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49 votes
49 votes

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?

  1. $20$
  2. $40$
  3. $160$
  4. $320$
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4 Comments

RTT is 80ms and not 82ms. This 80ms already includes transmission time for first packet + propagation time of first packet to reach receiver + propagation time of ACK receiver sends after receiving first packet.

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@rfzahid I think now i am satisfied ...thanks bro I have been searching since long time but not get what i wanted .....whenever RTT given Then use  2a formula else use 1+2a........because some people take RTT =2Tp thats why this confusion arises....But actually  RTT  should be Tx+2Tp........let me know if i am wrong.........thanku very much

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@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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10 Answers

60 votes
60 votes
Best answer
Round trip delay $= 80\ ms$.

Quoting from Wikipedia

the round-trip delay time (RTD) or round-trip time (RTT) is the length of time
it takes for a signal to be sent plus the length of time it takes for an acknowledgment
of that signal to be received.

Now, in many books including standard ones, they have used $\text{RTT}$ to mean just the $2\text{-way}$ propagation delay by considering the signal/packet as of the smallest possible quantity so that its transmission time is negligible. The given question is following the first definition as given by Wikipedia which is clear from the choices.

During this time the first $\text{ACK}$ arrives and so sender can continue sending frames.
So, for maximum utilization sender should have used the full bandwidth during this time.
i.e., it should have sent $128\text{ kbps}\times 80\ ms$ amount of data and a packet being of size $32\text{ bytes},$ we get

no. of packets $=\dfrac{128 \times 80}{32 \times 8} = 40.$
Correct Answer: $B$
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4 Comments

@pritishc it's transmission time that could be neglected, not propagation delay... Propagation delays always considered.

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A/C TO YOUR DIAGRAM 40 FRAMES WILL SENT WITHIN 80ms BUT YOU FORGET TO ADD THE 1st FRAME.

AT t=0 TRANSMISSION OF FIRST FRAME STARTS 

AT t=2 TRANSMISSION OF 2nd FRAME.

SIMILARLY, 

AT t=82 TRANSMISSION OF 41st FRAME STARTS.

SO, I THINK IT MUST BE 41.

PLEASE CHECK @

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can anyone please explain why RTT = 2TP in this qn https://gateoverflow.in/54975/isro2014-45

if we solve this by using RTT = 2 * TP then answer is 270.

and if we solve this by using RTT = TT + 2*Tp then answer should be 260.

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73 votes
73 votes

Answer: B

Round Trip Time $= 80ms$
Frame size $=32\times 8\text{ bits}$
Bandwidth $=128\text{ kbps}$
Transmission Time $=\dfrac{32\times 8}{128}\ ms = 2\ ms$
Let $n$ be the window size.

Utilization $=\dfrac{n}{1+2a}$

where $\large a = \dfrac{\text{Propagation Time}}{\text{Transmission Time}}$

$ =\dfrac{n}{1+\dfrac{2 \times 40}{2}}$

For maximum utilization: $n = 41$ which is close to option (B).

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4 Comments

reshown by
Here the bandwidth mentioned as 128kilo bytes per second....so dont we need to write in the denominator as 128×8....
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k.eswar prasanth No! 128 Kbps means 128 * 10^3 bits per second. 

 

if it's 128Mbps it would be 128*10^6 bits per second. 

There's no multiplying by 8 

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The tp you calculate is incorrect it should be 39 not 40 because Tt+2 Tp =80ms

as Tt=2ms so Tp =80-2/2 = 39 so it should be 1=n/1+39 so n=40 not 41 .
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11 votes
11 votes
RTT is 80 ms , irrespective of what T(p) & T(t) values are .

The acknowledgment is reaching at 80th ms and not 82nd ms (If calcualated T(t) values and added to twice of T(p) = 40 )

Here T(t) + 2 T(p) = 80 ms

 

So 128 kb  == 1 sec
      128 bits == 1 ms
       1ms == 128 bits
        80 ms == 80 * 128 bits

So no of frame required = ( 80 * 128 ) / 256
                                   = 40 frames (Window Size)
9 votes
9 votes

Here, we can use the concept of bandwidth delay product.

What is bandwidth delay product?

"It's the amount of data that left the sender before the first acknowledgement was received by the sender. That is, bandwidth * round trip time = the desired window size under perfect conditions. If the round trip time is measured from the last packet and the sender's outbound bandwidth is perfectly stable and fully used, then the measured window size exactly calculates the number of packets (data and ACKs together) in transit. If you want only one direction, divide the quantity by two."

Further reading: http://www.kehlet.cx/articles/99.htmlhttps://learningnetwork.cisco.com/thread/125627

The bandwidth delay product is given by:  $BW \times RTT$, which in this case is $128 \,\text{kbps} \times 80 \times 10^{-3} \text{s}$, which comes as $10240$ bits.

Therefore, number of packets can be given by $\frac{10240}{32 \times 8} = 40$ packets, which does not have any off by one error.

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