please all the questions ahove 41S .ill 88% 8:47 am https://interact2.csu.edu.au

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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

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