GATE Overflow Test Series | Computer Networks | Test 1 | Question: 20

Question

Host $P$ is sending data to host $Q$ over a full duplex link. $P$ and $Q$ are using the sliding window protocol for flow control. The send and receive window sizes are $4$ packets each. Data packets $\text{(sent only from}\; P\;\text{to}\; Q)$ are all $800\;\text{bytes}$ long and the link capacity is $160\;\text{Mbps}.$ Acknowledgment packets $\text{(sent only from}\; Y \;\text{to}\; X)$ are very small and require negligible transmission time. The propagation delay over the link is $120\;\mu s$. The maximum achievable throughput (in Mbps rounded to one decimal points) in this communication is __________

Answer 1

For maximum throughput, the sender must continuously send packets until ACK comes back (RTT).
    
    RTT $=$ Transmission Time $+\; 2$ Propagation Delay
    $\quad = \dfrac{800 \times 8}{160}+ 2 \times 120 \; \mu s = 280\; \mu s $    

Out of this $280\; \mu s$ only $40$ is frame transmission time. For maximum throughput, sender must sent frames for the remaining $280-40 = 240\; \mu s$ as well which means $240/40 = 6$ more frames. That is, with $1 + 6 = 7$ frames, full throughput is achieved. The window size given is $4.$ So, the efficiency will be $\dfrac{4 \times 40}{280} = 0.5714$
    
Achieved throughput $ = $ Efficiency $\times$ Bandwidth
    
$\qquad  = 0.5714 \times 160\;\text{Mbps} = 91.424\;\text{Mbps}$

Comments

Arjun sir

This question is creating confusion in many people head

Will you please take out time and make a little video on this topic only for 3 minutes only and share here

Request you

Thanks

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It's a SR protocol problem Sender window=Receiver Window

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Thank you so much. It seems like a previous year gate question

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A little typo, you wrote msec for Tp.