Addressing Modes- Q8

+2 votes

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

asked Jun 4 in Computer Organization by gbmentor (54,290 points)
reshown Jun 5 by gbmentor

3 Answers

+2 votes
 
Best answer
opcode (?) address 1 (9 bit) address2 (9 bit)

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.

answered Jun 5 by tskushagra-guptacse (11,400 points)
selected Jun 5 by vinayakjhapro18
0 votes

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 soumyadeeppro20 (1,720 points) 1 flag
some one please tell me how to beautiful equations
reference: https://gateoverflow.in/blog/2327/how-to-write-nice-answers-questions-in-go

use latex using fx(insert equation) button and insert code
Note: Here We have to add code without worrying about $$
0 votes
D is correct answer
answered Jun 5 by tsnikhilsharmagate2018 (19,690 points)
Answer:
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