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).
