Question
please all the questions ahove
41S .ill 88% 8:47 am https://interact2.csu.edu.au.. ?HELP I STUDENT PORTAL , pRINTING automatically lead to a fail grade for the subject irrespective of the marks obtained in all other assessments. Assessment item 4 back to top Assi ignment 2: MARIE and ISA Value: 15% Due Date: 30-Sep-2018 Return Date: 18-Oct-2018 Length Submission method options: Alternative submission method Task back to top A digital computer has a memory unit with 16 bits per word. The instruction set consists of 122 different operations. All instructions have an operation code part (opcode) and an address part (allowing for only one address). Each instruction is stored in one word of memory. [10 marks] 1. How many bits are needed for the opcode? 2. How many bits are left for the address part of 3. What is the maximum allowable size for 4. What is the largest unsigned binary number the instruction? memory? that can be accommodated in one word of memory?
Explanation / Answer
Q1)
Part 1)
To find the size of opcode we use information that there are 122 different operations
Opcode size = Ciel of log2(122) = 7 bits
Part 2)
Instruction Word = Opcode + Address
Instruction Word = 1 word = 16bit
Opcode = 7BIt
therfore Address = 16 - 7 = 9 BIt
Part 3)
there are 9 bits of address therefore memory has (2 power 9) = 512 address
One address has 16 bit memory = 2 Byte
Therefore total memory = 512 * 2Byte = 1024Byte
Part 4)
If there are 2 bits largest number is 3 = 22 - 1
If there are 3 bits largest number is 7 = 23 - 1
Similarly
If there are 16 bits largest number is 3 = 216 - 1