Answer 6 and 7

1 vote

Consider two computers A and B are

connected through a network of 30 Mbps. Assume the distance between them is 3000km and the signal propagation speed is same as the speed of light and the packet size is 12 KB.

connected through a network of 30 Mbps. Assume the distance between them is 3000km and the signal propagation speed is same as the speed of light and the packet size is 12 KB.

What is the minimum number of bits required for window to achieve 100% of utilization during GoBack-N and selective repeat protocol?

A . 5 and 6

B. 6 and 7

C. 6 and 6

D. 7 and 8

3 votes

BW = 30Mbps, Distance = 3000Km, Speed = 3*10^8 m/s, Data Length = 12*10^3*8bits

Transmission Delay (Tt) = Data Length / BW

= (12 * 10^3* 8) / (30 * 10^6) = 3.2 ms

Propagation Delay (Tp) = Distance / Speed

= (3000 * 10^3) / (3 * 10^8) = 10 ms

Efficiency = N / (1 + 2a), where N = Window Size, and a = Tp / Tt = 10 / 3.2 = 3.125

1 = N / (1 + 2 *3.125)

N = ceil(7.25) = 8

For GB-N, N should be 8(Sender) + 1(Receiver) = 9, No. of bits required is 4

For Selective Repeat, N should be 8(Sender) +8(Receiver) = 16, No. of bits required is 4

Please Correct me if I am wrong

Transmission Delay (Tt) = Data Length / BW

= (12 * 10^3* 8) / (30 * 10^6) = 3.2 ms

Propagation Delay (Tp) = Distance / Speed

= (3000 * 10^3) / (3 * 10^8) = 10 ms

Efficiency = N / (1 + 2a), where N = Window Size, and a = Tp / Tt = 10 / 3.2 = 3.125

1 = N / (1 + 2 *3.125)

N = ceil(7.25) = 8

For GB-N, N should be 8(Sender) + 1(Receiver) = 9, No. of bits required is 4

For Selective Repeat, N should be 8(Sender) +8(Receiver) = 16, No. of bits required is 4

Please Correct me if I am wrong

0

No. of bits for GB-N = 6 means sum of sender and receiver window is between 33 and 64, which i dont think is correct. Answer could be wrong too.

0

@Sumit Rana 1 data is taken in powers of 2 isn't it??

12KB = 12* 1024

and why you took ceil(7.25) can you please explain..?

0

12KB = 12 * 1024 * 8 bits is almost equivalent to 12 * 1000 * 8 bits

Sometime answer could be different but most of the times there is no much effect, you can try it by yourself.

Window size = 7.25, how can window be in fraction, and also window needs to be minimum 7.25 and needs to be integer. Thats why i took ceil.

Sometime answer could be different but most of the times there is no much effect, you can try it by yourself.

Window size = 7.25, how can window be in fraction, and also window needs to be minimum 7.25 and needs to be integer. Thats why i took ceil.

0

alright.!

No, i meant why not floor ? what i thought was max window size is 7.25 ie 7.

please check and tell.

No, i meant why not floor ? what i thought was max window size is 7.25 ie 7.

please check and tell.

2

window size 7.25 means it needs greater than 7 but only fraction more, so we need greater than 7 ,i.e. 8 because we cant provide fractional more window as window size is in integer, also if we will take Floor(7.25) which would be 7, then it would not satisfy the given condition

0

Actually, i also think it should be 7 as 7.25 is the max possible packets which can be transmitted.

Maximum whole packets which can be transmitted in the duration of (Transmission Delay +2 * Propagation Delay) would be floor(7.25) i.e 7.

Min sequence number required for GBN will be log ( 7+1 ) i.e 3.

And for SR it should be ceil(log(7*2)) i.e 4.

Please correct me if i am wrong.

Maximum whole packets which can be transmitted in the duration of (Transmission Delay +2 * Propagation Delay) would be floor(7.25) i.e 7.

Min sequence number required for GBN will be log ( 7+1 ) i.e 3.

And for SR it should be ceil(log(7*2)) i.e 4.

Please correct me if i am wrong.