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A message M = 11111111 after bit stuffing with End delimeter as 111 will be sent as____________?
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is it $\underbrace{110} \underbrace{110}\underbrace{110}\underbrace{110}$

$\text{110 will be treated as 11}$
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@Anand: 110110110110111...last bit should be 1
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Anand, not getting you... I think as per question last should be 111 and all double 1's (11) will converted to 110 appended by 111 at last...
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$111$ is delimeter, it means that on seeing $11$, a $0$ will be appended.

And at the reciever side $110$ will be decoded as $11$
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110 110 110 110 as after 11 we have to add 0 which will be removed at the receivers end.
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@sumit...but what is the use of end delimiter if it is not sent with message ??

What I want to say is 110 110 110 110 111 this is the message we sent to receiver after bit stuffing i.e. along with terminator....please correct me If I am wrong...

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Yes you are right we have to add the end delimiter in the end otherwise how the send will know about the stuffing we did.
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that's what I write in 2nd comment...
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110110110110.

It is very simple
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In such questions i never see end delimiter at the end ,after bit stuffing. Like 110110110110111.

May be because it is not the part of bit stuffing.

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Its just like flag instead of flag it says end delimeter so we have to add that in the last so that receiver understands that it is the ending of data of that frame so abbas is right
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Did u see any such question where after bit stuffing end delimiter is added in answer? I'm not talking about what is actual phenomenon or in reality how message is send.
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No, we don't have to add the end delimiter. I had gone through some previous year question and there was not even a single question where they have added the end delimiter.

It is nothing but like a flag. Do the bit stuffing and leave the rest.
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Yes. exactly the same thing i tried to say.

by Loyal (9k points)
End delimiter =111 (stuff bit after 11)

data send by sender=11111111

ans=after 11 put one 0.

data =110110110110.
by (77 points)

+1 vote
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