Chapter 5, Problem 19E

Explanation of Solution

Explanation:

“No”, the value of 16-bit instructions cannot be encoded into 32 registers.

Reason:

The total number of registers is 32 or 2⁵.

The value of 60 instructions with 2 register operands is $60\times (2^5)\times (2^5)$.

The value of 30 instructions with 1 register operand is $30\times (2^5)$.

The value of 3 instructions with one 10-bit address is $3\times 10$.

Calculation of the value of 5-bit register with 2 addresses instructions:

5-bit register with 2-address instructions is $5\times (2^5)\times (2^5)$.

Then, 45 1-address instructions can be $45\times (2^5)$.

From the above, the value of 26 instructions with zero operands can be $26\times 1$