# UGCNET-DEC2016-III: 26

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Station A uses 32 byte packets to transmit messages to station B using sliding window protocol. The round trip delay between A and B is 40 milliseconds and the bottleneck bandwidth on the path between A and B is 64 kbps. The optimal window size of A is ______

1. 20
2. 10
3. 30
4. 40
in Others

Bottleneck  bandwidth = 64 kbps=64x103 bps

Round trip delay = 40 ms =40x10-3 sec

Total data $40\times 64\times 10^3 \times 10^{-3} bits = 40\times \dfrac{64}{8} bytes =320 bytes$

1 packet size = 32 byte

No. of packets 320/32  = 10

Hence (2 ) is the ans

edited by
0

Why you multiply delay with rate

Here is the link which has different way to solve same kind of problem, that seems correct

http://www.geeksforgeeks.org/computer-networks-set-11/

transmission delay = Packet Length / Bandwidth = 32 Byte/ (64/8 KByte);

transmission delay = 4 ms;

1 Packet = Transmission delay

?(N) = Transmission delay + RTT( 2 * Propagation delay )

No of packets (N) = (Transmission delay + RTT) / Transmission delay = 1 + 2a, where a = Propagation delay / transmission delay

N = 1+ ((2 *  20)/4)

Therefore,

1+10 = 11 Packets.

Usually Some author ignore "1" in "1+2a" formulae..

Hence, Answer can be 10 or 11

$\underline{\textbf{Answer:}\Rightarrow}\;\textbf{(B)}\;10$

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

$\underline{\mathbf{Method:\; 1}\Rightarrow}$

Round Trip propogation delay $=40\;\text{ms}$

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

Transmission Time $=\mathbf{\dfrac{L}{B}} = \dfrac{32 \times 8}{64} \;\text{ms} = 4\;\text{ms}$

Let $\mathbf n$ be the window size.
$\text{Utilizaton} =\mathbf{\dfrac{n}{1+2a}}$, where $\mathrm {\mathbf a = \dfrac{Propogation\; time}{transmission\; time}} = \dfrac{\mathrm n}{1+\dfrac{40}{4}}$

For maximum Utilization, Efficiency $= 1$

$\Rightarrow 1 = \dfrac{\mathrm n}{1+\dfrac{40}{4}}\\ \Rightarrow\mathrm n = 11$

Which is close to option $\mathbf B$

Or, we can do it another way as well. Like ignoring $\mathbf 1$ from $\mathbf{1+2a}$ which few authors do as well.

$\therefore\;\mathbf B$ is the correct answer.

$\underline{\textbf{Method:}\; \mathbf 2\Rightarrow}$

Round Trip propagation delay $=40\;\text{ms}$

Size of frame $= 32\times 8$

Bandwidth $=64\;\text{kbps}$

Total Data $= 40 \times 64\;\text{bits} = 40 \times \dfrac{64}{8}\;\text{Bytes}= 320 \;\text{Bytes}$

Size of single packet $= 32 \; \text{Bytes}$

Number of packets $= \dfrac{\text{Total Data}}{\text{Frame Size}}=\dfrac{320}{32} = 10 \;\text{Packets}$

$\therefore \mathbf{B}$ is the correct option.

edited by
0
why u took bandwidth as $128$ ? It's given as $64 kbps$ right ?
0
Edits
0
How is optimal window size related to maximum utilization efficiency??
0

@ Optimal window size will provide the maximum efficiency and the maximum efficiency that can be achieved is $\mathbf {100\%}$ or we can say $\mathbf {n = 1}$

Total data 40x64x103 x10-3 =  =320 bytes.

No. of packets 320/32  = 10

ANS: 2

Length=32*8bits

Bandwidth=128*10^3bps

Transmission time =frame length/ Bandwidth

So ,Transmission time=4ms

RTT=40ms

No of frames to achieve 100% utilization= RTT/Transmission time

So 40/2=20

20 is the optimal window size

If anyone is following the NPTEL course(Computer Networks and Internet Protocol), the course discusses the concept of BDP(Bandwidth Delay product) that can also be used to calculate the window size.

Given, RTT = 40ms, therefore one side time = 40/2 = 20ms

Hence,

BDP = 64kbps * 20ms = 1280 bits

packet size = 32byte = 256 bits,

number_of_packets = 1280/256 = 5

WindowSize = 2*no_of_packets + 1

w = 2*5 + 1 = 11 (we add +1 as the ACK is sent only when the first first segment is received)

Hence, (b) 10 is the correct option (closest option)

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