Review Fig 4 15 page 320 Fig 4

Multicycle Implementation • Realistic Approach • Enables time sharing functional units during an instruction

Multicycle Implementation Assumptions: • Each STEP in the execution will take 1 clock cycle

Multicycle Implementation Assumptions: • • Each STEP in the execution will take 1 clock

Instruction register PC Address Memory Data IR Instruction or data Memory data register Data

Processor Design Processor (Instruction Set Architecture) Datapath Control Mem Reg File ALU Mux Reg

Processor Design Processor (Instruction Set Architecture) Datapath Control Register Transfer Sequence Mem Reg File

STATE S 0 S 1 S 2 S 3 S 4 Register Transfers M[PC]

MIPS - Lite Consider the following instructions for implementation INSTRUCTION OP R type add

Basic steps all instructions execute 1. Access the Instruction from Memory 2. Decode Instruction

For Multicycle Design Define S = S 0 , S 1 , S 2

Load Word & Store Word ( I – type ) lw rt, imm 16

lw rt, imm 16 (rs) # load word M[ R[rs] + sign_ext(imm 16) ]

sw rt, imm 16 ( rs) # store word R[rt] M[ R[rs] + sign_ext(imm

R – Arithmetic – Logic Instruction op rd, rs, rt R[rs] op R[rt] R

beq rs, rt, imm 16 op rs rt 6 5 5 I -type address/

beq rs, rt, imm 16 I -type Zero (PC+4) + Zero SUM[ (Sh. Lt

Jump Instruction j Label go to Label op address 2 address 6 bits 26

j Label go to Label op address 2 address 6 bits 26 bits Op

State Register Transfers S 0 M[PC] IR, PC + 4 PC, S 1 S

State Register Transfers S 0 M[PC] IR, PC + 4 PC, S 1 R[IR[25

What would “Halt” look like? S 0 S 10 M[PC] IR, PC + 4

What would “Halt” look like? S 0 S 1 M[PC] IR, PC + 4

What would “Start” look like? Start S 0 S, ‘boot’ PC

What if a Memory Read took 3 clock cycles? Instead of S 0 M[PC]

Slides: 60

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Review Fig 4. 15 page 320 / Fig. 4. 17 page 322 Review Timing

Multicycle Implementation • Realistic Approach • Enables time sharing functional units during an instruction ( Memory and ALU) • Enables instructions to be implemented with different number of clock cycles

Multicycle Implementation Assumptions: • Each STEP in the execution will take 1 clock cycle

Multicycle Implementation Assumptions: • • Each STEP in the execution will take 1 clock cycle Each of these operations require 1 clock cycle - Memory Access ( Idealized) - Register File Access - An ALU Operation

Instruction register PC Address Memory Data IR Instruction or data Memory data register Data A Register # ALU Registers Register # B Register # MDR IR timing requires a write control signal ALUOut

Processor Design Processor (Instruction Set Architecture) Datapath Control Mem Reg File ALU Mux Reg Gates Decoder Sequencer

Processor Design Processor (Instruction Set Architecture) Datapath Control Register Transfer Sequence Mem Reg File ALU Mux Reg Gates Decoder Sequencer

STATE S 0 S 1 S 2 S 3 S 4 Register Transfers M[PC] IR, PC+4 PC, R[IR[25 -21]] A, A + sign_ext(IR[15 -0]) ALUOut, M[ ALUOut] MDR, MDR R[ IR[20 -16]], NEXT STATE S 1 S 2 S 3 S 4 S 0 S S S

MIPS - Lite Consider the following instructions for implementation INSTRUCTION OP R type add 0 subtract 0 AND 0 OR 0 set on less than 0 load word 35 store word 43 branch equal 4 FUNCT 32 34 36 37 42 na na na

Basic steps all instructions execute 1. Access the Instruction from Memory 2. Decode Instruction and Access the Data from Registers 3. Perform the Instruction 4. Write the Result

For Multicycle Design Define S = S 0 , S 1 , S 2 , . . . , Sn Where S is the STATE of the processor. Sk denotes that S decodes to k

Load Word & Store Word ( I – type ) lw rt, imm 16 (rs) or sw rt, imm 16 ( rs) op rs rt address/ immediate 31, . . . 26, 25, . . . 21, 20, . . . 16, 15, . . . lw # load word M[ R[rs] + sign_ext(imm 16) ] sw R[rt] 0 R[rt] # store word M[ R[rs] + sign_ext(imm 16) ]

lw rt, imm 16 (rs) # load word M[ R[rs] + sign_ext(imm 16) ] R[rt] For Op = ‘LW’ S 0 M[PC] IR, PC+4 PC, Next State S 1 S

