HW 1 Data Arquitecture

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Florida International University *

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MISC

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Electrical Engineering

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Feb 20, 2024

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HW #1 Assume a color display using 8 bits for each of the primary colors (red, green, blue) per pixel and a frame size of 1280x1024. a) What is the minimum size in bytes of the frame buffer to store a frame (1 byte = 8 bits)? Answer: --Frame Size = 1280x1024 pixels --Each color contains 8 bits, and we have 3 different colors. 3 colors * 8 bits =24 bits --Minimum size in bytes of the frame buffer to store a frame: (24bits/8bits) = 3 bytes per frame. --1280*1024 pixels = 1310720 pixels --1310720 * 3 bytes= 3932160 bytes per frame Minimum size in bytes per frame is equal to 3932160. b) How long would it take, at a minimum, for the frame to be sent over a 100Mbps network? Answer: --Calculate the size of the frame in bits = (1280 * 1024pixels) (3*8 bits) = 31457280 bits --Now we need to divide bits by bits per second --100Mbps = 100,000,000 bits per second --31457280 bits 100,000,000 bits per second = 0.314573 seconds 2, Consider three different processors P1, P2, and P3 executing the same instruction set. P1 has a 3GHz clock rate and a CPI of 1.5. P2 has a 2.5GHz clock rate and a CPI of 1.0, P3 has a 4GHz and a CPI of 2.4. a) Which processor has the highest performance expressed in instructions per second? Answer: -- P1, P2, and P3 same instruction set -- P1 = 3GHz clock rate & CPI of 1.5 -- P2 = 2.5GHz clock rate & a CPI of 1 -- P3 = 4GHz clock rate & CPI 2.4 -- Highest CPI Performance? -- Calculate the Number of instructions per second --P1 (3*10^9) / (1.5 * 10^9) =2 IPS --P2 (2.5*10^9) / (1 * 10^9) = 2.5 IPS) --P3 (4 *10^9) / (2.4 * 10^9) = 1.666667 IPS b) If the processors each execute a program in 10 seconds, find the number of cycles and the number of instructions. Answer:

--P1 Clock Cycles = Clock Rate * Execution time = 3GHz * 10 sec = 30 billion cycles --CPI * Cycles = 1.5 * 30 billion cycles = 45 billion Instructions. Or 30 billion cycles / 1.5 CPI = 20 billion instructions --P2 Clock Cycles = Clock Rate * Execution time = 2.5GHz * 10 sec= 25 billion Cycles --CPI * Cycles = 1* 25 billion cycles = 25 billion Instructions. Or 25 billion cycles / 1 CPI = 25 billion instructions --P3 Clock Cycles = Clock Rate * Execution time =4GHz *10 sec = 40 billion Cycles. -- CPI * Cycles = 2.4 * 40 billion cycles = 96 billion Instructions Or 40 billion cycles / 2.4 =16,67 billion instructions c) We are trying to reduce the execution time by 20% but this leads to an increase of 10% in the CPI. What clock rate should we have to get this time reduction? Answer: New Execution time = Old Execution time * (1- time reduction %) New CPI = Old CPI * (1 + CPI Increase Percentage) New Clock Rate = (Instruction count * old CPI) / (New Execution time *10^9) New Execution time = 10secs *(1-0.20) = 8 secs New CPI = 1.5 (1+0.10) = 1.65 CPI New Clock Rate = (45 billion instructions * 1.5/ ( 8 secs *10^9)=

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