ISRO-DEC2017-36

Question

Station $A$ uses $32-byte$ packets to transmit messages to Station $B$ using a sliding window protocol.
The round trip time delay between $A$ and $B$ is $40\,ms$ and the bottleneck bandwidth on the path
$A$ and $B$ is $64\, kbps.$What is the optimal window size that $A$ should use?

• A. $5$
• B. $10$
• C. $40$
• D. $80$

Comments

it should be 11, but i will go with nearest option i.e n=10

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yes it should be 11.

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maximum bits transmited in Rtt = 64 Kbps * 40 msec = 2560

frame size = 32* 8 bits = 256

so size of sender window = = 2560 /256= 10

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 Anu007 Sir,

so size of sender window = = 2560 /256= 10

Here, units of denominator & numerator are not matching.

It should be

 so size of sender window = = 2621.44 /256= 10.24

Answer 1

• In one second link can send 64 kb
• In RTT link can send 64kb*40ms =  64* 1000* 40/1000 = 2560 bits = 320 bytes
• Size of one packet =32 bytes
• Size of optimal window =320/32 =10

Answer is B

PS:  kbps means 1000 bits per second, not 1024. Please refer 

  https://en.wikipedia.org/wiki/Data_rate_units#Decimal_multiples_of_bits

Answer 2

Transmission delay $T_t=\frac{32\times8}{64\times10^{3}}=$ 4 msec.

Assuming 100% bandwidth utilization, we get $1=\frac{N\times T_t}{RTT}$
$\rightarrow1=\frac{N\times 4}{40}\rightarrow N=10.$