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

More clarity for concept on this Question

Reference : Wikipedia & Andrew S. Tanenbaum

==> To achieve 100% throughput utilization, a sender's transmission window size ($W$) must be at least equal to the BDP. If $W < \text{BDP}$, the sender runs out of permitted data to send and sits idle while waiting for ACKs to return.

i.e $\text{Window size} \times \text{Packet size} \ge \text{BDP}$

Reference :  Tanenbaum

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Meaning of Bottleneck bandwidth 

References : Kurose ,A Top-Down Approach, 7th Edition Page (Delay ,loss chapter)

 

Bottleneck link as "the link along the end-to-end path with the smallest transmission rate." The end-to-end throughput is constrained to be no greater than the bottleneck bandwidth ($R = \min(R_1, R_2, \dots, R_N)$).

 

Meaning : No matter how fast other links on the path are , the maximum data rate the entire path can sustain is limited by the slowest link.

 

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The main thing is that the formula 1/(1+2a) is derived using the concept of round trip time. Then, if the question itself mentions the round trip time, how can we take the formula?
The efficiency is defined as time spent transmitting useful data/ total time taken , not 1/(1+2a) ,which is something we derived using the concept.

Answer 1

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: $B$

Comments

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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Because there it is said the round trip propagation delay , we often say 1 time propagation but there they have talked about only the prop time in both sending the data and receiving the ACK

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@Jyotish Ranjan

rtt = tt + 2tp = 80 ... so 2tp = 78 not 80 ... therefore 1+78/2 = 1+39 =40

Answer 2

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).

Comments

yes i am also getting 41 , but 41 is NOT in any options

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it can be trouble in NAT type questions

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thx @Ritik gupta

Answer 3

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)

Answer 4

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.html, https://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.

Answer 5

It is already mentioned in the question that bottleneck bandwidth is used and initial bandwidth is not given ie. no use to calculate TT.

So, no. of packets = (128×80)/(32×8) = 40.      Ans.

Answer 6

ans b)

Comments

can any one explain it.?please ...

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@harsha RTT=80 therfore TP=40  as RTT=2TP

TT= L/B   L= 32*8 = 256 and BW= 128 therfore TT=2

window size = (1+2tp/tt)

1+2(40/2) = 41 which is very close to B)

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What if 41 was in options along with 40?