1+\rho}




Lq=P0(λ/μ)2ρ2!(1ρ)2=2ρ31ρ2L_q={P_0(\lambda/ \mu)^2\rho \over 2!(1-\rho)^2}={2\rho^3 \over 1-\rho^2}

Wq=Lqλ=2ρ3λ(1ρ2)=ρ2μ(1ρ2)W_q={L_q \over \lambda}={2\rho^3 \over \lambda(1-\rho^2)}={\rho^2 \over \mu(1-\rho^2)}

W=Wq+1μ=1μ(1ρ2)W=W_q+{1 \over \mu}={1 \over \mu(1-\rho^2)}

λ=10,μ=12,ρ=102(12)=512\lambda=10, \mu=12, \rho={10 \over 2(12)}={5 \over 12}

Wq=(512)212(1(512)2)=2514280.017507W_q={({5 \over 12})^2 \over 12(1-({5 \over 12})^2)}={25 \over 1428}\approx0.017507

W=251428+112=121190.100840W={25 \over 1428}+{1 \over 12}={12 \over 119}\approx0.100840


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