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+1 vote
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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?

  1. $5$
  2. $10$
  3. $40$
  4. $80$
asked in Computer Networks by Boss (18.3k points)
recategorized by | 669 views
+1

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

0
yes it should be 11.
0
maximum bits transmited in Rtt = 64 Kbps * 40 msec = 2560

frame size = 32* 8 bits = 256

so size of sender window = = 2560 /256= 10
0

 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

2 Answers

+6 votes
  • 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

answered by Boss (34.7k points)
0 votes
Transmission delay $T_t=\frac{32\times8}{64\times10^{3}}=$ 4 msec.

Assuming 100% bandwidth utilization, we get $1=\frac{N\times T_t}{RTT}$
$\rightarrow1=\frac{N\times 4}{40}\rightarrow N=10.$
answered by Active (2k points)


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