Homework Answers
The word size in the given instruction set architecture is 16 bit
The ISA uses 3 adress instrcution i.e. instruction of the form:
ADD R1 R2 R3
Where one destination operand and two source operands are explicitly specified
There are total 8 registers in the architecture
No. of bits required to uniquely identify each register is 3 (as 2^3 = 8)
As the ISA uses 3 address instruction, each instruction must allot 3*3 = 9 bits for 3 addresses
Remaining 16-9 = 7 bits can be utilized for opcode
So the microprocessor can have a maximum of 2^7 = 128 instructions
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please answer all of 5,thank you!
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