CS/COE0447 Computer Organization & Assembly Language

1. CS/COE0447Computer Organization & Assembly Language Chapter 5 Part 3

2. Single-cycle Implementation of MIPS • Our first implementation of MIPS used a single long clock cycle for every instruction • Every instruction began on one up (or, down) clock edge and ended on the next up (or, down) clock edge • This approach is not practical as it is much slower than a multicycle implementation where different instruction classes can take different numbers of cycles • in a single-cycle implementation every instruction must take the same amount of time as the slowest instruction • in a multicycle implementation this problem is avoided by allowing quicker instructions to use fewer cycles • Even though the single-cycle approach is not practical it was simpler and useful to understand first • Now we are covering a multicycle implementation of MIPS

3. A Multi-cycle Datapath • A single memory unit for both instructions and data • Single ALU rather than ALU & two adders • Registers added after every major functional unit to hold the output until it is used in a subsequent clock cycle

4. Multi-Cycle ControlWhat we need to cover • Adding registers after every functional unit • Need to modify the “instruction execution” slides to reflect this • Breaking instruction execution down into cycles • What can be done during the same cycle? What requires a cycle? • Need to modify the “instruction execution” slides again • Timing • Control signal values • What they are per cycle, per instruction • Finite state machine which determines signals based on instruction type + which cycle it is • Putting it all together

5. Execution: single-cycle (reminder) • add • Fetch instruction and add 4 to PC add $t2,$t1,$t0 • Read two source registers $t1 and $t0 • Add two values $t1 + $t0 • Store result to the destination register $t1 + $t0  $t2

6. A Multi-cycle Datapath • For add: • Instruction is stored in the instruction register (IR) • Values read from rs and rt are stored in A and B • Result of ALU is stored in ALUOut

7. Multi-Cycle Execution: R-type • Instruction fetch • IR <= Memory[PC]; sub $t0,$t1,$t2 • PC <= PC + 4; • Decodeinstruction/register read • A <= Reg[IR[25:21]]; rs • B <= Reg[IR[20:16]]; rt • ALUOut <= PC + (sign-extend(IR[15:0])<<2); later • Execution • ALUOut <= A op B; op = add, sub, and, or,… • Completion • Reg[IR[15:11]] <= ALUOut; $t0 <=ALU result

8. Execution: single-cycle (reminder) • lw (load word) • Fetch instruction and add 4 to PC lw $t0,-12($t1) • Read the base register $t1 • Sign-extend the immediate offset fff4  fffffff4 • Add two values to get address X =fffffff4 + $t1 • Access data memory with the computed address M[X] • Store the memory data to the destination register $t0

9. A Multi-cycle Datapath • For lw: lw $t0, -12($t1) • Instruction is stored in the IR • Contents of rs stored in A $t1 • Output of ALU (address of memory location to be read) stored in ALUOut • Value read from memory is stored in the memory data register (MDR)

10. Multi-cycle Execution: lw • Instruction fetch • IR <= Memory[PC]; lw $t0,-12($t1) • PC <= PC + 4; • InstructionDecode/register read • A <= Reg[IR[25:21]]; rs • B <= Reg[IR[20:16]]; • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • Execution • ALUOut <= A + sign-extend(IR[15:0]); $t1 +-12 (sign extended) • Memory Access • MDR <= Memory[ALUOut]; M[$t1 + -12] • Write-back • Load: Reg[IR[20:16]] <= MDR; $t0 <= M[$t1 + -12]

11. Execution: single-cycle (reminder) • sw • Fetch instruction and add 4 to PC sw $t0,-4($t1) • Read the base register $t1 • Read the source register $t0 • Sign-extend the immediate offset fffc  fffffffc • Add two values to get address X =fffffffc + $t1 • Store the contents of the source register to the computed address $t0  Memory[X]

12. A Multi-cycle Datapath • For sw: sw $t0, -12($t1) • Instruction is stored in the IR • Contents of rs stored in A $t1 • Output of ALU (address of memory location to be written) stored in ALUOut

13. Multi-cycle Execution: sw • Instruction fetch • IR <= Memory[PC]; sw $t0,-12($t1) • PC <= PC + 4; • Decode/register read • A <= Reg[IR[25:21]]; rs • B <= Reg[IR[20:16]]; rt • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • Execution • ALUOut <= A + sign-extend(IR[15:0]); $t1 + -12 (sign extended) • Memory Access • Memory[ALUOut] <= B; M[$t1 + -12] <= $t0

14. Execution: single-cycle (reminder) • beq • Fetch instruction and add 4 to PC beq $t0,$t1,L • Assume that L is +3 instructions away • Read two source registers $t0,$t1 • Sign Extend the immediate, and shift it left by 2 • 0x0003  0x0000000c • Perform the test, and update the PC if it is true • If $t0 == $t1, the PC = PC + 0x0000000c • [we will follow what Mars does, so this is not Immediate == 0x0002; PC = PC + 4 + 0x00000008]

