CSEE W3827 Fundamentals of Computer Systems Homework Assignment 5 Solutions

CSEE W3827 Fundamentals of Computer Systems Homework Assignment 5 Solutions

CSEE W3827
Fundamentals of Computer Systems
Homework Assignment 5
Solutions
Prof. Martha A. Kim
Columbia University
Due November 13, 2014 at 10:10 AM
Write your name and UNI on your solutions
Show your work for each problem; we are more interested in how you
get the answer than whether you get the right answer.
1. (30 pts.) List the wire names in the datapath below for which the value is ignored
when executing the given instruction.
31:26
5:0
MemtoReg
Control
MemWrite
Unit
Branch
ALUControl2:0
Op
ALUSrc
Funct RegDst
PCSrc
RegWrite
CLK
0
1
PC'
PC
A
RD
Instr
Instruction
Memory
25:21
20:16
A1
CLK
WE3
RD1
RD2
A2
RD2
A3
Register
WD3
File
20:16
+
15:11
WriteReg4:0
PCPlus4
WriteData
WE
A
RD
Data
Memory
WD
ReadData
0
1
0
1
SignImm
15:0
0 SrcB
1
ALUResult
Sign Extend
<<2
+
4
Zero
SrcA
ALU
CLK
PCBranch
Result
(a) add $t0, $t0, $t1
PCBranch, SignImm, Zero, ReadData
(b) lw $t0, 4($a0)
PCBranch, WriteData/RD2
2. Assuming the following dynamic instruction frequency for a program running on the
single-cycle MIPS processor...
add
addi
beq
lw
sw
15%
25%
10%
30%
20%
(a) (5 pts.) ... in what fraction of all cycles is the data memory used (either read
or written)?
Only for loads and stores, so 50%.
(b) (5 pts.) ... in what fraction of cycles is the sign extend circuit needed?
It is used in all but the add, so 85%.
3. Assuming the single cycle datapath with the components and critical path
described on the slides (mips-uarch.pdf, slide 17) as a baseline, analyze the impact
of the following changes.
(a) (10 pts.) If the ALU were made 5x faster, how much faster would the
processor become? (Give a percent.)
The 200ps ALU becomes 40ps for a 160ps savings.
Alternate frequency interpretation accepted:
160
925
= 17.3%.
1
1
925−160 − 925
1
925
= 20.9%
(b) (20 pts.) If you could double the size of the MIPS register file, causing a 100ps
increase in register file read time and a 15% reduction in the number of
instructions (thanks to fewer loads and stores), should you make this change?
Provide quantitative support for your answer. In
Seconds
Progrm
•
•
=
nstrs
Progrm
·
Cyces
nstr
·
Seconds
,
Cyce
this change would
ntrs
decrease Progrm
by 15%, and
100ps
Seconds
increase Cyce by 925ps = 10.8%
so the benefits outweigh the costs for a net benefit.
4. (30 pts.) A “stuck at” fault is a fabrication error where a wire is stuck at some
constant value no matter what. Below, design a small MIPS function called
stuckatzero to detect whether the Branch control wire is stuck at 0. Your function
should take no arguments and return 1 if Branch is stuck at 0 and 0 if the wire is
behaving properly.
Note: The functon should respect calling conventions, and use only instructions
supported by our single cycle processor (lw, sw, beq, addi, add, slt, and,
or, addu, subu, and slt) plus jr to return from the function.
If Branch is stuck at zero, then PCSrc will also be stuck at zero, and the PC register
will always increment to PC+4. If Branch is not stuck, then beq will work.
stuckatzero:
beq $0, $0, OK
addi $v0, $0, 1
jr $ra
OK:
addu $v0, $0, $0
jr $ra
# if Branch working, will be taken
# else fall through, so v0 = 1
# v0 = 0