GATE CSE 2006 | Question: 44

Question

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

• A. $20$
• B. $40$
• C. $160$
• D. $320$

Comments

@rfzahid

No, RTT simply means 2*Tp.

 
What you are trying to mention here is “Total Cycle Time” but not RTT.

 

Efficiency = Useful Time / Total Cycle Time .

 

but not,

Efficiency = Useful Time / RTT .

 

Note:  Total Cycle Time and RTT are Both Different.

 

So, This Question requires to select the Closest Answer. Those were the days when gate questions were little ambiguous, nowadays it is not the case, so no need to worry :-)

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GO to the Root of Concept: Video Explanation with timestamp: Sliding Window PYQs| GATE 2009 Q57, 58| 2006 Q46, 64| 2003, 2014, 2015, 2004, 2016, 2007| With NOTES

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if the transmission time was considered 2 ms then the propagation time will be 78 ms
and the answer will still be 40

Answer 1

Basically, Acknowledgement will be received by A will be 80 millisecond. Now,

A can send 128kb in 1 sec  (bandwidth),

In 1 sec ----→ 128kb

    1 sec ---→ (128kb/(32*8 bits))   packets  ( As each packet of size 32*8 bits)

    solving above, A can send 500 packet per second  

So, In 80ms  ---→ 500* 80*10^-3  = 40 packets

So, A can send 40 packets before it receives an acknowledgement.

So, Optimal window size for will be 40

Answer 2

we can send as frames as possible in sliding window protocol in a round trip time

T $_{t}$ =   $_{\tfrac{size of frame}{bandwidth}}$

T$_{t}$ =$_{\tfrac{32*8 bits}{128*10$^{3}$ bps}}$ = 2msec

no.of frames=$_{\tfrac{RTT}{T_{t} }}$= $_{\tfrac{80 msec}{20 msec}}$= 40 frames

Answer 3

Round trip time = 2*Tp =  80 * 10^-3 s

Length of packet (L) = 32 * 8 bits

Bandwidth = 128 * 10³ bits/s

Transmission time(T_t) = L / bandwidth

                                   = ( 32 * 8 ) / ( 128 * 10³)

                                   = 2 * 10^-3 s

Optimal window size = 1 + 2 * (T_p / T_t )

                                  = 41

There for the answer will be 40.

Answer 4

I m trying to tell all of u with some example lets forget 40 or 41 here probelm is not prpblem os what should be logic , so if calculate max window size we try to calculate max pkt we can send in time of 1st pkt reach at destination so according to this rtt includes both transmisson time as well as propogation time . Here rtt is given so we have not to think extra propagation time all r included in it so if u calculate u will find ans