ISRO2018-44

Question

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

1. $20$
2. $40$
3. $160$
4. $320$

Answer 1

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

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

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.

Answer 2

OPTION B

Answer 3

By the given data first we have to calculate transmission time Tt = 32*8/128 kbps = 2msec. and propagation time became half of the RTT i.s 40msec.
now optimal window size  = 2 *40/2 = 40.

Comments

Could you explain how you concluded what the formual for optimal window size is ??

Answer 4

Ans:B

Since using sliding window protocol we can send multiple(n) packets in a RTT.

To use the time most efficiently we have to send as many packets as possible in the available time that is try to maximize the efficiency

given

$RTT=80ms$,

$packet size=32byte=32*8bits$

$BW=128kbps$

$Efficiency=\frac{n*T_t}{RTT}=1$

then

$n=\frac{RTT}{T_t}$

$T_t=\frac{L}{BW}=\frac{32*8}{128*10^3}=\frac{2}{10^3}s=2ms$

$n=\frac{80ms}{2ms}=40$