15. A Multi-cycle Datapath • For beq beq $t0,$t1,label • Instruction stored in IR • Registers rs and rt are stored in A and B • Result of ALU (rs – rt) is stored in ALUOut

16. Multi-cycle execution: beq • Instruction fetch • IR <= Memory[PC]; beq $t0,$t1,label • PC <= PC + 4; • Decode/register read • A <= Reg[IR[25:21]]; rs • B <= Reg[IR[20:16]]; rt • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • PC + #bytes away label is (negative for backward branches, positive for forward branches) • Execution • if (A == B) then PC <= ALUOut; • if $t0 == $t1 perform branch • Note: the ALU is used to evaluate A == B; we’ll see that this does not clash with the use of the ALU above.

17. Execution: single-cycle (reminder) • j • Fetch instruction and add 4 to PC • Take the 26-bit immediate field • Shift left by 2 (to make 28-bit immediate) • Get 4 bits from the current PC and attach to the left of the immediate • Assign the value to PC

18. A Multi-cycle Datapath • For j • No accesses to registers or memory; no need for ALU

19. Multi-cycle execution: j • Instruction fetch • IR <= Memory[PC]; j label • PC <= PC + 4; • Decode/register read • A <= Reg[IR[25:21]]; • B <= Reg[IR[20:16]]; • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • Execution • PC <= {PC[31:28],IR[25:0],”00”};

20. Multi-Cycle ControlWhat we need to cover • Adding registers after every functional unit • Need to modify the “instruction execution” slides to reflect this • Breaking instruction execution down into cycles  • What can be done during the same cycle? What requires a cycle? • Need to modify the “instruction execution” slides again • Timing • Control signal values • What they are per cycle, per instruction • Finite state machine which determines signals based on instruction type + which cycle it is • Putting it all together

21. Multicycle Approach • Break up the instructions into steps • each step takes one clock cycle • balance the amount of work to be done in each step/cycle so that they are about equal • restrict each cycle to use at most once each major functional unit so that such units do not have to be replicated • functional units can be shared between different cycles within one instruction • Between steps/cycles • At the end of one cycle store data to be used in later cycles of thesame instruction • need to introduce additional internal (programmer-invisible) registers for this purpose • Data to be used in later instructions are stored in programmer-visible state elements: the register file, PC, memory

22. Operations • These take time: • Memory (read/write); register file (read/write); ALU operations • The other connections and logical elements have no latency (for our purposes)

23. Operations • Before: we had separate memories for instructions and data, and we had extra adders for incrementing the PC and calculating the branch address. Now we have just one memory and just one ALU.

24. Five Execution Steps • Each takes one cycle • In one cycle, there can be at most one memory access, at most one register access, and at most one ALU operation • But, you can have a memory access, an ALU op, and/or a register access, as long as there is no contention for resources • Changes to registers are made at the end of the clock cycle

25. Step 1: Instruction Fetch • Access memory w/ PC to fetch instruction and store it in Instruction Register (IR) • Increment PC by 4 • We can do this because the ALU is not being used for something else this cycle

26. Step 2: Decode and Reg. Read • Read registers rs and rt • We read both of them regardless of necessity • Compute the branch address in case the instruction is a branch • We can do this because the ALU is not busy • ALUOut will keep the target address

27. Step 3: Various Actions • ALU performs one of three functions based on instruction type • Memory reference • ALUOut <= A + sign-extend(IR[15:0]); • R-type • ALUOut <= A op B; • Branch: • if (A==B) PC <= ALUOut; • Jump: • PC <= {PC[31:28],IR[25:0],2’b00};

28. Step 4: Memory Access… • If the instruction is memory reference • MDR <= Memory[ALUOut]; // if it is a load • Memory[ALUOut] <= B; // if it is a store • Store is complete! • If the instruction is R-type • Reg[IR[15:11]] <= ALUOut; • Now the instruction is complete!

29. Step 5: Register Write Back • Only the lw instruction reaches this step • Reg[IR[20:16]] <= MDR;

30. Summary of Instruction Execution Step 1: IF 2: ID 3: EX 4: MEM 5: WB

31. 4 Multicycle Execution Step (1):Instruction Fetch IR = Memory[PC]; PC = PC + 4; PC + 4

32. Reg[rs] PC + 4 Reg[rt] Multicycle Execution Step (2):Instruction Decode & Register Fetch A = Reg[IR[25-21]]; (A = Reg[rs]) B = Reg[IR[20-15]]; (B = Reg[rt]) ALUOut = (PC + sign-extend(IR[15-0]) << 2) Branch Target Address

