Proof of Theorem 11

The donor queue can be seen as an M/GI/1 queue with generalized vacations, where a vacation starts when the number of donor jobs becomes zero and ends when the donor server starts working on donor jobs. In an M/GI/1 queue with generalized vacations, the mean response time has the following expression [57]:

$$\mbox{{\bf\sf E}}\left[ T \right] = \mbox{{\bf\sf E}}\left[ ... ... A(A-1) \right]}{2\lambda_D\mbox{{\bf\sf E}}\left[ A \right]},$$ (A.2)

where $A$ denotes the number of jobs that arrive during a vacation period. Observe that the first two terms constitute the mean response time in a corresponding M/GI/1 queue (without vacation). Therefore, it suffices to analyze $\mbox{{\bf\sf E}}\left[ A \right]$ and $\mbox{{\bf\sf E}}\left[ A(A-1) \right]$.

There are two types of vacations, depending on how the vacation ends. The first type of vacation ends when a donor job arrives at an empty donor queue while the donor server is staying at the donor queue. In this case, $A=1$. The second type of vacation ends when a donor job arrives at a donor queue with $N_D^{th}-1$ jobs while the donor server is staying at the beneficiary queue or in the process of switching to the beneficiary queue. In this case, $A=N_D^{th}+B$, where $B$ is the number of donor job arrivals during $K_{ba}$. Let $p$ be the probability that a vacation is of the second type. Then,

$$\mbox{{\bf\sf E}}\left[ A \right] = (N_D^{th} + \lambda_D\mbox{{\bf\sf E}}\left[ K_{ba} \right])p + (1-p)$$ (A.3)

$$\mbox{{\bf\sf E}}\left[ A(A-1) \right] = \left(N_D^{th}(N_D^{th}-1) + 2N_D^{th}\lambda_D\mbox{{\bf\sf E}}\... ...[ K_{ba} \right] + \lambda_D^2\mbox{{\bf\sf E}}\left[ K_{ba}^2 \right]\right)p.$$ (A.4)

All that remains is to derive $p$. Observe that the first type of vacation starts when a donor job arrives while the donor server is at the donor queue and the number of donor jobs is zero. Also, observe that the second type of vacation starts when a donor job arrives while the donor server is either at the beneficiary queue or in the process of switching to the beneficiary queue and the number of donor jobs is $N_D^{th}-1$. Since the arrival process is Poisson, $p$ is given by the expression (6.1). The theorem now follows from (A.2)-(A.4).     width 1ex height 1ex depth 0pt