Madeeasy[CN-Flow control]

Question

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. What is minimum number of bits required for window to achieve 100% utilization during GBN and SR protocol ?

Comments

Please see my answer.

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@Satbir so madeeasy key is wrong..!

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YES

Answer 1

Bandwidth $(B) = 30$ Mbps $= 30 * 10^6$ bps

Distance$(d) = 3000$ km $= 3*10^6$ m

$v= 3*10^8 $ m/s

$f = 12$ KB $= 12 * 10^3 *8 $ b

$T_t = \frac { f}{B} = \frac { 12*8*10^3}{30*10^6} = 3.2 $ ms

$T_p = \frac{d}{v} = \frac{3*10^6}{3*10^8} = 10$ ms

$\eta = \frac { N *T_t} { T_t + 2T_p} $

$\implies 1 = \frac { N *3.2} {3.2 + 20} $

$\implies 23.2 = N *3.2$

$\implies N = \frac{ 23.2}{3.2} = 7.25=8 $

For GBN

$N \leq 2^k -1 \implies 8+1 \leq 2^k \implies k =4$

For SR

$N \leq 2^{k-1} \implies 2^3 \leq 2^{k-1} \implies k =4 $

Hence we need $4$ bits.

Answer 2

according to me

w=51

than SR = 5 & GBN = 4

but made easy test series show

GBN=6 & SR=7

Comments

u must have not converted Bytes into bits

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i think first we have to convert packet size from Byte to Bit, correct me if i am wrong.