Question #267136

A bank has one drive-in counter. It is estimated that cars arrive according to Poisson distribution at



the rate of 2 every 5 minutes and that there is enough space to accommodate a line of 10 cars. Other



arriving cars can wait outside this space, if necessary. It takes 1.5 minutes on an average to serve a



customer, but the service time actually varies according to an exponential distribution. You are



required to find:



a) The probability of time, the facility remains idle.



b) The expected number of customers waiting but currently not being served at a particular point



of time.



c) The expected time a customer spends in the system



d) The probability that the waiting line will exceed the capacity of the space leading to the drive-



in counter

1
Expert's answer
2021-11-17T14:40:53-0500

Mean arrival rate:

λ=2/5\lambda=2/5

Mean service rate:

1/μ=1.51/\mu=1.5

μ=1/1.5\mu=1/1.5


a)

The probability of time, the facility remains idle:

1λ/μ=12/7.5=0.7333=73.33%1-\lambda/\mu=1-2/7.5=0.7333=73.33\%


b)

The expected number of customers waiting:

λ2μ(μλ)=0.420.67(0.670.4)=0.9\frac{\lambda^2}{\mu(\mu-\lambda)}=\frac{0.4^2}{0.67(0.67-0.4)}=0.9


c)

expected time a customer spends in the system:

1μλ=10.670.4=3.7\frac{1}{\mu-\lambda}=\frac{1}{0.67-0.4}=3.7 min


d)

the probability that n customers waiting in a queue for service:

pn=(1λ/μ)(λ/μ)np_n=(1-\lambda/\mu)(\lambda/\mu)^n

then probability that the waiting line will exceed 10 cars:

p(n>10)=1i=010pip(n>10)=1-\displaystyle{\sum^{10}_{i=0}}p_i


p0=0.4,p1=0.24,p2=0.144,p3=0.0864,p4=0.0519,p5=0.0311p_0=0.4,p_1=0.24,p_2=0.144,p_3=0.0864,p_4=0.0519,p_5=0.0311

p6=0.0187,p7=0.0112,p8=0.0067,p9=0.0040,p10=0.0024p_6=0.0187,p_7=0.0112,p_8=0.0067,p_9=0.0040,p_{10}=0.0024


p(n>10)=10.9964=0.0036p(n>10)=1-0.9964=0.0036


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