GATE CSE 2009 | Question: 57, ISRO2016-75

Question

Frames of $\text{1000 bits}$ are sent over a $10^6$ $\text{bps}$ duplex link between two hosts. The propagation time is $\text{25 ms}$. Frames are to be transmitted into this link to maximally pack them in transit (within the link).

What is the minimum number of bits $(I)$ that will be required to represent the sequence numbers distinctly? Assume that no time gap needs to be given between transmission of two frames.

• A. $I=2$
• B. $I=3$
• C. $I=4$
• D. $I=5$

Comments

Imagine the network as a pipe, the length of the pipe is the propagation delay, and diameter of pipe as the bandwidth.

At any time it can contain $\frac{10^{6} \text{ bits}}{\text{ sec}} \times \frac{25}{1000}\text{ sec} = 25,000 \text{ bits} \implies 25 \text{ frames}.$

Each of these $25$ frames must have different sequence number, so, we need $\lceil\log_{2} 25\rceil=5\;\text{bits}.$

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Cool

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GO to the Root of Concept: Video Explanation with timestamp: Efficiency,Throughput| Relation Between WindowSize & Sequence No. Sliding Window | With NOTES

Answer 1

L=1000b,B=1000000b/s

Tt = L/B = 1ms

Tp = 25 ms

a  = Tp/Tt

For maximising the efficiency,no of frames that must be transmitted is 1+2a

thus no. of frames is 51

so no of bits required to represent 51 is 5(ans).

Answer 2

I think this the correct way. Please check if incorrect

Answer 3

Since, it is duplex link Therefore PIGGYBACKING is used.

Formula to be used here is
$\eta = \frac{w*2tx}{2tx + 2tp}$
where n = 1 here

Since in Piggybacking with frame ack. is also sent so txa should not be ignored here and txa=tx in piggybacking.

Therefore, Useful time = 2tx and total time is 2tx+2tp.

Use this formula you will get the answer.

Answer 4

maximally pack them in transit 

Bits will only stay  in transit for max 25ms, because that's the propagation delay.

 

In $1$ second —> $10^6$ bits are passed through the link at max.

In $25 ms$ —> $25m * 10^6$ bits are passed through the link at max.

=> In $ 25 ms$ —> $25 * 10^3$ bits are passed through the link at max.

=> In $25ms$ —> $25 packets$ are passed through the link at max..

 

To uniquely identify 25 packets., we need 5 bits.

 

Option D