Gate[2006] Small change

Question

Gate 2006 –

Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The round trip delay between A and B is 80 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps . What is the optimal window size that A should use?

https://gateoverflow.in/1820/gate2006-44

Changed -

Station A uses 32 byte packets to transmit messages to Station B using a sliding window protocol. The Propagation delay between A and B is 40 milliseconds and the bottleneck bandwidth on the path between A and B is 128 kbps . What is the optimal window size that A should use?

Comments

@Shaik Masthan

@tusharp

@Utkarsh Joshi

In first case RTT is considered as RTT = Tt + 2Tp = 80ms

RTT * BW = Ws * L ==> Ws = [RTT * BW] / L ==> Ws = 40

 

In second case Tp = 40ms ..Tt = L/B = [32*8] / [128k] = 2ms

RTT = Tt + 2Tp = 82ms

RTT * BW = Ws * L ==> Ws = [RTT * BW] / L ==> Ws = 41

.is this right ..

we should consider RTT as Tt + 2Tp in LAN always ?

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First of all it is not for LAN.It is pipelining to avoid wastage od stop and wait. LAN doesn't use acknowledgement. TT is transmission time and RTT is taken as 2*tp because we don't consider any other processing simply Tp for Src to dest and Tp for Ack dest to src. To answer such question we have to look at journey of first packet. Don't blindly follow Tt + 2*Tp. It is valid for stop and wait only.

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By LAN I means to say distance is less between nodes not shared channel network[ethernet]

if Tt<<Tp then we can consider RTT = 2TP.

Why it is for stop and wait only :( RTT means time to sent and get back ack. RTT is for frame then why only for stop and wait it is Tt + 2Tp..in general for any frame to go and ack to come back we neeNd that much time in any protocol at DLL.seems you havn't check this here it is not Stop and wait btw https://gateoverflow.in/1820/gate2006-44