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If the Sender Window size is 128 using selective repeat ARQ. Then the sequence number of frame to be send after sending 400 th frame is ?
asked in Computer Networks by Loyal (8.6k points) | 372 views
0

Na462  what's the answer given ??

16 ??

0
144 ?
0

Rutvik Reshamwala how ??

128 * 3 = 384

384 + 16 = 400

as we know sequence number starts from 0 - 127

therefore after 400th frames sequence number should be 16

 

 

+4
For selective repeat ARQ,number of sequence numbers=2N since sender window size=receiver window size.Hence 0-255 would be consumed for sending 256 frames.Next 0-143 would be consumed for sending 144 frames hence total number of frames sent=256+144=400.Hence the sequence number for next frame should be 144.
0

yes you're right

thank you  Rutvik Reshamwala

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Rutvik Reshamwala , Brother 

Here in question it is given that 128 is window size so 2^(n-1) = 128 , n = No. of bits for Sequence number = 8

So total sequence number possible = 0 to 255.

So the sequence number of Frame = 400 % 256 = 144

So the frame after that should be 145 ????

I am confused :(

0
Brother in such question if i have k bit sequence number then the numbering of frames will always be from 0 to 2^k - 1, now its according to the protocol how much frame i can send at a time which constitutes the window size right ?
+1

Na462

as we know that sequence number is start from 0 to 255

So 400 % 256 = 144  [It means 144 th frames you send but sequence number of  144th frames should be 143 ]

after that 145 th frames should be send and sequence number should be 144

i:e ist frame is send but the sequence number of ist frame is '0'

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Ohk silly hell mistake magma is my below comment also correct about k bit sequence number ?
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in SR there is k bits sequence number

then senders window size = 2 n-1

and receiver windows size  = 2 n-1

total sequence number = 2n-1+2n-1  = 2(0 to 2 n-1 sequences we use)

yes you're right    Na462

0
if it were a go back n then ?
0

In case of GBN

in GBN there is n bits sequence number

then senders window size = 2 n - 1

and receiver windows size  = 1

total sequence number = 2n - 1 + 1   = 2(0 to 2 n- 1 sequences we use)

0
Here in this case available sequence number = 128 + 128 = 256 that's why (0 to 255)

for goback n Available sequence number = 128 + 1 = 129 so (0 to 128)

because sender WS = 2^k - 1 and receiver is 1.

Am i correct ?
0

Na462  yes you are right

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@magma please can you solve it for GBN and tell the answer. using same data!
+1
yeah sure ,

In GBN

No of sequence number = Sender's window size + 1 (receiver window size)

                                          = 128 +1

                                           = 129  [sequences  0 to 128]

No of sequence number consumed after sending the 400 frames  = 400 % 129 = 13

it means  0 to 12 sequences  we used already

after this we send a frame with sequence no : 13

understood ??
0
Yes magma! Perfect, thank you! :)

1 Answer

+2 votes

For selective repeat ARQ, we know that sender window size=receiver window size

and total no number of sequence numbers=2N (where N is the sender window size)

Hence 0-255 would be consumed for sending 256 frames.Next 0-143 would be consumed for sending 144 frames hence total number of frames sent=256+144=400.Hence the sequence number for next frame should be 144.

the sequence no of 400th frame will be 143 

but they are asking 

sequence number of frame to be send after sending 400 th frame is ?

so answer will be 144
 

answered by Loyal (8.3k points)
edited by

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