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 150 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. (a) How many bits are needed for the opcode? (b) How many bits are left for the address part of the instruction? (c) What is the maximum allowable size for memory? (d) What is the largest unsigned binary number that can be accommodated in one word of memory.

Explanation / Answer

For 32 bits per word and the instruction set consists of 150 different operations

So 150 different unique instructions

So 2^8 =256 in these set only we have a chance of getting our operation because if we go for 2^7 =128 which is less than 150 so no chance of getting instructions

a) soo the bits that are needed for thw opcode is "8" bits because 2^8 possible opcodes

B) the bits that are left for the address part of the instruction are "24" bits

Because given 32 - 8(opcode bits) =24 bits (left for address part)

C) 2^24 maximum allowable size for memory because with 24bits you can address at most 2^24 words ans each word is 32 bit long

D) 2^32-1 largest unsigned binary number

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