A digital computer has a memory unit with 24 bits per word and an instruction se

Question

A digital computer has a memory unit with 24 bits per word and an instruction set with 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. How many bits are needed for the opcode? How many bits are left for the address part of the instruction? What is the maximum allowable size for memory? What is the largest unsigned binary number (integer) that can be handled in one word of memory? What is the smallest signed binary number (integer) that can be handled in one word of memory?

Explanation / Answer

(A) Consider 7 bits 27=128 bits it is not accomidate 150 different instructions.
Consider 8 bits 28=256 bits are required.

(B) The no of bits required by opcode is 8bits, the remaining bits for an address part of the instruction is 24-8=16 bits

(C) The maxmum allowable size of memory is 216 address.

(D) The largest unsigned number that can handle one word of memory 2N-1= 224-1.
(E) The smallest signed number that can handle one word of memory -8,388,608.

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