GATE CSE 2003 | Question: 84

Question

Host $A$ is sending data to host $B$ over a full duplex link. $A$ and $B$ are using the sliding window protocol for flow control. The send and receive window sizes are $5$ packets each. Data packets (sent only from $A$ to $B$) are all $1000$ bytes long and the transmission time for such a packet is $50$ $μs$. Acknowledgment packets (sent only from $B$ to $A$) are very small and require negligible transmission time. The propagation delay over the link is $200$ $μs$. What is the maximum achievable throughput in this communication?

• A. $7.69 \times 10^6$ Bps
• B. $11.11 \times 10^6$ Bps
• C. $12.33 \times 10^6$ Bps
• D. $15.00 \times 10^6$ Bps

Comments

How half & full duplex system affects the SR protocol?i am not getting this. @Pranavpurkar

---

Because of full duplex link ,while sending the window of packets to the Receiver over a channel, the Destination can also send Acknowledgement packet to the Sender over the same channel

@Nisha Bharti 

---

GO to the Root of Concept: Video Explanation with timestamp: Sliding Window PYQs| GATE 2009 Q57, 58| 2006 Q46, 64| 2003, 2014, 2015, 2004, 2016, 2007| With NOTES

Answer 1

The efficiency $η$ is given as $\frac{N}{1+2a}$

And $Throughput$ $=$ $η*B$

Substitute all the values, and see the image, you will get $11.11 * 10^{6} bytes persecond$

Answer 2

we know, throughput = (total data size)/(Tf + 2Tp)

given: no. of packets= 5

           packet size = 1000 bytes

so,total data size = 5 * 1000 = 5000 bytes

now, Tf = 5*50 microsec (as, total 5 packets are there and each takes 50 microsec)

here, actually 2Tp = 200 microsec is given

reasons:

* it says propogation time over the link means it covers the ack delay as well.

* it's also given in the question that Tf(ack) is negligible....these points sud b kept in mind while solving this ques.

now, throughput = 5000/(250+200) = 11.11 * 10^6 bytes per sec.


one more point i wud clear here...u guys look confused in bits and bytes....

b     indicates bits

and

B     inficates bytes...👍

Answer 3

Answer is (B)

max throughput =  Max data that can be sent / Total time taken

Total time = TR + 2TP = 20 + 2x 200 = 450 Microsecond = 450/10^6 sec.

Given that Sender and Receiver window size is equal hence send  5 Packet each of 1000 bytes.

Therefore we get:     ((5 x 1000) x 10^-6)/450 = 11.11 x 10^6.