HW1

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University of Central Florida *

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4798

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Computer Science

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Apr 3, 2024

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EEL 4798 - Big Data Instructor: Dr. Jun Wang Homework 1 1. (15 points) Draw a table to explain the difference among data-intensive scalable computing, cloud computing, HPC, AIoT (Artificial Intelligence of Things) in the following aspects: System Architectures, Application, and Programming Model. Data-Intensive Scalable Computing Cloud Computing HPC (High- Performance Computing) AIoT (Artificial Intelligence of Things) System Architectures Distributed systems, clusters Virtualized infrastructure Supercomputers, clusters Edge devices, IoT devices Applications Big Data analytics, Data mining Web applications, Storage Scientific simulations, Research Edge AI, Smart devices, IoT applications Programming Model MapReduce, Spark, Hadoop APIs, Virtual machines MPI, OpenMP, CUDA Edge computing frameworks, TensorFlow Lite 2. (15 points) Explain the differences among the three NAND flash memory technologies: SLC, MLC, and QLC. Discuss their applications. Single-Level Cell (SLC): Stores one bit of data per cell, making it super durable, speedy, and reliable. It's often used in crucial applications like industrial machines, cars, and military gear, as well as in high- performance storage gadgets where we really need things to be tough and fast. Multi-Level Cell (MLC): Stores two bits of data per cell, giving a good mix of durability, speed, and cost. It's not as tough as SLC, but it's faster than Quad-Level Cell (QLC). MLC applications include regular computer SSDs, business data storage, and some industrial gear where we want a balance between price and performance. Quad-Level Cell (QLC): Stores four bits of data per cell, making it the cheapest option for a lot of storage space. However, it's not as tough and is slightly slower than both SLC and MLC. QLC applications include everyday computer tasks, cheap SSDs for normal use, and storage where super-fast performance isn't the main goal, but having lots of space is important.
3. (20 points) SPEC ratios are shown for the Pentium and the Pentium Pro (Pentium+) processors Clock Rate Pentium SPECint Pentium+ SPECint Pentium SPECfp Pentium+ SPECfp 100 MHz 3.2 N/A 2.6 N/A 150 MHZ 4.3 6.0 3.0 5.1 200 MHZ 5.5 8.0 3.8 6.8 What can we learn from this information? 1). SPECint shows Pentium+ is 1.4 to 1.45 times faster than Pentium. How about SPECfp results? To find out how many times faster Pentium+ is compared to Pentium for SPECfp, we can use the formula: Performance Ratio = SPECfp of Pentium + ¿ SPECfp of Pentium = 6.8 3.8 = 1.789 ¿ So, Pentium+ is approximately 1.789 times faster than Pentium in terms of SPECfp. 2). Clock rate of the Pentium doubles from 100 MHz to 200 MHz, are both the SPECint and SPECfp performance doubled correspondingly? Simply doubling the clock rate doesn't mean you'll get double the performance in the real world. The boost in performance depends on how the processor is built, how efficiently it works, and any tweaks in the pipeline. For SPECint, Pentium goes from a 3.2 at 100 MHz to a 5.5 at 200 MHz. So, it has roughly doubled its performance. But for SPECfp at 100 MHz, Pentium scores 2.6, and at 200 MHz, it's 3.8. It's an improvement, but not necessarily double. Basically, turning up the clock rate does give you a decent performance bump, but it's not a straight 2x boost. There are too many factors in the mix, making it a bit more complicated than just doubling everything.
4. (15 points) We are using Amdahl’s Law and Gustafson’s Law to calculate overall speedup and scaled speedup. Given a 10% serial section of a program, calculate both speedups for a 64-processor machine. S Ahmdal = 1 ( Fs + Fp P ) = 1 ( 0.10 + 1 0.10 64 ) = 9.92 So, according to Amdahl's Law, the overall speedup is approximately 9.92 for a 64-processor machine. SGustafson = P Fs ( P 1 ) = 64 0.10 ∗( 64 1 )= 63.1 So, according to Gustafson's Law, the scaled speedup is approximately 63.1 for a 64-processor machine. 5. (35 points) Textbook (A Quantitative Approach to Computer Architecture, 5th edition) Page 63: Exercise 1.4 1.4 [20-min/10-min/20-min] <1.5> Figure 1.23 presents the power consumption of several computer system components. In this exercise, we will explore how the hard drive affects power consumption for the system. a. [20] <1.5> Assuming the maximum load for each component, and a power supply efficiency of 80%, what wattage must the server’s power supply deliver to a system with an Intel Pentium 4 chip, 2 GB 240- pin Kingston DRAM, and one 7200 rpm hard drive? Total Max Load = ( 1 66 ) + ( 2 2.3 ) + ( 1 7.9 ) = 78.5 W Efficiency = Total Max Load P → P = 78.5 0.8 = 98.125 W The server should supply 98.125 W of power.
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b. [10] <1.5> How much power will the 7200 rpm disk drive consume if it is idle roughly 60% of the time? Total Power = ( 0.6 4 ) + ( 0.4 7.9 ) = 5.56 W The power consumed is 5.56 W. c . [20] <1.5> Given that the time to read data off a 7200 rpm disk drive will be roughly 75% of a 5400 rpm disk, at what idle time of the 7200 rpm disk will the power consumption be equal, on average, for the two disks? 7200 RPM Power = ( 4 ) ( 1 x ) + ( 7.9 x ) = 4 + 3.9 x 5400 RPM Power = ( 2.9 ) ( 1 x 0.75 ) + ( 7 x 0.75 ) = 2.9 + 4.1 0.75 x 7200 RPM Power = 5400 RPM Power 4 + 3.9 x = 2.9 + 4.1 0.75 x x = 0.825 1.175 = 0.7 The idle time of the 7200 RPM drive is determined by 1 0.7 = 0.3 .