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+18 votes

On a wireless link, the probability of packet error is $0.2$. A stop-and-wait protocol is used to transfer data across the link. The channel condition is assumed to be independent of transmission to transmission. What is the average number of transmission attempts required to transfer $100$ packets?

  1. $100$
  2. $125$
  3. $150$
  4. $200$
asked in Computer Networks by Boss (19.1k points)
edited by | 1.9k views

5 Answers

+22 votes
Best answer
Consider that we have to send $N$ packets and $p$ is the error probability rate. Error rate $p$ implies that if we are sending $N$ packets then $N\times p$ packets will be lost and thus we have to resend those $N\times p$ packets. But the error is still there, so again while resending those $N\times p$ packets, $N\times p\times p$ will be further lost and so on. Hence, this forms a series as follows:

$N + N\times p + N \times p^2 + \ldots$
$=N(1 + p + p^2+\ldots$
$=\dfrac{N}{1-p} \text{(Sum to infinite GP series)}$

Now we are having $N=100$ and $p=0.2,$ which implies $125$ packets have to be sent on average.
answered by Active (1.7k points)
selected by
Yes same implies for 100 packers. Instead of using formula we can think of like for sending 5 packets we will have 1 packet extra as error rate is 20% similarly to send 10 packets we will send 2 packets extra and so on for 100 packets 20 packets extra so total 120 packets but to send those extra 20 packets we have to send extra 4 packets as they also have error now to send extra 4 packet we have error rate of 20 percent which will involve sending 1 more packet hence total 120+4+1  i.e 125 packets !!
+21 votes
Total number of re-transmissions for one frame, in general, is $\dfrac{1}{(1-p)}$
where $p$ is the probability of error.

So here it would be

for one frame $\dfrac{1}{(1-0.2)}$

So for $100$ frames $\dfrac{100}{(0.8)}=125$
answered by Active (2.5k points)
edited by
I am not getting this 1/(1-p). . plz explain
//I think this is how it was obtained:


//Assume that it took x attempts to transmit 100 packets,

// one successful attempt = 1 successful packet transfer.


 => x - (20/100)x = 100

      ^         ^                 ^

      |            |                  |---------------- total packets transmitted.    

      |            |------------   attempts that failed.

   total attempts.                    |      



                        successful attempts

 => x = 100 / (1 - .2)                                      

 i.e. x = n / (1 - p)
+15 votes
Let random variable X is 1 . (A function which map a packet to a real number 1)

And probability of it's success is given is 0.8..

E = 0.8 * 1 + 0.8 * 1 +... n_packest = 100   (we expect 100 packets to get transmitted successfully)

=> 0.8 * 1 * n_packet = 100

=> n_packet = 125...

So, if we transmit 125 packets then we can expect 100 packets got transmitted successfully!! And this is average number of transmission attempts required!!


So, option (B)
answered by Active (5k points)
0 votes

Using Binomial distribution we can solve this question.

If we send N packets then, expected number of successful(error free) transmission will be  N*0.8 (Expectation of binomial distribution=NP, where N= no. of trials and P=probability of success) .

We need to send 100 packets successfully, So,  N *0.8 =100 .... or, N= 100/0.8 = 125  Answer B

Hence if we send 125 packets then , expected number of error free packets will be 100.

answered by Active (2k points)
0 votes
Probability of packet error is 0.2 which means 2 out of 10 packets contain error.

Therefore, for 8 successful packets, no. Of transmissions needed=10

For 100 packets, no. Of transmissions= (10*100)/8 = 125
answered by (117 points)

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