Testbook Computer networks

Question

A sender is sending data to the receiver over a half-duplex link.Transmission time for data packets sent from sender to receiver is 80 μs. The propagation delay over the link is 280 μs. What frame size is required for the sliding window protocol with a window size of 8 packets to achieve throughput of 40 x 10⁶ bits per second.

A) 1800 B

B) 225 B

C) 250 bits

D) 2000 bits

I am getting answer as 400 B 

Comments

https://www.youtube.com/watch?v=1DvXxrzq0Wg&t=0s&index=13&list=PLEbnTDJUr_IegfoqO4iPnPYQui46QqT0j

was referring this video as we have only sender transmitting the data(i.e. data is transmitted in only one direction)

so we efficiency = N/P

p=1+2Tp*B/L

Tp*B will be 2Tp*B in case of full duplex transmission 

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Brother I just leave this question with my single doubt

here they mention half duplex...which means one way transmission possible right ??

window size of 8 packets --- > it means in 1 RTT i can send 8 packets right ??

which means it follow GBN or selective repeat

and we know that GBN is only work in Full duplex channel sinceduring the forward transmission, backward ACKs are also in progress.

No it means in this case ..In 1 RTT we just send only 1 packet which is equal to stop and wait

 

 

I'm getting confused...and I didn't waste my tym in it

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Magma in this question, https://gateoverflow.in/967/gate2003-84

as the link is given as full duplex, so we can send data packets and recieve acks at the same time. So before getting a ack for a packet sent by the sender, total T_t+2T_p+T_ack time elaspes during which the sender can continue sending packets continuously.. In the question it is mentioned that T_ack is negligible. So as we are transmitting 5*1000 B in T_t+2T_p time, so throughput(number of bytes sent per second)= 5*1000/(T_t+2T_p)= 11.11*10⁶ Bps.