Addressing Modes- Q1

+2 votes

Consider 1 GHz clock frequency processor, uses different operand accessing models shown below:

Operand Accessing Mode Frequency(%)
Register 10
Immediate 20
Direct 30
Memory Indirect 20
Indexed 20

 

Assume that 2 memory cycles consumed for each memory reference, 3 cycles consumed for arithmetic computation and 1 cycle consumed when an operand is in register(s) instruction itself. What is the average operand fetch rate (in million words/sec) of the processor

A. 342.82

B. 344.82

C. 355.82

D. 346.82

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

2 Answers

+1 vote
 
Best answer

\\ cycle time =1/clock frequency=1/1GHz=1ns\\ \\ (0.1*1+0.2*1+0.3*2+0.2*4+0.2*6)*1ns=2.9 ns\\ \\ 1\ operand \ (1 word ) =2.9ns\\ number \ of\ words \rightarrow in 1\ sec\\ \\ 1/2.9ns=0.34482*10^9=344.82*10^6=344.82\ million\ words/sec

Therefore ans should be B.

answered Jun 5 by tskushagra-guptacse (11,400 points)
selected Jun 5 by vinayakjhapro18
how immediate  takes 1 register cycle at the time of operand fetch ?
The question must have mentioned properly. But i have considered because of this line:::
1 cycle consumed when an operand is in register(s) instruction itself.
I think after register there should be an "AND".
how indexed takes 6 clock cycle. i am not able to understand plz explain
Index addressing mode contain 1 register ref. and 1 alu and 1mem ref. So that 1*1+1*3+1*2
–1 vote
Effective no of memory cycles = 10 % (3) + 20 % (1) + 30% (1*2) + 20% (2*2) + 20% (3 +2) = 2.9 ns

 

Therefore,avg operand fetch rate= 1/ effective no of memory cycles= 1/ 2.9 = 0.3448275 * 10 ^9 = 334.82 milion words/sec
answered Jun 5 by soumyadeeppro20 (1,720 points)
by mistake you wrote 334.82
how tell me ??
Answer:
...