Compute 16-bit One's Complement Sum (memo)

 

Compute 16-bit One's Complement Sum

Date: 03/20/2002 at 13:54:13
From: John
Subject: How to compute 16-bit one's complement sum

I would like to know how to compute one's complement sum. In my TCP/IP 
book it says, "to compute the IP checksum for outgoing datagram, the 
value of checksum field is set to zero, then the 16-bit one's 
complement sum of the header is calculated (i.e. the entire header is 
considered a sequence of 16-bit words)" - but I do not know how to 
compute the one's complement sum. Please help.

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Date: 03/21/2002 at 01:27:19
From: Doctor Twe
Subject: Re: How to compute 16-bit one's complement sum

Hi John - thanks for writing to Dr. Math.

What that means is:

1. Add the 16-bit values up. Each time a carry-out (17th bit) is 
produced, swing that bit around and add it back into the LSb (one's 
digit). This is somewhat erroneously referred to as "one's complement 
addition."

2. Once all the values are added in this manner, invert all the bits 
in the result. A binary value that has all the bits of another binary 
value inverted is called its "one's complement," or simply its 
"complement."

For example, suppose our header has the following data. I separate the 
data into groups of 4 bits only for readability. (I know that headers 
are really longer than 8 bytes, but this will demonstrate the 
process):

     1000 0110 0101 1110
     1010 1100 0110 0000
     0111 0001 0010 1010
     1000 0001 1011 0101

First, we add the 16-bit values 2 at a time:

         1000 0110 0101 1110   First 16-bit value 
     +   1010 1100 0110 0000   Second 16-bit value
       ---------------------
       1 0011 0010 1011 1110   Produced a carry-out, which gets added
     +  \----------------> 1   back into LBb
       ---------------------
         0011 0010 1011 1111
     +   0111 0001 0010 1010   Third 16-bit value
       ---------------------
       0 1010 0011 1110 1001   No carry to swing around (**)
     +   1000 0001 1011 0101   Fourth 16-bit value
       ---------------------
       1 0010 0101 1001 1110   Produced a carry-out, which gets added
     +  \----------------> 1   back into LBb
       ---------------------
         0010 0101 1001 1111   Our "one's complement sum"

(**) Note that we could "swing around" the carry-out of 0, but adding 
0 back into the LSb has no affect on the sum. (But technically, that's 
what the checksum generator does.)

Of course, we'd continue to add the rest of the header data in 16-bit 
segments until all data were included. 

Then we have to take the one's complement of the sum. We do this by 
simply inverting all the bits in the final result from above:

         0010 0101 1001 1111   Our "one's complement sum"

         1101 1010 0110 0000   The "one's complement"

So the checksum stored in the header would be 1101 1010 0110 0000.

I hope this helps. If you have any more questions, write back.

Refer to http://mathforum.org/library/drmath/view/54379.html and http://en.wikipedia.org/wiki/Header_checksum