134 views

closed as a duplicate of: tanenbaum
closed | 134 views
0
given 26 minute for this question but not solved

how weak my concept
0

I am getting 2.97%

0

Check my approach @ srestha mam

+1 vote
Error Rate = 1\100 ie, so out of  100 frames only 1 frame is retransmited.

Header = 40 bits., so  For 100 Frames = 4000 bits(40*100)

Total useful Data = 3960 *100 = 396000 bits

Now one Frame is Retransmitted = 3960(data) + 40 bits(header) => 4000 bits

Also NAK is sent = 40 bits

=> 4000 + 4000 + 40 / 396000 + 4000 + 4000 + 40

=> 0.0198

so in percentage % => 0.0198 * 100 => 1.98 %
0
What is the significance of sequence number bits here?
0
with 8  bit sequence number , sender window size = 2^(8-1) = 2^7 = 128  bt for simplicity i m sending 100 frames ... bt u can send 128  frames even and tht answer will be more appropriate
0
@Manas-What would be total transmissions for 128 frames?
+1
See I did like this. Let me know if I am in the correct direction.

The sequence number bits $m=8$

Protocol is Selective repeat, so $W_s=128$

Now since this protocol, unlike GBN, allows out of order packets, NAK will be generated only for corrupted packets.

The error rate for a data packet=1%.

So, the number of NAK generated(hence the number of retransmission needed)=$\lceil 0.01 \times 128 \rceil=2$

So, in total sender transmitted 130 frames.

Total data transmitted=$(130 \times 4000)+(2 \times 40)(NAK)=520080$

Overhead data sent=$(128 \times 40)+(2 \times 4000)(retransmissions)+(2 \times40)(NAK)=13200$

Bandwidth wasted=$\frac{13200}{520080}=2.53\,percent$

I am getting another answer :(
+1
@ayush first of all NAK generated  = floor(0.01*128) = 1

so only 1 retransmission needed

Now caclculate u will get the right answer
0

+1 vote
1
2
+1 vote