# In a computer instruction format, the instruction length is 10 bits and the size of an addressfield is 3 bits. The system architect has already designed FIFTEEN 2-address instructions andSEVEN 1-address instructions. How many 0-address instruction still possible to accommodatefor the instruction set architecture?

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In a computer instruction format, the instruction length is 10 bits and the size of an address
field is 3 bits. The system architect has already designed FIFTEEN 2-address instructions and
for the instruction set architecture?

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Step 1

Given an instruction format with instruction length of 10 bits and size of address filed is 3 bits. The total number of possible combinations with 10 bits is equal to 210=1024.

Fifteen 2-address instructions and seven 1-address instructions need to be assigned in this system architecture which can be done as –

The given address field is 3 bits which is 23 = 8.

Step 2

15 two-address instruction will be consisting of an opcode along with two address registers. Four bits from MSB(most significant bit) will be used for the opcode and remaining 6 bits will be used for 2 address registers(having 3 bits each).

The 15 two-address instructions require “15 x 23 x 23 = 960” bit pattern.

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