Question

3. Problem 5. 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. 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

Answer:

a) 8 bits b) 16 bits. c) 2¹⁶ * 24 bits d) 2²⁴ -1

Explanation:

a) In order to be able to accommodate 150 different instructions, the number of bits needed must be equal to the minimum power of 2 that satisfies this equation:

2n > 150, i.e. n=8.  

b) If the total number of bits for a word is 24, and 8 are used for the op code, there are 16 bits left for the address part.

c) If the address part has 16 bits, this means that the total addressable space is just 2¹⁶, so the maximum allowable size is 2¹⁶ * 24 bits.

d) As we have 24 bits to be filled, the largest unsigned binary number is just 2²⁴ – 1 .

(As we need to leave a position for all zeros).