CS 3853 Computer Architecture Chapter 3 Section 4 Example Answers

Modification 5: DIV.D uses F2 instead of F0 and there are 2 multiply functional units

Instruction		Reservation
Station		Execution
Unit		 Issue
Cycle		Ex Start
Cycle		Ex End
Cycle		 Memory
Cycle		CDB
cycle		Write
dest
L.D    F6,32(R2)Load1ALU12234F6
L.D    F2,44(R3)Load2ALU23345F2
MUL.DF0,F2,F4Mult1fmul1361516F0
SUB.DF8,F2,F6Add1fadd4678F8
DIV.DF10,F2,F6Mult2fmul2564546F10
ADD.DF6,F8,F2Add2fadd691011F6
Note: The DIV.D issues in cycle 5 and both operands are available for it to start execution on cycle 6.