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

More clarity for concept on this Question

Reference : Wikipedia & Andrew S. Tanenbaum

==> To achieve 100% throughput utilization, a sender's transmission window size ($W$) must be at least equal to the BDP. If $W < \text{BDP}$, the sender runs out of permitted data to send and sits idle while waiting for ACKs to return.

i.e $\text{Window size} \times \text{Packet size} \ge \text{BDP}$

Reference :  Tanenbaum

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Meaning of Bottleneck bandwidth 

References : Kurose ,A Top-Down Approach, 7th Edition Page (Delay ,loss chapter)

 

Bottleneck link as "the link along the end-to-end path with the smallest transmission rate." The end-to-end throughput is constrained to be no greater than the bottleneck bandwidth ($R = \min(R_1, R_2, \dots, R_N)$).

 

Meaning : No matter how fast other links on the path are , the maximum data rate the entire path can sustain is limited by the slowest link.

 

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The main thing is that the formula 1/(1+2a) is derived using the concept of round trip time. Then, if the question itself mentions the round trip time, how can we take the formula?
The efficiency is defined as time spent transmitting useful data/ total time taken , not 1/(1+2a) ,which is something we derived using the concept.

Answer 1

Hope This Helps !🦚 

Answer 2

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 3

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

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

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.

Answer 4

Round trip time = 2*Tp =  80 * 10^-3 s

Length of packet (L) = 32 * 8 bits

Bandwidth = 128 * 10³ bits/s

Transmission time(T_t) = L / bandwidth

                                   = ( 32 * 8 ) / ( 128 * 10³)

                                   = 2 * 10^-3 s

Optimal window size = 1 + 2 * (T_p / T_t )

                                  = 41

There for the answer will be 40.

Answer 5

Basically, Acknowledgement will be received by A will be 80 millisecond. Now,

A can send 128kb in 1 sec  (bandwidth),

In 1 sec ----→ 128kb

    1 sec ---→ (128kb/(32*8 bits))   packets  ( As each packet of size 32*8 bits)

    solving above, A can send 500 packet per second  

So, In 80ms  ---→ 500* 80*10^-3  = 40 packets

So, A can send 40 packets before it receives an acknowledgement.

So, Optimal window size for will be 40

Answer 6

we can send as frames as possible in sliding window protocol in a round trip time

T $_{t}$ =   $_{\tfrac{size of frame}{bandwidth}}$

T$_{t}$ =$_{\tfrac{32*8 bits}{128*10$^{3}$ bps}}$ = 2msec

no.of frames=$_{\tfrac{RTT}{T_{t} }}$= $_{\tfrac{80 msec}{20 msec}}$= 40 frames