Average frames to sent?

Hello Everyone, I am stuck on a problem.

-- Question Source - Tannenbaum .

Question : An upper-layer packet is split into 10 frames, each of which has 80% chance of arriving undamaged. If no error control is done by the data-link layer, how many times the message must be sent on average to get the entire thing through?

I am not able to think of a solution, the formula given in the solutions online looks like this $E = \sum_{1}^{inf}ip(1 -p)^{i - 1}$ , where p is the probability of sending all the packets without error. I was thinking of a proof using Bernoulli trials but got stuck.

If you can provide a proof for the formula above or through Bernoulli Trials. Please do so. I am stuck on this problem from last week.

Easy Math Editor

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Comments

Shivam, this is you. From future and here is your understanding of the problem after reading quite a lot on Expected Values and Probablility.

Lets say you treat the number of packets sent out of 10 original packets as X. Now you want to find the average packets that you must sent so that you can get all the 10 frames at the receiver.

Solution : Using Linearity of Expectation. lets say p(n) is the average frames you have to sent to get the n frames through. So if you want that n+1 frames are sent through. Then this is the recursive formula at hand. p(n+1) = 0.8(p(n) + 1) + 0.2(p(n))

Base Case p(1) = 1

def another_solve(n):
    # if n == 1:
    #     return 1
    # v =  (another_solve(n-1) + 1)*0.8 + 0.2(another_solve(n-1))
    # return v
    dp = [0]*n
    dp[1] = 1
    for i in range(2, n):
        dp[i] = (dp[i-1] + 1)*0.8 + 0.2*dp[i-1]

    return dp

shivam jalotra - 1 year, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$