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Pls verify that Ans. would be 45 or 46 ?

@Magma Mam what will be ans if it is half duplex?

RTT  =  90 x 10-3 s

Bandwidth =  256 Kbps  =  215 kiloByte/sec

IN case of Half duplex , capacity of channel  = ( 2*Tp*Bandwidth) /2

= (Round-Trip time)*Bandwidth /2

= 90 x 10-3 * 215 /2

= 0.09 x 214 Byte

0.09 x 214 Byte of data send by a sender to receiver

packet size =  64 Byte

Number of windows require =  0.09 x 214 Byte / 64  =  23.04 = 24 windows

Optimal window size is 1 greater than number of frames in 1 RTT  = 1+24 = 25

25 ans

Mam i think in half duplex it will behave like stop and wait because at a time only one packet can be on link i.e. either sender packet or receiver packet. I don't understand why have you calculated like that.

Bandwidth delay product= BW*RTT

=256*90

optimal window size= bd product/packet size =(256*90)/(64*8)

=45

this is best approach to solve this kind of questions

reshown
For optimial window size sender should tranmit data to the link untill it get the ack for the first packet. So in the entire RTT sender should transmit. RTT = 90ms while Tt(tranmission time) = 2ms.

So no of packet sender can transmit in one RTT = 90ms/2ms = 45.

I think this is simple approch :P

Can you please tell me why we are not considering the transmission time delay for sender?

This is what I am saying:

Total time for one packet sending = Tt + Tp + Tpack  (others are negligible).

= 2 + 90

= 92 mSec.

In 92 mSec how many packets can we transmit? That would be 92 / 2 = 46.

Well , i will not give much Emphasis on this approach but you can do this if you dont remember the formula or confused in Middle..Dimension checking.

clearly we require here is sender Window size which should be obviously in bits , so $Bandwidth: Kb/Sec$  $RTT: Sec$

so Answer: Bandwidth*RTT/ packet Size. :)
it will be 46
by

I also think same but i would be 45 .
Pls check this : https://gateoverflow.in/1820/gate2006-44
It should not be 46 you are getting this because you are considering round trip time = 2 * (Propogation delay).

while RTT = Tt (data) + Tp (data) + Tt (ack) + Tp (ack). Transmisstion time for ack is consider neglible here.

Moreever for optimial window size sender should tranmit data to the link untill it get the ack for the first packet. So in the entire RTT sender should transmit. RTT = 90ms while Tt(tranmission time) = 2ms.

So no of packet sender can transmit in one RTT = 90ms/2ms = 45.

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