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 ?
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