GATE CSE 2006 | Question: 44

Question

Station $A$ uses $32\; \text{byte}$ packets to transmit messages to Station $B$ using a sliding window protocol. The round trip delay between A and $B$ is $80\; \text{milliseconds}$ and the bottleneck bandwidth on the path between $A$ and $B$ is $128\; \text{kbps}$ . What is the optimal window size that $A$ should use?

• A. $20$
• B. $40$
• C. $160$
• D. $320$

Comments

@rfzahid

No, RTT simply means 2*Tp.

 
What you are trying to mention here is “Total Cycle Time” but not RTT.

 

Efficiency = Useful Time / Total Cycle Time .

 

but not,

Efficiency = Useful Time / RTT .

 

Note:  Total Cycle Time and RTT are Both Different.

 

So, This Question requires to select the Closest Answer. Those were the days when gate questions were little ambiguous, nowadays it is not the case, so no need to worry :-)

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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

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if the transmission time was considered 2 ms then the propagation time will be 78 ms
and the answer will still be 40

Answer 1

It is already mentioned in the question that bottleneck bandwidth is used and initial bandwidth is not given ie. no use to calculate TT.

So, no. of packets = (128×80)/(32×8) = 40.      Ans.

Answer 2

ans b)

Comments

can any one explain it.?please ...

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@harsha RTT=80 therfore TP=40  as RTT=2TP

TT= L/B   L= 32*8 = 256 and BW= 128 therfore TT=2

window size = (1+2tp/tt)

1+2(40/2) = 41 which is very close to B)

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What if 41 was in options along with 40?

Answer 3

L = 32 Byte = 32*8 = 256 bits

B/w= 128 Kbps

Tt= L/B =2 ms

tp= RTT/2 = 40 ms

optimum window size = 1+2a

= 1+2*40/2 = 41 

Answer 4

$\underline{\textbf{Answer:}\Rightarrow}\;\mathbf {(B)}$

$\underline{\textbf{Explanation:}\Rightarrow}$

Round Trip propogation delay $=80\;\text{ms}$

Frame Size $=32\times 8\;\text{bits}$

Transmission Time $=\mathbf{\dfrac{L}{B}} = \dfrac{32 \times 8}{128} \;\text{ms} = 2\;\text{ms}$

Let $\mathbf n$ be the window size.

Utilization $=\mathbf{\dfrac{n}{1+2a}}$, where $\mathrm {a = \dfrac{Propogation\; time}{transmission\; time}} = \dfrac{\mathrm n}{1+\dfrac{80}{2}}$

For maximum Utilization Efficiency $ = 1$

$\Rightarrow 1 = \dfrac{\mathrm n}{1+\dfrac{80}{2}}\\ \Rightarrow\mathrm n = 41$

Which is close to option $\mathbf{ (B)}$

$\therefore\;\mathbf {(B)}$ is the correct answer.

Comments

Why we are not converting 128 kbps to bytes per sec by multiplying it with 2^10 and then multiplying it with 8 to convert it into bits. Please explain.