Consider two systems: M/M/1 and M/M/1/k. There are two types of costs associated: (i) a rejection...

Consider two systems: M/M/1 and M/M/1/k. There are two types of costs associated: (i) a rejection cost r (cost per rejection); and (ii) a holding-cost rate h (cost per unit time per customer in system). Note that there is no rejection in M/M/1. Assuming ρ<1 for both systems, what is the condition for k that makes the M/M/1/k better than M/M/1. Simplify your answer as far as possible

Expert Solution

Lets start with the basic.

QUEUEING MODELS -IS created to predict the the queue length & waiting time. queues are part of ones regular life and they are one of the most unpleasant experience of life .this mostly happens at the telephone exchange,supermarket,petrol station . queueing theory is the mathematical study of waiting lines & considered as branch of OPERATION RESEARCH ,with the help of it many business decisions are taken regarding providing sevices.QUEUEING THEORY is propounded in a research conducted by AGNER KRARUP ERLANG.

lets see the two theories

M/M/1 -- this queue represent the queues length in a system having a single server .it is one of the basic models where their is only one server available to provide service to as many customer available.the model name is written in kendall's notation .where arrivals are determined by poisson process and job service by exponential distribution [i.e is the probability distribution that describes the time between events in a poisson point process.
M/M/1/K-- here K represents the capacity of M/M/1 system ,that is the maximum number of customers in the system including the one under service this can easily get associated with the birth death process with rates .it should b noted that the system is stable for any p>0 when k is fixed .second condition is when K - INFINITE the stability condition will be p<1 ,since the distribution of M/M/1/K coverges to the distribution of M/M/1 .

keosha answered 11 months ago

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