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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 ago in Computer Networks by Loyal (5.6k points) | 36 views
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Na462  what's the answer given ??

16 ??

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144 ?
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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

 

 

+2
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.
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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 :(

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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

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if it were a go back n then ?
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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)

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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 ?
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Na462  yes you are right

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