1. Computer Organization Lab 1 Soufiane berouel

2. Formulas to Remember CPU Time = CPU Clock Cycles x Clock Cycle Time CPU Clock Cycles = Instruction Count x Cycles per Instruction Clock Cycle Time = 1 / Clock Cycle Rate Amdahl’s Law – Overall Speedup = 1 / [(1 – ∑ FE i ) + ∑ (FE i / SE i )] – FE is the proportion affected, SE is the speed up for the specific proportion

3. Exercise 1 For a color display using 8 bits for each primary color (R, G, B) per pixel and with a resolution of 1280 x 800 pixels, what should be the size (in bytes) of the frame buffer to store a frame?

4. Exercise 1 (sol) Each frame requires 1280 x 800 x 3 = 3072000 ~ 3 Mbytes

5. Exercise 2 Consider 3 processors P1, P2 and P3 with clock rates and CPI given below, are running the same program: clock rate CPI P1 2 GHz 1.5 P2 1.5 GHz 1.0 P3 3 GHz 2.5 Which processor will have the best performance?

6. Exercise 2 (sol) Suppose the program has N instructions. Time taken to execute on P1 is = 1.5 N / (2 x 10 9 ) = 0.75 N x 10 -9 Time taken to execute on P2 is = N/ (1.5 x 10 9 ) = 0.66 N x 10 -9 Time taken to execute on P3 is = 2.5 N/ (3 x 10 9 ) = 0.83 N x 10 -9 P2 has the best performance (since it takes the least time to execute).

7. Exercise 3 Consider 3 processors P1, P2 and P3 with same instruction set with clock rates and CPI given below: clock rate CPI P1 2 GHz 1.5 P2 1.5 GHz 1.0 P3 3 GHz 2.5 If the processors each execute a program in 10 seconds, find the number of cycles and the number of instructions.

8. Exercise 3 (sol) For P1 – 10 (execution time) = 1.5 (CPI) x N 1 / (2 x 10 9 ) – So N 1 (number of instructions) = 1.33 x 10 10 – Number of Cycles = 1.5 x N 1 = 2 x 10 10 …Do the same for P2 and P3

9. Exercise 4 Given below are the number of instructions of a program: arithmetic store load branch 500 50 100 50 Assuming the instructions take 1, 5, 5 and 2 cycles, what is the execution time in a 2 GHz processor?

10. Exercise 4 (sol) Execution time = cycle time x CPI x no. of inst – Cycle time = 1/(2 x 10 9 ) – CPI = (500 x 1 + 50 x 5 + 100 x 5 + 50 x 2) / (500 + 50 + 100 + 50) – Thus, Execution Time = 1350 / 700 x 700 / (2 x 10 -9 ) So the Execution time = 675 x 10 -9 sec

11. Exercise 5 Compilers have a profound impact on the performance of an application on a given processor. This problem will explore the impact compilers have on execution time: compiler A compiler B no instructions exec. Time no. instructions exec. Time a) 1.0 x 10 9 1 s 1.2 x 10 9 1.4 s b) 1.4 x 10 9 0.8 s 1.2 x 10 9 0.7 s Find the average CPI for programs a and b, on different compilers, given that the processor has a cycle time of 1 ns.

12. Exercise 5 (sol) Execution Time = CPI x cycle time x no. of inst -cycle time = 1ns = 10 -9 s For program a on Compiler A -CPI = 1 / (10 -9 x 10 9 ) = 1 For program a on Compiler B – CPI = 1.4 / (10 -9 x 1.2 x 10 9 ) = 1.667 Do the same for program b…

13. Exercise 6 - a Three enhancements with the following speedups are proposed for a new architecture: Speedup 1 = 30 Speedup 2 = 20 Speedup 3 = 10 Only one enhancement is usable at a time. a. If enhancements 1 and 2 are each usable for 30% of the time, what fraction of the time must enhancement 3 be used to achieve an overall speedup of 10?

14. Exercise 6 - a (sol) By applying Amdahl’s Law to three enhancements we find Overall Speedup = 1 / [(1 – ∑ FE i ) + ∑ (FE i / SE i )] = 1 / [1 – (FE 1 + FE 2 + FE 3 ) + ((FE 1 /SE 1 ) + (FE 2 /SE 2 ) + (FE 3 /SE 3 ))] By replacing every known variable by its value in the equation, we get: 10 = 1 / [1 – (0.3 + 0.3 + FE 3 ) + ((0.3/30) + (0.3/20) + (FE 3 /10))] Solving for FE3 we get 1 = 10 x (1 - 0.3 - 0.3 + 0.3/30 + 0.3/20 - FE 3 + FE 3 /10) = 4.25 – 9FE 3 Which means FE 3 = 3.25 / 9 = 0.361111… Thus, Enhancement 3 must be used around 36% of the time.

15. Exercise 6 - b b. Assume for some benchmark, the fraction of use is 15% for each of enhancements 1 and 2 and 70% for enhancement 3. We want to maximize performance. If only one enhancement can be implemented, which should it be? If two enhancements can be implemented, which should be chosen?

16. Exercise 6 – b (sol) When we can choose between enhancements, we pick the one or combination that offers the highest speedup. Following that logic, and if we can pick only one enhancement, Speedup E1 = 1 / [(1 – FE 1 ) + (FE 1 /SE 1 )] = 1 / [(1 – 0.15) + (0.15/30))] = 1.169 Speedup E2 = 1 / [(1 – FE 2 ) + (FE 2 /SE 2 )] = 1 / [(1 – 0.15) + (0.15/20))] = 1.166 Speedup E3 = 1 / [(1 – FE 3 ) + (FE 3 /SE 3 )] = 1 / [(1 – 0.70) + (0.70/10))] = 2.703 Since the Speedup provided by enhancement 3 is the highest, we choose it.

17. Exercise 6 – b (sol) But if two enhancements can be implemented, we will need to find the best combination which means finding the speedup for all combinations and comparing them Speedup E1 and E2 = 1 / [(1 – FE 1 – FE 2 ) + (FE 1 /SE 1 + FE 2 /SE 2 )] = 1 / [(1 – 0.15 – 0.15) + (0.15/30 + 0.15/20))] = 1.403 Speedup E1 and E3 = 1 / [(1 – FE 1 – FE 3 ) + (FE 1 /SE 1 + FE 3 /SE 3 )] = 1 / [(1 – 0.15 – 0.70) + (0.15/30 + 0.70/10))] = 4.444 Speedup E2 and E3 = 1 / [(1 – FE 2 – FE 3 ) + (FE 2 /SE 2 + FE 3 /SE 3 )] = 1 / [(1 – 0.15 – 0.70) + (0.15/20 + 0.70/10))] = 4.396 Since the Speedup provided by the combination of enhancements 1 and 3 is the highest, we choose them.