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.

---

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

---

 

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

 

we can solve like this??

 

Comments

we

Answer 2

.