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

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+1 vote

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?

- $5$
- $10$
- $40$
- $80$

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

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

+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: k*bps** means 1000 bits per second, not 1024. Please refer *

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

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