33. Reg[rs] Mem. Address PC + 4 Reg[rt] Multicycle Execution Step (3):Memory Reference Instructions ALUOut = A + sign-extend(IR[15-0]);

34. Reg[rs] R-Type Result PC + 4 Reg[rt] Multicycle Execution Step (3):ALU Instruction (R-Type) ALUOut = A op B

35. Branch Target Address Reg[rs] Reg[rt] Multicycle Execution Step (3):Branch Instructions if (A == B) PC = ALUOut; Branch Target Address

36. Branch Target Address Reg[rs] Reg[rt] Multicycle Execution Step (3):Jump Instruction PC = PC[31-28] concat (IR[25-0] << 2) Jump Address

37. Mem. Address Reg[rs] PC + 4 Reg[rt] Multicycle Execution Step (4):Memory Access - Read (lw) MDR = Memory[ALUOut]; Mem. Data

38. Reg[rs] PC + 4 Reg[rt] Multicycle Execution Step (4):Memory Access - Write (sw) Memory[ALUOut] = B;

39. Reg[rs] R-Type Result PC + 4 Reg[rt] Multicycle Execution Step (4):ALU Instruction (R-Type) Reg[IR[15:11]] = ALUOUT

40. Reg[rs] Mem. Address PC + 4 Mem. Data Reg[rt] Multicycle Execution Step (5):Memory Read Completion (lw) Reg[IR[20-16]] = MDR;

41. For Reference • The next 5 slides give the steps, one slide per instruction

42. Multi-Cycle Execution: R-type • Instruction fetch • IR <= Memory[PC]; sub $t0,$t1,$t2 • PC <= PC + 4; • Decode instruction/register read • A <= Reg[IR[25:21]]; rs • B <= Reg[IR[20:16]]; rt • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • Execution • ALUOut <= A op B; op = add, sub, and, or,… • Completion • Reg[IR[15:11]] <= ALUOut; $t0 <=ALU result

43. Multi-cycle Execution: lw • Instruction fetch • IR <= Memory[PC]; lw $t0,-12($t1) • PC <= PC + 4; • Instruction Decode/register read • A <= Reg[IR[25:21]]; rs • B <= Reg[IR[20:16]]; • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • Execution • ALUOut <= A + sign-extend(IR[15:0]); $t1 +-12 (sign extended) • Memory Access • MDR <= Memory[ALUOut]; M[$t1 + -12] • Write-back • Load: Reg[IR[20:16]] <= MDR; $t0 <= M[$t1 + -12]

44. Multi-cycle Execution: sw • Instruction fetch • IR <= Memory[PC]; sw $t0,-12($t1) • PC <= PC + 4; • Decode/register read • A <= Reg[IR[25:21]]; rs • B <= Reg[IR[20:16]]; rt • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • Execution • ALUOut <= A + sign-extend(IR[15:0]); $t1 + -12 (sign extended) • Memory Access • Memory[ALUOut] <= B; M[$t1 + -12] <= $t0

45. Multi-cycle execution: beq • Instruction fetch • IR <= Memory[PC]; beq $t0,$t1,label • PC <= PC + 4; • Decode/register read • A <= Reg[IR[25:21]]; rs • B <= Reg[IR[20:16]]; rt • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • Execution • if (A == B) then PC <= ALUOut; • if $t0 == $t1 perform branch

46. Multi-cycle execution: j • Instruction fetch • IR <= Memory[PC]; j label • PC <= PC + 4; • Decode/register read • A <= Reg[IR[25:21]]; • B <= Reg[IR[20:16]]; • ALUOut <= PC + (sign-extend(IR[15:0])<<2); • Execution • PC <= {PC[31:28],IR[25:0],”00”};

47. Example: CPI in a multicycle CPU • Assume • the control design of the previous slides • An instruction mix of 22% loads, 11% stores, 49% R-type operations, 16% branches, and 2% jumps • What is the CPI assuming each step requires 1 clock cycle? • Solution: • Number of clock cycles from previous slide for each instruction class: • loads 5, stores 4, R-type instructions 4, branches 3, jumps 3 • CPI = CPU clock cycles / instruction count =  (instruction countclass i  CPIclass i) / instruction count =  (instruction countclass I / instruction count)  CPIclass I = 0.22  5 + 0.11  4 + 0.49  4 + 0.16  3 + 0.02  3 = 4.04

48. Multi-Cycle ControlWhat we need to cover • Adding registers after every functional unit • Need to modify the “instruction execution” slides to reflect this • Breaking instruction execution down into cycles • What can be done during the same cycle? What requires a cycle? • Need to modify the “instruction execution” slides again • Timing • Control signal values • What they are per cycle, per instruction • Finite state machine which determines signals based on instruction type + which cycle it is • Putting it all together

49. A (Refined) Datapath fig 5.26

50. Datapath w/ Control Signals Fig 5.27