3

Worksheet # 2 Draft Solution (Cont'd)

3.2 The execution profile without bypassing and forwarding is as follows:

Clock Cycle #

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Instruction #

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Clock Cycle #

18

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34

Instruction #

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6

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		Clock Cycle #
		35	36	37	38	39
I #	10	W
I #	11	S	I	X	M	W

The execution profile with bypassing and forwarding is as follows:

Clock Cycle #

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Instruction #

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		Clock Cycle #
		18	19	20	21	22	23	24	25
Instr. #	8	M	W
	9	I	X	M	W
	10	S	F	I	X	M	W
	11			S	F	I	X	M	W

4.1 The execution profile without bypassing and forwarding is as follows:

Clock Cycle #

1

2

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Instruction #

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Clock Cycle #

18

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Instruction #

6

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The execution profile with bypassing and forwarding is as follows:

Clock Cycle #

1

2

3

4

5

6

7

8

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18

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Instruction #

1

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11

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4.2 The original code fragment was:

LD R1, 100(R1);
LD R2, 200(R1);
DADDI R1, R2, #3000;
DSUB R2, R4, R1;	That instruction stores a value in R2 that will be destroyed by the next instruction. So, it can be removed.
AND R2, R0, R2;	That instruction puts zero in R2. We can use R0 instead. So, it can be removed.
SD R2, 400(R2);	It can changed to SD R0, 400(R0)
LD R3, 100(R0);	That instruction loads a value to R3 that will be destroyed by the next instruction. So, it can be removed.
XOR R3, R3, R3;	That instruction puts zero in R3. We can use R0 instead. So, it can be removed.
SD R3, 0(R3);	It can be changed to SD R0, 0(R0).
LD R3, 0(R2);	Actually, that loads R3 with zero. We can use R0 instead. So, it can be removed.
SD R3, 100(R3);	It can be changed to DS R0, 100(R0).

The optimized code is:

LD R1, 100(R1);

LD R2, 200(R1);

DADDI R1, R2, #3000;

SD R0, 400(R0);

SD R0, 0(R0);

SD R0, 100(R0);

Working problem 3.2 on the new code without bypassing and forwarding, we obtain:

Clock Cycle #

1

2

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18

19

Instruction #

1

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6

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Working problem 3.2 on the new code with bypassing and forwarding, we obtain:

Clock Cycle #

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Instruction #

1

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6

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Working problem 4.1 on the new code without bypassing and forwarding, we obtain:

Clock Cycle #

1

2

3

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9

10

11

12

13

14

Instruction #

1

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6

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Working problem 4.1 on the new code with bypassing and forwarding, we obtain:

Clock Cycle #

1

2

3

4

5

6

7

8

9

10

11

12

Instruction #

1

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3

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4.3

For problem 3.2 without bypassing:

For problem 3.2 with bypassing:

For problem 4.1 without bypassing:

For problem 4.1 with bypassing:

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