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The memory unit of a computer has 256k words of 32 bits each. The computer has an instruction format with 4 fields: an opcode field; a mode field to specify 1 of 7 addressing modes; a register address field to specify 1 0f 60 registers; and a memory address field. Assume an instruction is 32 bits long. Answer the following:

A: How large must the mode field be?

B: How large must the register field be?

C: How large must the address field be?

D: How large is the opcode field?

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  1. 18 July, 06:23
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    A. 3 bits required

    B. 6 bits are required

    c. 8 bits are required

    d. 15 bits are required

    Explanation:

    A. In the address mode selector, we need to specify 1 out of 7 addressing modes. Therefore, there must be 3 bits (2³ = 8)

    B. There are 60 address registers. Therefore the bits to represent 60 numbers are 6 bits. (2^6 = 64)

    C. As indicated, the memory bit word required is 8 bits.

    D. Total Bits = Opcode + Address mode + Register Add + Memory ADD

    32 bits = Opcode + 3 bits + 6 bits + 8 bits

    Opcode = 32 bits - 17 bits = 15 bits.

    Therefore 15 bits are required for Opcode field.
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