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Computer Organization Lab 1 Soufiane berouel 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 – ∑ FEi) + ∑ (FEi / SEi)] – FE is the proportion affected, SE is the speed up for the specific proportion 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? Exercise 1 (sol) • Each frame requires 1280 x 800 x 3 = 3072000 ~ 3 Mbytes 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? Exercise 2 (sol) • Suppose the program has N instructions. • Time taken to execute on P1 is = 1.5 N / (2 x 109) = 0.75 N x 10-9 • Time taken to execute on P2 is = N/ (1.5 x 109) = 0.66 N x 10-9 • Time taken to execute on P3 is = 2.5 N/ (3 x 109) = 0.83 N x 10-9 • P2 has the best performance (since it takes the least time to execute). Exercise 3 • Consider 3 processors P1, P2 and P3 with same instruction set with clock rates and CPI given below: P1 P2 P3 clock rate CPI 2 GHz 1.5 GHz 3 GHz 1.5 1.0 2.5 • If the processors each execute a program in 10 seconds, find the number of cycles and the number of instructions. Exercise 3 (sol) • For P1 – 10 (execution time) = 1.5 (CPI) x N1/ (2 x 109) – So N1 (number of instructions) = 1.33 x 1010 – Number of Cycles = 1.5 x N1 = 2 x 1010 • …Do the same for P2 and P3 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? Exercise 4 (sol) • Execution time = cycle time x CPI x no. of inst – Cycle time = 1/(2 x 109) – 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 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 a) b) 1.0 x 109 1.4 x 109 exec. Time 1s 0.8 s no. instructions 1.2 x 109 1.2 x 109 exec. Time 1.4 s 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. 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 109 ) = 1 • For program a on Compiler B – CPI = 1.4 / (10-9 x 1.2 x 109 ) = 1.667 • Do the same for program b… Exercise 6 - a • Three enhancements with the following speedups are proposed for a new architecture: Speedup1 = 30 Speedup2 = 20 Speedup3 = 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? Exercise 6 - a (sol) • By applying Amdahl’s Law to three enhancements we find Overall Speedup = 1 / [(1 – ∑ FEi) + ∑ (FEi / SEi)] = 1 / [1 – (FE1 + FE2 + FE3) + ((FE1/SE1) + (FE2/SE2) + (FE3/SE3))] • By replacing every known variable by its value in the equation, we get: 10 = 1 / [1 – (0.3 + 0.3 + FE3) + ((0.3/30) + (0.3/20) + (FE3/10))] • Solving for FE3 we get 1 = 10 x (1 - 0.3 - 0.3 + 0.3/30 + 0.3/20 - FE3 + FE3/10) = 4.25 – 9FE3 • Which means FE3 = 3.25 / 9 = 0.361111… • Thus, Enhancement 3 must be used around 36% of the time. 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? 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, SpeedupE1 = 1 / [(1 – FE1) + (FE1/SE1)] = 1 / [(1 – 0.15) + (0.15/30))] = 1.169 SpeedupE2 = 1 / [(1 – FE2) + (FE2/SE2)] = 1 / [(1 – 0.15) + (0.15/20))] = 1.166 SpeedupE3 = 1 / [(1 – FE3) + (FE3/SE3)] = 1 / [(1 – 0.70) + (0.70/10))] = 2.703 • Since the Speedup provided by enhancement 3 is the highest, we choose it. 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 SpeedupE1 and E2 = 1 / [(1 – FE1 – FE2) + (FE1/SE1 + FE2/SE2)] = 1 / [(1 – 0.15 – 0.15) + (0.15/30 + 0.15/20))] = 1.403 SpeedupE1 and E3 = 1 / [(1 – FE1 – FE3) + (FE1/SE1 + FE3/SE3)] = 1 / [(1 – 0.15 – 0.70) + (0.15/30 + 0.70/10))] = 4.444 SpeedupE2 and E3 = 1 / [(1 – FE2 – FE3) + (FE2/SE2 + FE3/SE3)] = 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.