1. 1 CS/COE0447 Computer Organization & Assembly Language Chapter 5 Part 3

2. 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. 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. 4 Multi-Cycle Control What 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. 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. 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. 7 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);  later Execution –ALUOut <= A op B; op = add, sub, and, or,… Completion –Reg[IR[15:11]] <= ALUOut; $t0 <= ALU result

8. 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. 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. 10 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]

11. 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. 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. 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. 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. 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. 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. 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. 18 A Multi-cycle Datapath For j No accesses to registers or memory; no need for ALU

19. 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. 20 Multi-Cycle Control What 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. 21 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 the same 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 Multicycle Approach

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

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

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

32. 32 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 Reg[rs] Reg[rt] PC + 4

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

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

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

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

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

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

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

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

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

42. 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. 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. 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. 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. 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. 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 count class i  CPI class i ) / instruction count =  (instruction count class I / instruction count)  CPI class I = 0.22  5 + 0.11  4 + 0.49  4 + 0.16  3 + 0.02  3 = 4.04

48. 48 Multi-Cycle Control What 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. 49 A (Refined) Datapath fig 5.26

50. 50 Datapath w/ Control Signals Fig 5.27

51. 51 Final Version w/ Control Fig 5.28

52. 52 Multicycle Control Step (1): Fetch IR = Memory[PC]; PC = PC + 4; 1 0 1 0 1 0 X 0 X 0 010 1

53. 53 Multicycle Control 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); 0 0 X 0 0 X 3 0 X X 010 0

54. 54 0 X Multicycle Control Step (3): Memory Reference Instructions ALUOut = A + sign-extend(IR[15-0]); X 2 0 0 X 0 1 X 010 0

55. 55 Multicycle Control Step (3): ALU Instruction (R-Type) ALUOut = A op B; 0 X X 0 0 0 X 0 1 X ??? 0

56. 56 1 if Zero=1 Multicycle Control Step (3): Branch Instructions if (A == B) PC = ALUOut; 0 X X 0 0 X 0 1 1 011 0

57. 57 Multicycle Execution Step (3): Jump Instruction PC = PC[21-28] concat (IR[25-0] << 2); 0 X X X 0 1 X 0 X 2 XXX 0

58. 58 Multicycle Control Step (4): Memory Access - Read ( lw ) MDR = Memory[ALUOut]; 0 X X X 1 0 1 0 X X XXX 0

59. 59 Multicycle Execution Steps (4) Memory Access - Write (sw) Memory[ALUOut] = B; 0 X X X 0 0 1 1 X X XXX 0

60. 60 1 0 0 X 0 X 0 XXX X X 1 1 55 RD1 RD2 RN1RN2WN WD RegWrite Registers Operation ALU 3 E X T N D 1632 Zero RD WD MemRead Memory ADDR MemWrite 5 Instruction I 32 ALUSrcB <<2 PC 4 RegDst 5 I R M D R M U X 0 1 2 3 M U X 0 1 M U X 0 1 A B ALU OUT 0 1 2 M U X <<2 CONCAT 2832 M U X 0 1 ALUSrcA jmpaddr I[25:0] rd MUX 01 rtrs immediate PCSource MemtoReg IorD PCWr* IRWrite Multicycle Control Step (4): ALU Instruction (R-Type) Reg[IR[15:11]] = ALUOut; (Reg[Rd] = ALUOut)

61. 61 Multicycle Execution Steps (5) Memory Read Completion (lw) Reg[IR[20-16]] = MDR; 1 0 0 X 0 0 X 0 X X XXX 0 55 RD1 RD2 RN1RN2WN WD RegWrite Registers Operation ALU 3 E X T N D 1632 Zero RD WD MemRead Memory ADDR MemWrite 5 Instruction I 32 ALUSrcB <<2 PC 4 RegDst 5 I R M D R M U X 0 1 2 3 M U X 0 1 M U X 0 1 A B ALU OUT 0 1 2 M U X <<2 CONCAT 2832 M U X 0 1 ALUSrcA jmpaddr I[25:0] rd MUX 01 rtrs immediate PCSource MemtoReg IorD PCWr* IRWrite

62. 62 Multi-Cycle Control What 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: Registers/memory updated at the beginning of the next clock cycle 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

63. 63 Fig 5.28 For reference

64. 64 State Diagram, Big Picture

65. 65 Handling Memory Instructions

66. 66 R-type Instruction

67. 67 Branch and Jump

68. 68 A FSM State Diagram

69. 69 FSM Implementation

70. 70 Example: Load (1) 0 10 0 1 01 00 1

71. 71 Example: Load (2) 0 00 11 rs rt

72. 72 Example: Load (3) 10 1 00

73. 73 Example: Load (4) 1 10

74. 74 Example: Load (5) 1 1 0

75. 75 Example: Jump (1) 0 10 0 1 01 00 1

76. 76 Example: Jump (2) 0 00 11

77. 77 Example: Jump (3) 1 10

78. 78 To Summarize… From several building blocks, we constructed a datapath for a subset of the MIPS instruction set First, we analyzed instructions for functional requirements Second, we connected buildings blocks in a way to accommodate instructions Third, we refined the datapath and added controls

79. 79 To Summarize… We looked at how an instruction is executed on the datapath in a pictorial way We looked at control signals connected to functional blocks in our datapath We analyzed how execution steps of an instruction change the control signals

80. 80 To Summarize… We compared a single-cycle implementation and a multi-cycle implementation of our datapath We analyzed multi-cycle execution of instructions We refined multi-cycle datapath We designed multi-cycle control

81. 81 To Summarize… We looked at the multi-cycle control scheme in detail Multi-cycle control can be implemented using FSM FSM is composed of some combinational logic and memory element

82. 82 Summary Techniques described in this chapter to design datapaths and control are at the core of all modern computer architecture Multicycle datapaths offer two great advantages over single-cycle –functional units can be reused within a single instruction if they are accessed in different cycles – reducing the need to replicate expensive logic –instructions with shorter execution paths can complete quicker by consuming fewer cycles Modern computers, in fact, take the multicycle paradigm to a higher level to achieve greater instruction throughput: –pipelining (later class) where multiple instructions execute simultaneously by having cycles of different instructions overlap in the datapath –the MIPS architecture was designed to be pipelined