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CPS3340 - Computer Architecture
Assignment 1: Computer Performance
Solution
- Conversion of number representations. Show the conversion procedure. a. Convert the following decimal numbers to binary, octal, and hexadecimal. (10 pts)
174, 3781 Answer: 10101110b = 256 8 = AE 16 Answer: 111011000101b = 7305 8 = EC5 16
b. Convert the following numbers to decimal. (10 pts)
(^10001011) b, 234 8 , FEA 16 100010112 = 1x2^7 + 1x2^3 + 1x2^1 + 1x2^0 = 139 2348 = 2x8^2 + 3x8^1 + 4x8^0 = 156 FEA 16 = 15x16^2 + 14x16^1 + 10x16^0 = 4074
- Consider the following two processors P1 and P2 executing the same instruction set with the clock rates and CPIs specified in the following table.
Processors P1 P Clock rate 2GHz 3GHz CPI 1.0 2.
a) Which processor has the highest performance? How much faster than the other two processors? (10 pts)
P1: Cycle time 500ps, CPU time: IC500ps1.0=500psIC P2: Cycle time 333ps, CPU time: IC333ps2.5=832.5psIC P1 is faster How much faster? Performance(P1)/Performance(P2)=832.5/500=1.
b) If the processors each execute a program in 100 seconds, find the number of cycles and the number of instructions for each processor. (10 pts)
of cycles
P1: 100s/500x10-12s=2x10^11 P2: 100s/333x10-12s=3x10^11
of instructions
P1: 2x10^11 /1.0=2x10^11 P2: 3x10^11 /2.5=1.2x10^11
c) For processor P2, we are trying to reduce the time by 40% but this leads to an increase of 20% in CPI, what clock rate should we have to achieve this time reduction? (10 pts)
CPU time = IC832.5(1-40%) = ICCycletime1.2*2. Solving this equation, Cycle time =166.5ps Clock rate=1/166.5ps=6.006GHz
- Consider two different implementations of the same ISA. There are four classes of instructions, Arithmetic, Store, Load, and Branch. The clock rate and CPI of each implementation are given in the following table.
Clock rate CPI Arithmetic CPI Store CPI Load CPI Branch P1 2.0 GHz 1 2 3 4 P2 2.5 GHz 2 2 2 2
a) Given a program with 10^6 instructions divided into classes as follows: 10% Arithmetic, 20% Store, 50% Load, and 20% Branch, which implementation is faster? (10 pts)
P1: Cycle Time= 1/2.0x10^9 = 500 ps 106 * (110%+220%+350%+420%)500x10-12^ s= 1.4x10-3^ s P2: Cycle time= 1/2.5x10^9 = 400 ps 106 * (210%+220%+250%+220%)400x10-12^ s= 0.8x10-3^ s P2 is faster
b) What is the global CPI for each implementation? (10 pts)
P1: total number of clock cycles: 10^6 * (110%+220%+350%+420%) total number of instructions: 10^6 global CPI = 2.
P2: total number of clock cycles: 10^6 * (210%+220%+250%+220%) total number of instructions: 10^6 global CPI = 2.
a) For P1, if we can improve the performance of the Branch instructions by reducing its CPI by half using a branch predictor, what is the speed up of the program? (10 pts)
