Question

This problem is to obtain checksum similar to that in UDP and TCP discussed in the lecture notes. Suppose the sender performs checksum on two 8-bit bytes A = 0101 1011 and B = 1100 1001, and then sends the both bytes along with the checksum to the receiver. What is the checksum of the two bytes?

Answer 1

The checksum for two 8-bit bytes A and B is calculated by performing binary addition, handling end-around carry, and then taking the bitwise complement of the result. The checksum for bytes A = 0101 1011 and B = 1100 1001 is 1111 1010.

Step-by-step explanation:

The checksum in protocols like UDP and TCP is used to detect errors in transmitted data. To calculate the checksum of the two 8-bit bytes A = 0101 1011 and B = 1100 1001, you first add them together with binary addition:

A = 0101 1011
B = 1100 1001
Sum = 1 0000 0100

Since we are only using 8-bit numbers, the leftmost '1' represents a carry-out and must be added back to the sum. This is known as end-around carry.

New Sum after adding carry-out:
0000 0100 (Sum, without the carry-out)
+ 0000 0001 (Carry-out)
=0000 0101

The checksum is the bitwise complement of this sum, which is 1111 1010. So, the checksum that the sender would send along with the two bytes to the receiver is 1111 1010.