Hello, this is a problem from course Probability and Random Processes. Please do
ID: 3073374 • Letter: H
Question
Hello, this is a problem from course Probability and Random Processes.
Please do the problem step by step and take a sharp picture of the solution. Thank you.
4. A computer network consists of M terminals (numbered from 1 to M) all sharing a common channel (e.g. wire or optical fiber). Time is divided into slots of duration T seconds. Terminal i will transmit information during each slot on the network with probability p and be idle (not transmit) with probability 1 - p. All terminals act independently. If more than one terminal transmits during a particular time slot a "collision" is said to occur (in which case the information is lost) If exactly one terminal transmits a "successful" transmission is said to take place (a) Calculate the probability of a successful transmission. (Make a plot of this as a function of n for M 20 .) (b) Determine the value of p that makes the successful transmission probability maxi- mum and determine the corresponding maximum successful transmission probability Evaluate the maximum successfull probability for M = 10, 100, 1000 (c) Calculate the probability of a collisionExplanation / Answer
ambiguous true - to go by the shortest route suceeds if there is no corruption on 3 transfers so probability of success is 0.8^3 but how is the receiver to know that it is correct? the longer way has 5 transfers so 0.8^5 probability of success - add them up but only if there is some inbuilt way to check that the data is right without comparing the two paths : if you have to compare both routes to figure out whether the data is OK ( get agreement) then the probability of success is 0.8^3*0.8^5 assuming that each terminal will only pass it on once and that the probability of the same sort of corruption is zero : if you allow the data to ring around for ever then you need to put additional constraints on the time taken to pass it on in each direction or ensure that at least the order is maintained ( I think ) it gets way too confusing for me at that point.
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