Instruction register PC Address Memory Data IR Instruction or data Memory data register Data A Register # ALU Registers Register # B Register # MDR IR timing requires a write control signal ALUOut

lw rt, imm 16 (rs) # load word M[ R[rs] + sign_ext(imm 16) ] R[rt] For Op = ‘LW’ S 0 M[PC] IR, PC+4 S 1 R[IR[25 -21]] A, PC, S 1 S 2 S S

Instruction register PC Address Memory Data IR Instruction or data Memory data register Data A Register # ALU Registers Register # B Register # MDR IR timing requires a write control signal ALUOut

lw rt, imm 16 (rs) # load word M[ R[rs] + sign_ext(imm 16) ] R[rt] For Op = ‘LW’ S 0 M[PC] IR, PC+4 PC, S 1 R[IR[25 -21]] A, S 2 A + sign_ext(IR[15 -0]) ALUOut, S 1 S S 2 S S 3 S

lw rt, imm 16 (rs) # load word M[ R[rs] + sign_ext(imm 16) ] R[rt] For Op = ‘LW’ S 0 M[PC] IR, PC+4 PC, S 1 R[IR[25 -21]] A, S 2 A + sign_ext(IR[15 -0]) ALUOut, S 3 M[ ALUOut] MDR, S 1 S S 2 S S 3 S S 4 S

lw rt, imm 16 (rs) # load word M[ R[rs] + sign_ext(imm 16) ] R[rt] For Op = ‘LW’ S 0 M[PC] IR, PC+4 PC, S 1 R[IR[25 -21]] A, S 2 A + sign_ext(IR[15 -0]) ALUOut, S 3 M[ ALUOut] MDR, S 4 MDR R[ IR[20 -16]], S 1 S 2 S 3 S 4 S 0 S S S

lw rt, imm 16 (rs) # load word M[ R[rs] + sign_ext(imm 16) ] R[rt] For Op = ‘LW’ S 0 M[PC] IR, PC+4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B S 2 A + sign_ext(IR[15 -0]) ALUOut, S 3 M[ ALUOut] MDR, S 4 MDR R[ IR[20 -16]], sw rt, imm 16 ( rs) R[rt] # store word M[ R[rs] + sign_ext(imm 16) ] What’s different? S 1 S S 2 S 3 S S 4 S S 0 S

sw rt, imm 16 ( rs) # store word R[rt] M[ R[rs] + sign_ext(imm 16) ] For Op = ‘SW’ S 0 M[PC] IR, PC+4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B S 2 A + sign_ext(IR[15 -0]) ALUOut, S 1 S S 2 S ? S

sw rt, imm 16 ( rs) # store word R[rt] M[ R[rs] + sign_ext(imm 16) ] For Op = ‘SW’ S 0 M[PC] IR, PC+4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B S 2 A + sign_ext(IR[15 -0]) ALUOut, S 5 B M[ ALUOut], S 1 S 2 S 5 S S 0

Instruction register PC Address Memory Data IR Instruction or data Memory data register Data A Register # ALU Registers Register # B Register # MDR IR timing requires a write control signal ALUOut

R – Arithmetic – Logic Instruction op rd, rs, rt R[rs] op R[rt] R [rd] op rs rt rd 6 5 5 5 shamt funct 5 6

R – Arithmetic – Logic Instruction op rd, rs, rt R[rs] op R[rt] R [rd] Op = R-type S 0 M[PC] IR, PC + 4 PC, S 1 S

R – Arithmetic – Logic Instruction op rd, rs, rt R[rs] op R[rt] R [rd] Op = R-type S 0 S 1 M[PC] IR, PC + 4 PC, R[IR[25 -21]] A, R[IR[20 -16]] B, S 1 S 6 S S

R – Arithmetic – Logic Instruction op rd, rs, rt R[rs] op R[rt] R [rd] Op = R-type S 0 S 1 S 6 M[PC] IR, PC + 4 PC, R[IR[25 -21]] A, R[IR[20 -16]] A op B ALUOut, B, S 1 S 6 S 7 S S S

R – Arithmetic – Logic Instruction op rd, rs, rt R[rs] op R[rt] R [rd] Op = R-type S 0 S 1 S 6 S 7 M[PC] IR, PC + 4 PC, R[IR[25 -21]] A, R[IR[20 -16]] A op B ALUOut, ALUOut R[IR[15 -11]], B, S 1 S 6 S 7 S 0 S S

