How many bits are needed for the opcode

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Reference no: EM131832532

Assignment

Question 1

Suppose a disk drive has the following characteristics:

• 23 surfaces
• 512 tracks per surface
• 64 sectors per track
• 32 bytes/sector
• Track-to-track seek time of 10 milliseconds
• Rotational speed of 9600 RPM

I. What is the capacity of the drive?
II. What is the rotation delay?
III. What is the access time?

Question 2

A digital computer has a memory unit with 32 bits per word. The instruction set consists of 232 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.

I. How many bits are needed for the opcode?
II. How many bits are needed to specify the register?
III. How many bits are left for the address part of the instruction?
IV. What is the maximum allowable size for memory?
V. What is the largest unsigned binary number that can be accommodated in one word of memory?

Question 3

In a computer instruction format, the instruction length is 13 bits and the size of an address field is 5 bits. The system architect has already designed FIVE 2-address instructions and TWENTY 1-address instructions. How many 0-address instruction still possible to accommodate for the instruction set architecture?

Question 4

Write code to implement the below expression:

A = (X+Y) * (Y-Z) / (X * Y) on 2, 1 and 0 address machines.

Reference no: EM131832532

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