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

Answer 1

simply:

L = 1000 bits (frame size)

Tp = 25 ms  (propagation delay)

B = 10^6 bps  (Bandwidth)

Tt = L/B = 1000/1000000 = 1 ms for 1 frame

in question given there is no any gap between two frames

so , i frame travel from sender to receiver there are 25 frames comes on link for fully utilize the link

Here 25 frames on link , 

so no. of required bits = ceil(log25)) = 5 

correct option (C)

Answer 2

Actually, the question is asking the bits needed or frames needed so as to pack the link. We don't need to think more beyond this.  

Now, $1$st bit will take $25$ms to reach other host. So, to fill the link we need frames occupying the link all this while which is $25$ms. Since $1$ frame is being transmitted in $1$ms, a host need to fire at least $25$ frames at once to keep the link fully occupied.  Therefore, $5$ bits needed.

Answer 3

Here transmission time Tt = L/B = 1000/10^6 = 1 ms

i.e  to transmit 1 frame we need 1 ms.

Now we are given that propagation time Tp=25 ms i.e in Tp time we can send 1*25 = 25 frames only.

Therefore no of bits required for sequence no = ceil( log2(25) ) = 5 bits.

 

Answer 4

A possible ans:

Every 1 ms we hv a new frame on the line. (BW 10^6 bps and frame size 1000 b).

See it as the first bit of the frame, the first bit touches the other end at 25 ms delay.

How many frames have we pushed on the line during this 25 ms as we were pushing 1 frame each 1 ms?

25 frames on the line at a time.

To identify 25 frames, we need 5 bits.