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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. $200$

edited | 2.8k views

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.

Correct Answer: $B$
by Active (1.9k points)
edited
+4
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 !!
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if the prob of packet error is changed to 0.35 instead of 0.2;

then what will be the avg no. of transmission??

1538.46 or 1539

mainly my doubt is avg no. of transmission should be a perfect int or it can be in float also??

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is this correct??

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@Gate Fever

We have to retransmit complete packet not fraction right?

@Bikram sir please confirm transmissions are 1538.46 or 1539?

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i think it is 1539

u can check the comments given below(at bottom)
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@AakS Why are we not considering the error possibilities in case of ACKs sent in response to the packet transfers? It could very well be the case that the ACK for a packet gets lost and we need to resend the packet again.

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 $\dfrac{100}{(0.8)}=125$
by Active (2.5k points)
edited
0
I am not getting this 1/(1-p). . plz explain
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//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)
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)
by Active (5k points)
+1
$E[100*X] = 100*E[X]$
I found this much simpler.

$N=>$   Total packets to be sent

$0.8*N=>$   Total packet to be sent successfully , each with probability of   $0.8$

So,

$0.8*N=100$

$N=125$

So total 125 attempts are required to send 100 packets if probability of success of each packet is $0.8$
by Active (5.1k points)
0

if the prob of packet error is changed to 0.35 instead of 0.2;

then what will be the avg no. of transmission??

1538.46 or 1539

+1

it will be "ceil " value i guess coz i think fractional part means some of the data it still need to be transferred hence 1 more packet will be required. ( That one packet will also may be have 0.35 error but i guess we dont consider it for last packet, exact reason i i nt know )

0

actually i asked it bcoz of this!

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0
i dont think u mentioned him correctly

his name is not coming in blue color
+3
as per my knowledge, we have to take ceil value but not fractional value

" it will be "ceil " value i guess coz i think fractional part means some of the data it still need to be transferred hence 1 more packet will be required. " --- absolutely correct as per my knowledge
0
ok,thanks
+1 vote

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.

by Active (3k points)
+1 vote

Given-

• Probability of packet error = 0.2
• We have to transfer 100 packets

Now,

• When we transfer 100 packets, number of packets in which error will occur = 0.2 x 100 = 20.
• Then, these 20 packets will have to be retransmitted.
• When we retransmit 20 packets, number of packets in which error will occur = 0.2 x 20 = 4.
• Then, these 4 packets will have to be retransmitted.
• When we retransmit 4 packets, number of packets in which error will occur = 0.2 x 4 = 0.8 ≅ 1.
• Then, this 1 packet will have to be retransmitted.

From here, average number of transmission attempts required = 100 + 20 + 4 + 1 = 125.

Option (B) is correct.

by (53 points)
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
by (277 points)

Let x packets need to be sent.

We know that, of them x, 0.2 X x packets will get discarded.

So we want that ,   x - 0.2x = 100 (i.e., we want of them total x , 100 packets remain to be correctly delivered)

therefore x = 125

by (369 points)