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Station $A$ uses $32$ byte packets t 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. $40$

### 1 comment

option B is correct

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

### 1 comment

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
by

B = 64 Kbps

RTT = 40 ms

No of bits which can be transmitted in 1 sec are

1 sec ---> 64 * 10³ bits

No of bits which can be transmitted in RTT

40 * 10^(-3) ---> 64 * 40 bits

Size of packet = 32 B

Size of window = (64*40)/(32*8) = 10

Hence option 2) is correct

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

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

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

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.

by

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

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