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).
The number of bits that are needed for this operation code (opcode) is 8 bits.
Binary encoding can be defined as a standard technique that is designed and developed for converting data in plain text (source alphabets) to a form that is easily used by different operating systems (OS), especially through the use of a binary digit (bit) or two-symbol system.
The number of bits that are needed for this operation code (opcode) is given by this mathematical expression:
[tex]2^n > 150\\\\2^8 > 150\\\\256 > 150[/tex]
n = 8 bits.
b. To determine the number of bits that are left for the address part of the instruction:
[tex]Address = I-O\\\\Address =24-8[/tex]
Address = 16 bits.
c. To determine the maximum allowable size for memory:
Based on the memory unit, this is given by [tex]2^{16}\times 24[/tex]
d. The largest unsigned binary number that can be accommodated in one word of memory is given by: [tex]2^{24}-1[/tex]
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