Question

4. UDP and TCP use 1's complement for their checksums. Suppose you have the following two 16-bit words:

0101 0101 0101 0101 and 1010 1010 1010 1011.

4.a. (2 pts.) What is the 1's complement of the sum of these words? Show all work.

4.b. (2 pts.) With the 1's complement scheme, how does the receiver detect errors?

4.c. (2 pts.) Is it possible that a 4-bit error will go undetected? If so, provide an example.

4.d. (2 pts.) Is it possible that a 1-bit error will go undetected? If so, provide an example.4.

Answer 1

Answer is as follows :

We have two binary numbers say

A = 0101 0101 0101 0101

B = 1010 1010 1010 1011

So answers are as follows :

4.a)

We have to calculate sum of both A and B as follows :

0101 0101 0101 0101 + 1010 1010 1010 1011 = 1 0000 0000 0000 0000 (1 is carry so add it again we get) = 0000 0000 0000 0001

So the 1's complement is inverted bits of calculated sum i.e. 1111 1111 1111 1110

So our 1's complement checksum is 1111 1111 1111 1110

4.b)

At the receiver end, recevier add all the three i.e. A , B and Checksum. If the result comes 0 than there is error in the data. if it don't contains 0 than data is correct

4.c)

No, 4-bit error will not get undetected by this scheme.

4.d)

Yes, 1-bit error is detectable with this scheme.

Let us show,

We have

A = 0101 0101 0101 0101

B = 1010 1010 1010 1011

and calculated checksum is : 0000 0000 0000 0001

Let us suppose sender send this, but to transmission data A get changed 1-bit and data send to receiver is 0101 0101 0101 0100. (last bold bit is changed).

So the data recived by receiver is

A = 0101 0101 0101 0100

B = 1010 1010 1010 1011

and calculated checksum is : 1111 1111 1111 1110

Now we add these all as

0101 0101 0101 0100 + 1010 1010 1010 1011 = 1111 1111 1111 1111

Now checksum is added into this 1111 1111 1111 1111 + 1111 1111 1111 1110 = 11111 1111 1111 1101

We get a 0 in this, so there is error. we can detect it.

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