mean no of transmission?? 13 or 200

Question

Consider a wireless link where the probability of packet error is 0.6 . To transfer data across the links,Stop and Wait protocol is used.The channel condition is assumed to be independent from transmission to transmission.The average number of transmissin attempts required to transfer X packet is 500. The value of X is?

Comments

1250??

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Since in the question only stop and wait is mentioned. So we will assume it's just a simple stop and wait. I mean it is not using sliding window.

In simple stop and wait, we will not begin new transmission until we have transfered previous packet

Probability of successful transmission = 1-0.6 = 0.4

so number of transmission required to transfer 1 packet = 1/0.4 = 2.5

So in 500 transmission we can transmit = 500/2.5 = 200 packets

Answer 1

500 + 0.6*500 + 0*6*0.6*500 ..............

So 500*(1/1-r) =  500*1/0.4 = 1250

Comments

wrong

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$500 = \frac{X}{1-r}$

@bad_engineer read ques carefully