CO-Grand Test 1-Q13

+1 vote

Consider the following instruction sequence for a hypothetical RISC processor.

                                         T. R_1 \leftarrow R_2+R_3

                                         U. R_4 \leftarrow R_5+R_6

                                         V. R_5 \leftarrow R_7+R_8

                                        W. R_9 \leftarrow R_5+R_1

                                        X. R_{10} \leftarrow R_4+R_1

                                        Y.R_{11}\leftarrow R_{10}+R_1

                                        Z. R_{9} \leftarrow R_1+R_4

Which of the following is a possible legal execution order for this instruction on an out-of-order processor without register renaming?

(A) T,U,X,V,W,Z,Y

(B) T,U,X,V,Z,W,Y

(C) T,V,U,X,W,Y,Z

(D) U,T,V,Y,X,W,Z

 

asked Jun 25 in Computer Organization by gbeditor (23,750 points)
edited Jun 26 by gbeditor

3 Answers

–1 vote
 
Best answer

(A)

T. R_1 \leftarrow R_2+R_3

U. R_4 \leftarrow R_5+R_6    Here is a dependency between U and X so there must be 2 stall

X. R_{10} \leftarrow R_4+R_1

V. R_5 \leftarrow R_7+R_8    Here is a dependency between V and W so there must be 2 stall

W. R_9 \leftarrow R_5+R_1

Z. R_{9} \leftarrow R_1+R_4

Y.R_{11}\leftarrow R_{10}+R_1

(B)

T. R_1 \leftarrow R_2+R_3

U. R_4 \leftarrow R_5+R_6    Here is a dependency between U and X so there must be 2 stall

X. R_{10} \leftarrow R_4+R_1

V. R_5 \leftarrow R_7+R_8   Dependency between V and W there is one stall

Z. R_{9} \leftarrow R_1+R_4

W. R_9 \leftarrow R_5+R_1

Y.R_{11}\leftarrow R_{10}+R_1

(C)

T. R_1 \leftarrow R_2+R_3

V. R_5 \leftarrow R_7+R_8  Here is a dependency between V and U so 2 stalls

U. R_4 \leftarrow R_5+R_6   Here is a dependency between U and X so 2 stalls 

X. R_{10} \leftarrow R_4+R_1  Here is a dependency between X and Y so 1 stall

W. R_9 \leftarrow R_5+R_1

Y.R_{11}\leftarrow R_{10}+R_1

Z. R_{9} \leftarrow R_1+R_4

(D)

U. R_4 \leftarrow R_5+R_6

T. R_1 \leftarrow R_2+R_3   Dependency between T and Y so 1 stall

V. R_5 \leftarrow R_7+R_8

Y.R_{11}\leftarrow R_{10}+R_1     

X. R_{10} \leftarrow R_4+R_1        

W. R_9 \leftarrow R_5+R_1 

Z. R_{9} \leftarrow R_1+R_4   

So from all of them, answer should be D because of less number of stalls        

answered Jun 27 by 23rishiyadavpro20 (6,850 points)
selected Jul 2 by gbmentor
+3 votes
Y is not there in test time.
answered Jun 26 by tsnikhilsharmagate2018 (31,640 points)
Giving marks for all, for ranking, will put the right choice later
0 votes

Answer is (A) because if we execute the above instructions in the orignal order and order which is given in the options we can clearly see in Option A the results of the registers after the complete execution of the sequence is equal to the result which would have been obtained if we have executed in the original order.

Registers------> R1 R2 R3 R4 R5 R6 R7 R8 R9 R10 R11
Original sequence R2+R3 R2 R3 R5+R6 R7+R8 R6 R7 R8 R2+R3+R5+R6 R5+R6+R2+R3 2*(R2+R3)+R5+R6
Option A R2+R3 R2 R3 R5+R6 R7+R8 R6 R7 R8 R2+R3+R5+R6 R5+R6+R2+R3 2*(R2+R3)+R5+R6
Option B R2+R3 R2 R3 R5+R6 R7+R8 R6 R7 R8 R2+R3+R7+R8 R5+R6+R2+R3 2*(R2+R3)+R5+R6
Option C R2+R3 R2 R3 R6+R7+R8 R7+R8 R6 R7 R8 R2+R3+R6+R7+R8 R2+R3+R6+R7+R8 2*(R2+R3)+R6+R7+R8
Option D R2+R3 R2 R3 R5+R6 R7+R8 R6 R7 R8 R2+R3+R5+R6 R2+R3+R5+R6 R2+R3+R10

 

answered Nov 22 by tsshibam-senapati (1,340 points)
Answer:
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