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

Question

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?

Explanation / Answer

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I. We should be able to hold 232 instructions, so we'll look for the power of 2 greater than 232. Hence the size needed for opcode will be 8 ( 28=256).

II. Registers should be able to hold the address of opcode, hence we look for power of 2 greater or equal to 8 which is 3. Hence the size of register is 3 ( 23=8).

III. we are given memory unit with 32 bits and we have register of 3 bits and opcode of 8 bits. Therefore bits left for address are 32 - 8 - 3 = 21.

IV. Max memory is equal to 2 to the power of address bits. Therefore size of max memory is 221.

V. Largest unsigned binary number possible is 223-1.

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