CS3410 HW1 Review
2014, 2, 21
Agenda
• We will go through the HW1 questions
together
• TAs will then walk around to help
Question 1: Karnaugh Map
c
a
b
c
out
0
0
0
0
0
0
1
1
0
1
0
0
0
1
1
1
1
0
0
1
1
0
1
1
1
1
0
0
1
1
1
0
ab
• Sum of products:
out  abc  abc  abc  abc
• Karnaugh map minimization:
00
01
11
10
0
00
00
0
0
11
1
11
11
0
0
11
• Cover all 1’s
• Group adjacent blocks of 2n
1’s that yield a regular shape
• Encode common features
out  ab  ac
Rules for Karnaugh Map Minimization
• Minterms can overlap
• Minterms can span 1, 2, 4, 8 … cells
• The map can wrap around
c
ab
00
01
11
10
0
01
00
0
0
11
1
10
10
0
0
10
c
ab
10
00
01
11
0
01
01
0
0
10
1
10
10
0
0
10
Question 2: Numbers & Arithmetic
• Binary translation:
– Base conversion via repetitive division
• From binary to Hex and Oct
• Negating a number (2’s complement)
• Overflow
– Overflow happened iff
carry into msb != carry out of msb
Question 4: FSM
Question 4: FSM (cont.)
Spam filter
1
0
1
x1
x0
x
x2
Output
Okay
2
0
1
2
0
…..
Question 4: FSM (cont.)
Spam filter
1
0
1
2
x1
x0
x
x2
Output
Okay SPAM
0
1
2
0
…..
Question 4: FSM (cont.)
State (x1=0, x2=0)
and (x1=0, x2=1)
have exactly the
same transitions AND
output. So they are
NOT distinct states.
Current state
Input
Next state
Output
x1
x2
x0
x1’=x0
x2’=x1
0
0
0
0
0
okay
0
0
1
1
0
okay
0
0
2
2
0
okay
0
1
0
0
0
okay
0
1
1
1
0
okay
0
1
2
2
0
okay
0
2
0
0
0
spam
…
…
…
…
…
…
Question 7: Performance
• Instruction mix for some program P, assume:
– 25% load/store ( 3 cycles / instruction)
– 60% arithmetic ( 2 cycles / instruction)
– 15% branches ( 1 cycle / instruction)
• CPI:
– 3 * .25 + 2 * .60 + 1 * .15 = 2.1
• CPU Time = # Instructions x CPI x Clock Cycle Time
– Assuming 400k instructions, 30 MHz :
• 400k * 2.1 / 30 = 28000 µs (1 µs = 1 microsecond = 1/1M S)
Question 8
Registers and
Control are in
parallel
Question 8 (cont.)
• Refer to section 1.6 in the text book
• 4.3.1: The clock cycle time is determined by
the critical path (the load instruction)
• 4.3.2:
– Speedup =
Execution time (old)
Execution time (new)
– Execution time = cycle time * num of instructions
– Speedup < 1 means we are actually slowing down
Question 8 (cont.)
• 4.3.3:
Performance
– Cost-performance ratio (CPR) =
Cost
– The higher, the better
– This question asks for a “comparison” of CPR
CPR ratio=
CPR (old)
CPR (new)
=
Cost (new)
Cost (old)
*
Perf (old)
Perf (new)
1
speedup
Question 9/10/11: Assembler Code
• Writing the instructions in a human-readable
format
– http://www.cs.cornell.edu/courses/CS3410/2014s
p/MIPS_Vol2.pdf
• Core instruction set
– http://www.cs.cornell.edu/courses/CS3410/2014s
p/project/pa1/pa1.html
Assembler Code
• When writing the assembler code:
– Decide which register stores which variable
• Typically you should use $t0~$t9 and $s0~$s7 (You
don’t need to understand their difference now)
– Decide which instruction you want to use
• Get familiar with the core instruction set
• Get familiar with some basic patterns
Basic Assembler Coding Patterns
• Arithmetic
– C code:
a = b + c;
– Assembler:
#a: $s0, b: $s1, c:$s2
ADD $s0, $s1, $s2
Basic Assembler Coding Patterns
• Brunch
– C code:
– Assembler:
if(a < b)
//DO A...
else
//DO B...
#a:
SLT
BEQ
#DO
$s0, b: $s1
$t0, $s0, $s1
$t0, $zero, POINTB
A...
POINTB:
#DO B...
Basic Assembler Coding Patterns
• While loop
– C code:
while(a < b)
//Do something...
– Assembler:
#a: $s0, b: $s1
LOOP:
SLT $t0, $s0, $s1
BEQ $t0, $zero, EXIT
#Do something...
J LOOP
EXIT:
#Out of the loop...
Basic Assembler Coding Patterns
• Array access
– C code:
int myArray[10];
a = myArray[2];
– Assembler:
#a: $s0, myArray: $s1
LW $s0, 8($s1)
C Programming
• Have you tried the hello-world?
• Use csuglab machines. It is easier.
• How to read input from the terminal?
– scanf:
int scanf ( const char * format, ... );
– You need a buffer for it
Good Luck! Questions?