A. Assume a computer has a physical memory organized into 64-bit words. Give the word address and offset within the word for each of the following byte addresses:

0, 9, 27, 31,120, and 256.

b. Extend the above exercise by writing a computer program that computes the answer. The program should take a series of inputs that each consist of two values: a word size specified in bits and a byte address. For each input, the program should generate a word address and offset within the word. Note: although it is specified in bits, the word size must be a power of two bytes.

Check the explanation

Explanation:

For 9th byte , it is from 8*8 bit to 9"8th bit so each word consists of 64 bits , to find word address u have to divide 8*8 by 64.

Offset within word = 9*8modulo 64.

For 27th byte , word address = 8*27/64.

Offset within word = 27* 8 modulo 64

For 21th byte , word address = 8*31/64

Offset within the word = 31*8 modulo 64

For 120 , word address = 8*120/64

Offset within the word = 120*8 modulo 64.

solution:

this situation is called an ip conflict.

data from a server is sent in small bits called packets. in an ip conflict, some packets will end up in one computer while some other packets will end up in another computer.

so…overall none of the computers get the data as the original message or picture being sent by the server cannot be retrieved with only half the packets. so it will show up as an error.

thus the final answer is d.