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. $2^{14}$

B. $2^{32} - 200$

C. $2^{14} - 400$

D. $(2^{14} - 400) * 2^9$

reshown Jun 5

Total size of instruction=32 bit

Opcode size =32-18=14 bit

# of opcodes =2^14

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

Therefore ans should be D.

answered Jun 5 by (8,650 points)
selected Jun 5
–1 vote

Total no of 32 bit instructions= $_{2}32$

Total of 2 address instructions= $400*2^9 *2^9$

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

answered Jun 5 by (1,640 points) 1 flag
some one please tell me how to beautiful equations