Instruction register PC Address Memory Data IR Instruction or data Memory data register Data A Register # ALU Registers Register # B Register # MDR IR timing requires a write control signal ALUOut

beq rs, rt, imm 16 op rs rt 6 5 5 I -type address/ immediate 16 Zero =1 iff $rs - $rt = 0 Zero (PC+4) + Zero SUM[ (Sh. Lt 2[Sign_Ext(imm 16)])+PC+4] PC

beq rs, rt, imm 16 I -type Zero (PC+4) + Zero SUM[ (Sh. Lt 2[Sign_Ext(imm 16)])+PC+4] PC Op = ‘BEQ’ S 0 M[PC] IR, PC + 4 PC, S 1 S

beq rs, rt, imm 16 I -type Zero (PC+4) + Zero SUM[ (Sh. Lt 2[Sign_Ext(imm 16)])+PC+4] PC Op = ‘BEQ’ S 0 S 1 M[PC] IR, PC + 4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B, PC + shlt 2[sign_ext(IR[15 -0])] ALUOut, S 8 The transfers for S 1 are the same for all instructions except the next control state which depends on the op code S S

beq rs, rt, imm 16 I -type Zero (PC+4) + Zero SUM[ (Sh. Lt 2[Sign_Ext(imm 16)])+PC+4] PC Op = ‘BEQ’ S 0 S 1 S 8 M[PC] IR, PC + 4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B, PC + shlt 2[sign_ext(IR[15 -0])] ALUOut, S 8 Zero PC + Zero ALUOut PC, S 0 S S S

Instruction register PC Address Memory Data IR Instruction or data Memory data register Data A Register # ALU Registers Register # B Register # MDR IR timing requires a write control signal ALUOut

Jump Instruction j Label go to Label op address 2 address 6 bits 26 bits The complete 32 bit address is : address 4 bits 26 bits 00 2 bits Upper 4 bits of the Program Counter, PC jump uses word addresses address * 4 = address: 00 This is Pseudodirect Addressing. Note: 256 MB word boundaries

j Label go to Label op address 2 address 6 bits 26 bits Op = ‘J’ S 0 M[PC] IR, PC + 4 PC, S 1 S

j Label go to Label op address 2 address 6 bits 26 bits Op = ‘J’ S 0 S 1 M[PC] IR, PC + 4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B, PC + shlt 2[sign_ext(IR[15 -0])] ALUOut, S 9 S S

j Label go to Label op address 2 address 6 bits 26 bits Op = ‘J’ S 0 S 1 S 9 M[PC] IR, PC + 4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B, PC + shlt 2[sign_ext(IR[15 -0])] ALUOut, S 9 PC[31 -28] : shlt 2(IR[25 -0]) PC, S 0 S S S

Instruction register PC Address Memory Data IR Instruction or data Memory data register Data A Register # ALU Registers Register # B Register # MDR IR timing requires a write control signal ALUOut

State Register Transfers S 0 M[PC] IR, PC + 4 PC, S 1 S

State Register Transfers S 0 M[PC] IR, PC + 4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B, PC + shlt 2[sign_ext(IR[15 -0])] ALUOut, (‘LW’+’SW’)S 2 + ‘R’S 6 + ‘BEQ’S 8 + ‘J’S 9 S S

State Register Transfers S 0 M[PC] IR, PC + 4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B, PC + shlt 2[sign_ext(IR[15 -0])] ALUOut, (‘LW’+’SW’)S 2 + ‘R’S 6 + ‘BEQ’S 8 + ‘J’S 9 S 2 A + sign_ext(IR[15 -0]) ALUOut, ‘LW’S 3 + ‘SW’S 5 S 3 M[ ALUOut] MDR, S 4 MDR R[ IR[20 -16]], S 0 S 5 B M[ ALUOut], S S S 0

R – Arithmetic – Logic Instruction op rd, rs, rt R[rs] op R[rt] R [rd] op rs rt rd 6 5 5 5 shamt funct 5 6

beq rs, rt, imm 16 op rs rt 6 5 5 I -type address/ immediate 16 Zero =1 iff $rs - $rt = 0 Zero (PC+4) + Zero SUM[ (Sh. Lt 2[Sign_Ext(imm 16)])+PC+4] PC

What would “Halt” look like? S 0 S 10 M[PC] IR, PC + 4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B, PC + shlt 2[sign_ext(IR[15 -0])] ALUOut, (‘LW’+’SW’)S 2 + ‘R’S 6 + ‘BEQ’S 8 + ‘J’S 9 + ‘Halt’ S 10 S S

What would “Halt” look like? S 0 S 1 M[PC] IR, PC + 4 PC, S 1 R[IR[25 -21]] A, R[IR[20 -16]] B, PC + shlt 2[sign_ext(IR[15 -0])] ALUOut, (‘LW’+’SW’)S 2 + ‘R’S 6 + ‘BEQ’S 8 + ‘J’S 9 + ‘Halt’ S 10 S S S

What would “Start” look like? Start S 0 S, ‘boot’ PC

What if a Memory Read took 3 clock cycles? Instead of S 0 M[PC] It would be S 0 M[PC] S 1 S 2 IR, PC + 4 PC, S 1 S PC S 1 S 2 S 3 S S S IR PC + 4