THE GATEBOOK

Normalization Lectures

A computer has 32 bit instruction and 9 bit address. If there are 400 two address instructions then how many single address instructions can be formulated ?

A.

B.

C.

D.

Total size of instruction=32 bit

Opcode size =32-18=14 bit

# of opcodes =2^14

# of two address instruction=400

Therefore # of 1 address instruction= #of free opcodes*2^address bits

= (2^14-400) * 2^9

Therefore ans should be D.

Total no of 32 bit instructions=

Total of 2 address instructions=

Therefore no of 1 address instruction = (2^32 - 400*2^9 *2^9)/2^9= (2^14 - 400)*2^9

Therefore option D is right

PS : - Sorry for bad writing for the equations