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Q3(b)(ii) · Internet checksum

2023-01-27 · AI-only Sem V

Exact historical variant. Current-syllabus questions solved; excluded topics and source limitations stated.

Q3(b)(ii). Compute the Internet checksum of 10110100 01000110 and 11001000 01101110. Write the result as two 8-bit groups separated by one space. [6 marks]

Answer

Given

1011010001000110, 1100100001101110; word width = 16 bits.

To find

Sender checksum and receiver verification.

Formula and why

S=\operatorname{endAroundCarry}(\text{sum of words}),\qquad C=\operatorname{NOT}(S)LaTeX: S=\operatorname{endAroundCarry}(\text{sum of words}),\qquad C=\operatorname{NOT}(S)

One’s-complement addition returns an overflow carry to the least-significant end. Complement only after all words have been added.

Solution

Step 1. Add the words, folding each carry

WordDecimalRaw additionFolded additionRunning binary sum
1011010001000110461504615046150 + carry 0 = 461501011010001000110
1100100001101110513109746031924 + carry 1 = 319250111110010110101

Step 2. Complement every bit

Running sum = 0111110010110101. Change every 0 to 1 and every 1 to 0: checksum = 1000001101001010 (decimal 33610). In the requested two-byte format: 10000011 01001010.

Step 3. Receiver adds the data and checksum

31925 + 33610 = 65535 = 1111111111111111. All bits are 1; complement gives all zeros. Therefore this intact message passes the checksum check. A matching checksum detects no error; it does not prove that every possible corruption is absent.