Question

Suppose that a computer is connected to a local network. To send a message the computer is consuming the network for a time frame of a multi seconds exactly. With probability p = 0.8, the transmission is a success but , if the transmission does not work then the computer will make a new attempt to transmit after a random period of time has elapsed, and so on until the transmission has been successfully allocated.
What is the expected total time required to transmit a message if the random delay has a hope than 3 milliseconds?
What is the standard deviation of the total transmission time if the time between transmissions is an integer of milliseconds randomly selected between 1 and 5 inclusive (equal opportunities?)

Answer 1

Question #51211Suppose that a computer is connected to a local network. To send a message the computer is consuming the network for a time frame of a multi seconds exactly. With probability p = 0.8, the transmission is a success but , if the transmission does not work then the computer will make a new attempt to transmit after a random period of time has elapsed, and so on until the transmission has been successfully allocated. 1)What is the expected total time required to transmit a message if the random delay has a hope than 3 milliseconds? 2)What is the standard deviation of the total transmission time if the time between transmissions is an integer of milliseconds randomly selected between 1 and 5 inclusive (equal opportunities?).

Solution. 1) Here we are at the situation, either we a successful transmission immediately (this happens with the probability 0.8 ) or wait 3 seconds for the another attempt and then have a successful attempt (this happens with the probability $0.2\cdot 0.8$) ans so on. So, in fact, we have that the waiting time has the distribution of $3\xi$, where $\xi\sim Geom(0.2)$, that is $\mathsf{P}(\xi=k)=0.2^{k}\cdot 0.8$, $k\geq 0$.Hence $E3\xi=3\cdot\frac{0.2}{0.8}=3/4.2$) Here, we have more complicated situation, if the attempt is not successful, then we wait random time $\xi$, that is distributed as the following $\mathsf{P}(\xi=1)=\ldots=\mathsf{P}(\xi=5)=1/5$.Denote by $\{\xi_{i}\}_{i\geq 1}$ the sequence of iid r.v. that have the same distribution as $\xi$, let $\eta\sim Geom(0.2)$ be the r.v., that indicates the number of step when the attempt to transmit is successful. Then, $T=\sum\limits_{i=1}^{\eta}\xi_{i}$. It is well-known that $VarT=(E\xi)^{2}Var\eta+E\eta Var\xi$. It can be easily calculated that $E\eta=1/4,E\eta^{2}=15/4,Var\eta=59/16$ and $E\xi=3,E\xi^{2}=2,Var\xi=2$. So, $VarT=539/16\approx 33.6875$.

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