Hi members,
Are these problems so difficult that no members know the correct answers?
Hi Members,
Problem is as follows.
C arrives at a two-server post office to find A being served by server 1 and B by server 2. C will enter service as soon as either A or B departs. If the service time of server i are exponential random variables with rates $\mu_i=1,2 $, find
a)The probability that A is the first one to depart.
b)The probability that A is the last one to depart.
c)The expected time until C departs.
Solutions:
I don't know what are the correct solutions, but I solved them as follows.
a)$\frac{\mu_1}{\mu_1+\mu_2}$
b)$\frac{(\mu_2)^2}{(\mu_1+\mu_2)^2}$
c)$\frac{1}{\mu_1}\left( \frac{(\mu_1)^2}{(\mu_1+\mu_2)^2}\right) +\frac{1}{\mu_2}\left( \frac{(\mu_2)^2}{(\mu_1+\mu_2)^2}\right)$
Please reply me, if my solutions are wrong.Your comments about the correctness of answers are always welcome
Hi members,
This is an explanation to answer (a).
When C arrives at two server post office to find A being served by server 1, A's remaining service at server 1 is exponential($\mu_1$), because of memorylessness and C start service at server 1 that is exponential($\mu_2$).C's service will start only when A's remaining service at server 1 ends.
Therefore $P_A$ is the probability that an Exponential ($\mu_1$) random variable < an Exponential($\mu_2$) random variable, which is
$P_A =\frac{\mu_1}{\mu_1 +\mu_2}$
The probability you should be looking for is P = T*exp(-mu*t). In your problem mu is a rate of time usage, t is time, and P is the probability to stay there up to time t is the time the user is allowed. Pick it up from there. This should be the answer to your problem "exponential probability distribution".
Hi statisticians,
This is an explanation to answer b)
A will sti[[ be in the system if B and C will depart if their service time < A's remaining service time. Let us condition P(A in the system)=P(A in the system|B finishes before him)
$\frac{\mu_2}{\mu_1+\mu_2}$ *P(A in the system|C finish before A)$\frac{\mu_2}{\mu_1+\mu_2}$
Therefore
P(A is the last one to depart)=$\frac{(\mu_2)^2}{(\mu_1+\mu_2)^2}$
Hi Votan,
There are two important key properties of exponential probability distribution 1) Memoryless property 2)Let $(X_1,...,X_n)$ be independent random variables, with $X_i$ having an exponential distribution.Then the distribution of minimum($X_1,...,X_n$) is exponential$(\lambda_1 +...+\lambda_n)$ and the probability that the minimum is $X_i$ is $\frac{\lambda_i}{\lambda_1 +...+\lambda_n}$
To see how this property works together with memoryless property, this thread of exponential distribution problem would be a example.
Hi members,
To compute expected time C is in the system,we first divide up C's time in the system into
T=$\T_1$+R
Where $T_1$ is the time until the first thing that happens and R is the remaining time.The time until the first thing to happen is exponential($\mu_1$) or ($\mu_2$). $T_1 =\mu_1 +\mu_2$ and E($T_1)=\frac1\mu_1+\frac1\mu_2$
To compute E(R),we condition on which was the first thing to happen if the first thing to happen is A or B finished service before C which occurs with the probability $\frac12$ hence E(R)=$\frac12$
Please, reply me my answer is correct or wrong?