Question

4.

Find the checksum (one’s compliment) of the following set of data. (10 pts.)

Data 1: 01010101

Data 2: 11100001

Data 3: 11001100

5.

Consider a TCP connection between two hosts that are 500 miles away from each other

(propagation delay is 100 miles per msec.). Assume that there is no error or loss in this

communication and the receive window size is set to 64000 bytes. The size of data that needs

to be transmitted is 1.6 Mbytes. Assume that maximum size of each segment is 1600 bytes. If

this connection speed is 100 Mbps, what is the link utilization, ignoring the effects of the slow

start?

Answer 1

4. The different data are given below :

Data 1 : 01010101

Data 2 : 11100001

Data 3 : 11001100

All of the above are of size 8 bits.

It is not given what the size of the checksum should be so I am going to assume that it should be of 8 bits.

• First of all, we have to add these data segments.
• We have to calculate the result as 01010101 + 11100001 + 11001100
• We get the result as 01000000010.

The result is not of size 8 bits but rather of size 11 bits. In such a situation we have to wrao around the extra bits and add them to the result. So, we wrap around the extra 3 bits.

The result becomes 00000010 + 010 = 00000100

Finally, we take the 1's complement of this result. We get 11111011.

Thus, the checksum value is 11111011.

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