A digital computer has a memory unit with 24 bits per word. The instruction set

Question

A digital computer has a memory unit with 24 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. How many bits are needed for the opcode? How many bits are left for the address pan of the instruction? What is the maximum allowable size for memory? What is the largest unsigned binary number that can be accommodatcd in one word of memory?

Explanation / Answer

a. 8 bits [2^8 = 256 so minimum 8 bits to store opcode]

b. 16 bits [24-8 = 16 bits for address]

c. 2^16 = 65536 [as 16 bit for address]

d. 2^15 = 32768-1 = 32767

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