EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
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Expert Solution & Answer
Chapter 6, Problem P4P
a.
Explanation of Solution
Finding binary numbers from 1 through 10:
| Numbers | Binary numbers |
| 1 | 0000 0001 |
| 2 | 0000 0010 |
| 3 | 0000 0011 |
b.
Explanation of Solution
ASCII values for the letters from “B” through “K”:
| Letters | ASCII values | Binary number |
| B | 66 | 0100 0010 |
| C | 67 | 0100 0011 |
| D | 68 | 0100 0100 |
c.
Explanation of Solution
ASCII values for the letters from “b” through “k”:
| Letters | ASCII values | Binary number |
| b | 98 | 0110 0010 |
| c | 99 | 0110 0011 |
| d | 100 | 0110 0100 |
d.
Explanation of Solution
Steps to find internet checksum:
Step 1: Find ASCII value for the string.
Step 2: Find 8-bit binary number for the corresponding ASCII values.
Step 3: Add 16-bit numbers. If carry occurs, then wrap that around.
Step 4: Take 1’s complement for the final result to find out internet checksum.
For part (a) the internet checksum is calculated as follows:
Add all the 16-bit integers:
The above result doesn’t produce carry and therefore take 1’s complement.
1’s complement for “0001 1001 0001 1110” is “1110 0110 1110 0001”.
Therefore, internet checksum for the numbers 1 through 10 is “1110 0110 1110 0001”.
For part (b) the internet checksum is calculated as follows:
Add all the 16-bit integers:
The above result produces carry bit so carry forward it.
Now, take 1’s complement. 1’s complement for “0101 1111 0110 0100” is “1010 0000 1001 1011”.
Therefore, internet checksum for the letters “B” through “K” is “1010 0000 1001 1011”...
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Chapter 6 Solutions
EBK COMPUTER NETWORKING
Ch. 6 - Consider the transportation analogy in Section...Ch. 6 - If all the links in the Internet were to provide...Ch. 6 - Prob. R3RQCh. 6 - Prob. R4RQCh. 6 - Prob. R5RQCh. 6 - Prob. R6RQCh. 6 - Prob. R7RQCh. 6 - Prob. R8RQCh. 6 - Prob. R9RQCh. 6 - Prob. R10RQ
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