Question

A system works on a job using three machines. The first machine works for 2 hours and returns the answer that it cannot do the job. The second machine works for 3 hours and returns the same answer. The third machine works for 5 hours and returns the job done.

(a) Assuming that the system always selects the machine to give the job to with equal probabilities , what is the expected number of hours until it gets the job done?
(b) Assuming that after each try the system records which machine was used, and then choses from the unused machines with equal probabilities, what is the expected number of hours until it gets the job done?

Answer 1

a) here exected time E(T)= $\tiny \sum$ T*P(T )=(1/3)E(first machine)+(1/3)*E(Second machine)+(1/3)*E(third m/c)

E(T)=(1/3)*(2+E(T))+(1/3)*(3+E(T))+(1/3)*5

3*E(T)=10+2E(T)

E(T)=10 Hours

b)

here let number of Hours are T for which pmf is as follows:

P(T=5) =P(first machine is 3rd machine) =1/3

P(T=7)=P(first machine and then 3rd machine)=(1/3)*(1/2) =1/6

P(T=8)=P(second machine and then 3rd machine)=(1/3)*(1/2) =1/6

P(T=10)=P(first ; second and then third+second,first and then third m/c) =(1/3)*(1/2)+(1/3)*(!/2)=1/3

hence expectd time E(T)= $\tiny \sum$ T*P(T )=5*(1/3)+7*(1/6)+8*(1/6)+10*(1/3)=7.5 Hours