CS161L-Lab 3 The Data path Control and ALU Control Units Solved

 Introduction

For this lab you will be building the control units for a MIPS processor. See the next figure. 

There are two control units. The​ main control unit​ manages the datapath. It receives an opcode input from the currently executing instructions and based on this opcode it configures the datapath accordingly. A truth table for the unit functionality (shown below) can be found in the slides of CS161. The table (read vertically) shows the output for ​R-format, lw, sw ​and beq​ instructions, additionally, you will need to implement the immediate type (​addi,subi​) instructions. To do this, you will trace through the datapath (shown above) to determine which control lines will need to be set. 

 

Control 
Signal name 
R-format 
lw 
sw 
beq 
imm 
 

 

 

 

Inputs 
Op5
0
1
1
0
0
Op4
0
0
0
0
0
Op3
0
0
1
0
1
Op2
0
0
0
1
0
Op1
0
1
1
0
0
Op0
0
1
1
0
0
 

 

 

 

 

 

 

Outputs 
RegDst
1
0
X
X
 
ALUSrc
0
1
1
0
 
MemtoReg
0
1
X
X
 
RegWrite
1
1
0
0
 
MemRead
0
1
0
0
 
MemWrite
0
0
1
0
 
Branch
0
0
0
1
 
ALUOp1
1
0
0
0
 
ALUOp0
0
0
0
1
 
Table 1. ​The control function for the simple one-clock implementation. Read each column vertically. I.e. if the ​Op[5:0]​ is ​000000​, the ​RegDst​ would be ‘​1​’, ​ALUSrc​ would be ‘0​ ​’, etc. You will need to fill in the ​imm​ column. Based on Figure D.2.4 from the book. 

The second control unit manages the ​ALU​. It receives an ALU opcode from the datapath controller and the ‘​Funct Field​’ from the current instruction. With these, the ALU controller decides what operation the ALU is to perform. The following figures from the CS161 slides give an idea of the inputs and outputs of the ALU controller.

 
Input 
Output 
Instruction Opcode 
ALUOp 
Instruction operation 
Funct field 
Desired ALU action 
ALU select input 
lw
00
Load word
XXXXXX
add
0010
sw
00
Store word
XXXXXX
add
0010
beq
01
Branch equal
XXXXXX
subtract
0110
R-type
10
add
100000
add
0010
R-type
10
subtract
100010
subtract
0110
R-type
10
AND
100100
and
0000
R-type
10
OR
100101
or
0001
R-type
10
NOR
100111
nor
1100
R-type
10
Set on less than
101010
Set on less than
0111
ALU select bits based on ALUop, and Funct field

ALUOp 
Funct field 
 

Operation 
 
 
 
 
 
 
 
 
ALUOp1 
ALUOp0 
F5 
F4 
F3 
F2 
F1 
F0 
 
0
0
X
X
X
X
X
X
0010
X
1
X
X
X
X
X
X
0110
1
X
X
X
0
0
0
0
0010
1
X
X
X
0
0
1
0
0110
1
X
X
X
0
1
0
0
0000
1
X
X
X
0
1
0
1
0001
1
X
X
X
1
0
1
0
0111
1
X
X
X
0
1
1
1
1100
Truth Table for ALU Control 

Deliverables
For this lab, you are expected to build and test both the datapath using the template provided (​controlUnit.v​) and ALU control (​aluControlUnit.v​) units. The target processor architecture will only support a subset of the MIPS instructions, listed below. You only have to offer control for these instructions. Signal values can be found within your textbook (and in the images above).

 

add, addu, addi
sub, subu slt
not*, nor
or
and
lw, sw
beq
 

Notice that for the ​addu​ it is sufficient to generate the same control signals as the ​add operation. 

* ​not​ is ​not​ an instruction, it is a ​pseudo-op​, which means it can be implemented using other operations. Think about how you would implement it using the other operations.

Architecture Case Study
For the lab this week you are also expected to perform a simple case study. It is meant to show how important understanding a computer's architecture is, and the compiler is when developing efficient code. For this study, you are to compare and analyze the execution time of the two programs given ​here​. You should run a number of experiments varying the input size from 100 to 30,000.  Based on the results you are to write a report of your findings. The report should contain a graph of your data and a useful analysis of it. You should draw conclusions based on your findings. Reports that simply restate what is in the graph will not get credit.  To make it clear, make sure you used the concepts you have learned so far in 161 and 161L when explaining the differences in performance.  If a confusing or fuzzy explanation is given you will get low or no marks.  The report should be in PDF format.