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

This is not stop-and-wait. It is about maximally packing the link, so only one-way propagation delay matters. The factor 2 × Tp is only for protocols that wait for ACK before sending the next frame.

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No protocol is mentioned as well as asking sequence no.

2^n >= Tp/Tt

Tp/Tt = 25/1 = 25

Hence, if we take n = 5 then 32 >= 25 ........condition satified

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iarnav Thanks for ur explanation .

In Transit Vs Full RTT 

In Transit -->  This problem specifically asks to pack the link in transit (within the physical channel during propagation), which requires $T_p / T_t = 25$ frames.

Full RTT Windo w==> If the question instead asked for the optimal sender window size for 100% throughput across the full round-trip time without waiting for acknowledgments, it would be $1 + 2a = 1 + \frac{2 T_p}{T_t} = 51$ frames

Answer 1

For Half Duplex, Capacity of the channel = BW * T_p
So we get capacity = $10^6$ b/s * 25 * 10 ^ -3 s
                               = 25 * $10^3 bits$

so a channel can contain 25 * $10^3 bits$ but we are sending only frames of 1000 bits
Therefore, the total no. of frames we are sending with its full channel capacity = 25 * 1000/1000 = 25 frames.
For this much frames the seq.numbers nedded are 25.
And Seq.no.bits required are $Ceil(\log_{2}25)$ = 5 bits

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

Propagation time is given as 25 ms.

Bandwidth = 106 bps.

So, to fully utilize the channel, we must send 106 bits into the channel in a second, which will be 1000 frames per second as each frame is 1000 bits. Now, since the propagation time is 25 ms, to fully pack the link we need to send at least 1000 * 25 * 10-3 = 25 frames. Since, to represent 25 frames we need 25 distinct index numbers. Thus, minimum no. of bits required to do this will be 5 (2^5=32) as 4 will not be enough (2^4=16, not enough to represent 25 distinct no.)

Comments

@Arjun  Sir ,why we are not considering the other tp for acknowledgement. in this time, we can sent more 25 packets.

Answer 4

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 5

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

Answer 6

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.