Question

A digital computer has a memory unit with 26 bits per word. The instruction set consists of 756 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?

Answer 1

a. the number of bits needed for the opcode = 10

b. address part is = 16 bits

c. maximum allowable size of memory = 65536

d. Maximum memory size per word = 67108864

Step-by-step explanation:

The memory size of the instruction set is 756 which is less than 2^10 (that is 1024).

Since the word size is 26 bits and the instruction takes 10, the remaining 16 is for the address part which is 65536 memory address. The maximum number of memory size for a word is 67108864.