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GO Classes DPP | CN | SLIDING WINDOW

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6 6 votes

Server S is transmitting data to client C over a full-duplex connection. S and C use the sliding window protocol for flow control. The send and receive window sizes are $8$ packets each. Data packets (sent from S to C) are $1500$ bytes long, and the connection's bandwidth is $200$ Mbps. Acknowledgement packets (sent from C to S ) are very small and have a negligible transmission time. The propagation delay for the connection is $270 \mu \mathrm{~s}$. What is the maximum possible throughput for this data transmission?

  1. $125.50 \times 10^6 \mathrm{Bps}$
     
  2. $160.00 \times 10^6 \mathrm{Bps}$
     
  3. $177.78 \times 10^6 \mathrm{Bps}$
     
  4. $200.00 \times 10^6 \mathrm{Bps}$
  • 🚩 Edit necessary | 👮 Prashant-G | 💬 “In Computer Network Bydefult B - Bytes & b - bits so answer either be 20 X10⁶ Bps or you have to mention specifically B means bits here”

3 Answers

2 2 votes
  • Propagation Delay $\left(T_p\right)=270 \mu s$
     
  • Transmission Delay $\left(T_t\right)=60 \mu s$ (as calculated previously)

$$
a=\frac{T_p}{T_t}=\frac{270 \mu s}{60 \mu s}=4.5
$$


Next, we calculate the efficiency of the sliding window protocol. The formula for efficiency is:

$$
\eta=\frac{N}{1+2 a}
$$


Where:

  • $N$ is the window size $=8$

  • $a$ is the ratio we just calculated $=4.5$

$$
\eta=\frac{8}{1+2 \times 4.5}=\frac{8}{1+9}=\frac{8}{10}=0.8
$$

  • Efficiency $(\eta)=0.8$
     
  • Bandwidth $(B W)=200 \mathrm{Mbps}$

$$
\text { Throughput }=\eta \times B W=0.8 \times 200 \mathrm{Mbps}=160 \mathrm{Mbps}
$$

0 0 votes
throughput is number of bytes sending per second
for max
max number of bytes sending per second
in RTT time we are sending 8 packets of 1500 byte
and RTT time is =Tt+2Tp
Tt is 60 us here(use length /bw)
now
RTT is 600 us
so in 600 us we are sending 1500 bytes *8 packets so now find
throughput which is
 1500*8 bytes/600 us
which is
1500*8*8 bit /600 us
gives us 160*10'6 Bps
0 0 votes

“In Computer Network Bydefult B - Bytes & b - bits so answer either be 20 X10⁶ Bps or you have to mention specifically B means bits here”

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