Answer:
a. 7 bits b. 9 bits c. 1 kB d. 2¹⁶ - 1
Explanation:
a. How many bits are needed for the opcode?
Since there are 72 different operations, we require the number of bits that would contain 72 different operations. So, 2ⁿ ≥ 72
72 = 64 + 8 = 2⁶ + 8
Since n must be an integer value, the closest value of n that would contain 72 different operations is n = 7. So, 2⁷ = 128
So, we require 7 bits for the opcode.
b. How many bits are left for the address part of the instruction?
bits left = bits per word - opcode bit = 16 - 7 = 9 bits
c. What is the maximum allowable size for memory?
Since there are going to be 2⁹ bits to addresses each word and 16 bits for each word, the maximum allowable size for memory is thus 2⁹ × 16 = 512 × 16 = 8192 bits.
We convert this to bytes
8192 bits × 1 byte/8 bits = 1024 bytes = 1 kB
d. What is the largest unsigned binary number that can be accommodated in one word of memory?
Since the number go from 0 to 2¹⁶, the largest unsigned binary number that can be accommodated in one word of memory is thus
2¹⁶ - 1
