1. CS3410 HW1 Review 2014, 2, 21

2. Agenda We will go through the HW1 questions together TAs will then walk around to help

3. Question 1: Karnaugh Map abcout 0000 0011 0100 0111 1001 1011 1100 1110 0001 1101 00 01 11 10 0 1 c ab 0 1 0 1 1 1 0 0 Sum of products: Karnaugh map minimization: Cover all 1’s Group adjacent blocks of 2 n 1’s that yield a regular shape Encode common features

4. Rules for Karnaugh Map Minimization Minterms can overlap Minterms can span 1, 2, 4, 8 … cells The map can wrap around 0001 1101 00 01 11 10 0 1 c ab 1 0 0 0 1 0 0 0 0001 1101 10 00 01 11 0 1 c ab 1 0 1 0 0 0 0 0

5. 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

6. Question 4: FSM

7. x Output Okay x0 Spam filter x2 x1 Question 4: FSM (cont.) 01 012 ….. 0 1 2

8. x Output Okay x0 Spam filter x2 x1 Question 4: FSM (cont.) 01 012 ….. 0 1 2 SPAM

9. Question 4: FSM (cont.) Current stateInputNext stateOutput x1x2x0x1’=x0x2’=x1 00000okay 00110 00220 01000 01110 01220 02000spam ……………… State (x1=0, x2=0) and (x1=0, x2=1) have exactly the same transitions AND output. So they are NOT distinct states.

10. 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)

11. Question 8 Registers and Control are in parallel

12. 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 = cycle time * num of instructions – Speedup < 1 means we are actually slowing down Execution time (old) Execution time (new)

13. Question 8 (cont.) 4.3.3: – Cost-performance ratio (CPR) = – The higher, the better – This question asks for a “comparison” of CPR CPR ratio= = * Performance Cost CPR (old) CPR (new) Cost (new) Cost (old) Perf (old) Perf (new) 1 speedup

14. 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 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 http://www.cs.cornell.edu/courses/CS3410/2014s p/project/pa1/pa1.html

15. 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

16. Basic Assembler Coding Patterns Arithmetic – C code: – Assembler: a = b + c; #a: $s0, b: $s1, c:$s2 ADD $s0, $s1, $s2

17. Basic Assembler Coding Patterns Brunch – C code: – Assembler: if(a < b) //DO A... else //DO B... #a: $s0, b: $s1 SLT $t0, $s0, $s1 BEQ $t0, $zero, POINTB #DO A... POINTB: #DO B...

18. Basic Assembler Coding Patterns While loop – C code: – Assembler: while(a < b) //Do something... #a: $s0, b: $s1 LOOP: SLT $t0, $s0, $s1 BEQ $t0, $zero, EXIT #Do something... J LOOP EXIT: #Out of the loop...

19. Basic Assembler Coding Patterns Array access – C code: – Assembler: int myArray[10]; a = myArray[2]; #a: $s0, myArray: $s1 LW $s0, 8($s1)

20. C Programming Have you tried the hello-world? Use csuglab machines. It is easier. How to read input from the terminal? – scanf: – You need a buffer for it int scanf ( const char * format,... );

21. Good Luck! Questions?