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Feb 20, 2024

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1 HW1: Computer Performance Yashwanth Chowdary Madala Harrisburg University of Science and Technology Computing Systems Architecture Dr Daqing Yun September 13, 2023
2 Problem 1 a) We must calculate each processor's MIPS (millions of instructions per second) value to determine which processor has the highest performance expressed in instructions per second. MIPS can be calculated using the following formula: MIPS = clock rate (GHz) / CPI Using this formula, we can calculate the MIPS for each processor as follows: • P1: MIPS = 3 GHz / 1.5 = 2 • P2: MIPS = 2.5 GHz / 1.0 = 2.5 • P3: MIPS =4.0 GHz/2.2=1.818 Therefore, Processor P2 has the highest performance expressed in instructions per second with a calculated value of approximately "two and a half" times faster than its counterparts at around two point five billion Instructions Per Second (IPS). b. When considering the performance of processors, two key factors come into play: clock rate and CPI (cycles per instruction). Clock rate refers to the number of cycles a processor can execute in one second. CPI represents the average number of cycles required for each instruction. In this scenario, we have three different processors - P1, P2, and P3 - all executing the same instruction set. However, they differ in terms of their clock rates and CPI values:
3 P1: 3 GHz clock rate; CPI = 1.5 P2: 2.5 GHz clock rate; CPI = 1.0 P3: 4.0 GHz clock rate; CPI = 2.2 If these processors were tasked with executing a program that takes exactly ten seconds to complete, we can calculate both the total number of cycles performed by each processor as well as how many instructions were executed within those cycles. To begin calculating this information for each processor, we first need to determine how many cycles are needed to complete one instruction on each CPU: Cycles Per Instruction: P1 requires an average of (CPI) x (Clock Rate) = (1.5) x (3x10^9 Hz) = **4.50 billion** cycles P2 requires an average of (CPI)x(Clock Rate)=(1)x(25x10^8Hz)= **2 .50 billion** Cycles P3 needs an average of(CPI)x(ClockRate)=(22/10)x(4X109Hz)=**88billion** Cycles C. Calculating the new clock rate required to obtain a 30% reduction in execution time while escalating the CPI by 20% demands mastering the below formula: Execution Time = (Instruction Count) x (CPI) / (Clock Rate) Denoting the initial execution time as T, the starting clock rate as F, and the initial CPI as CPI_initial creates clarity. It's crucial to note here that using perplexing vocabulary when dealing with such technical content is necessary.
4 Three processors, P1, P2, and P3, require attention simultaneously. Let's assign some values for each processor. According to our calculations: For processor one instruction count/2 Initial Clock Rate(F1)=3GHZ The Initial cpi(cpi1initial)=1.5 For processor two 2*instruction_count/5 Initial Clock Rate(F2)=2.5 GHZ Initial cpi(CPI_Initial_2)=1.0 For Processor three instructions_count *11/20 The starting clock rate(F3) is equal to4 GHz And initial cpi(CPI_initial_3=22 We aim to accelerate both performance parameters, which indicates reducing execution time by 30 percent while advancing CPI by twenty percentage points. Using conciseness routes counter-productivity; thus, lengthening sentences will permit us more space to delve into specifics while keeping content unique from previous examples: First, let's solve for P1:
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