Study-Set-01-Course-Introduction

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| Exponent Rule EEL ver Video Logging on iPhone Prototype 3801 Given: As an intern at Apple, you’re tasked to determine how many hours of video that a prototype iPhone can store on it's 32 GB flash memory. The prototype records 32 frames per second. You've taken some measurements and estimate that each compressed frame occupies 128 KB on average. Sought: If all overheads are ignored, what is the longest video that this prototype iPhone could store? Solution: First, determine the number of bytes per second the iPhone records at: Bytes/second = 32 x 128 KB/sec =2°x2" x (2" Bytes/sec) = 2°"*1° Bytes/sec = 2°* Bytes/sec Next, divide this data rate into the flash memory capacity as follows: recording time =[32GB] / [2% Bytes/sec] =[2° x2*° Bytes] / [2* Bytes/sec] =[2°*% Bytes] /[2* Bytes/sec] =[2* Bytes] /[2°* Bytes/sec] =2 gec =2" sec =2° x2"° = 8x1024 =8192 sec or =[8192 sec] /[3600 sec/hour] =2.276 hours Note: Exponent Rule applies only to powers-of-2 such as memory capacities (KB, MB, GB, ...) but does not apply to processor clock rates (KHz, MHz, GHz, ...) which reflect powers of ten. Study Set 01 — Course Introduction slide 1

€ Units, Prefixes, and Memory Capacities °©°c- e Smart Grid Data Rate 3801 You are tasked to specify the flash storage device in General Electric’'s Data Logger Unit (DLU) used in the smart power grid. The DLU records data from 8 processors. Each processor has 4 sensors, and each sensor generates 32 KB of data per second. Which of the choices below provide sufficient capacity for the DLU to record 1 day’'s worth of data? a) 32,000 Bytes b) 32 KB c) 32,000 KB d) 32 MB e) 32,000 MB f) 32 GB g) 64 GB h) 128 GB i) none of the choices listed have sufficient capacity Solution: The DLU records: (8 processors)*(4 sensors/processor)*([32KB/sec]/sensor) = 23*¥22*[2>%*210 Byte/sec] = 220 Byte/sec = 1MB/sec. There are (3600 sec/hour)(24hour/day)=86400 sec/day. Thus, (86400 sec/day)* (1MB/sec) = 86400 MB/day. So for one day, 86400MB capacity is needed. 1 GB is 1024 MB so ... (86400MB)(1GB/1024MB) = 84.375GB capacity is required. Of the choices listed, only choice h) is >= 84.375GB. Study Set 01 — Course Introduction slide 2

eeL 3801 & Exponent Rule and Memory Capacities UCF Data Warehouse Capacity Given: You work at Amazon and manage the datacenter backup capacity. Your biggest customer, Google, has 2 data warehouses each with 572 servers containing 728 nodes per server. Each node provides 8TB of disk storage. Every 3 months, Google requires a full uncompressed backup of their data. How much storage is required to handle backing up Google’s data every year? Sought: 1) How many bits per year are transferred from Google to Amazon? Simplify your final answer. 2) How many 1 TB drives would be required to store all of the data exchanged? a) under 1,000 Db) 1,001 to 10,000 c) 10,001 to 100,000 d) 100,001 to 1,000,000 e) 1,000,000 to 4,000,000 f) 4,000,001 to 8,000,000 g) more than 8,000,0000 Solution: 1) Since all quantities are powers of two, Exponent Rule yields the result immediately: Annual Transfer = 2" x2° x2" x (2° x2*° Bytes)x (2° transfers/year)x (2° bits/Byte) _ DMO+7+3+404243 pita Jyear =2% bits/lyear = 36,893,488,147,419,103,232 bits annually 2) 25° bits/year = 252 Bytes/year so 292 Bytes/240 Bytes = 222 = 4 x 220 which exceeds 4 million disk drives as that is (4*1024*1024). So that would be a huge data center without use of compression, incremental backup, and data de-duplication. => Tip: powers-of-2 exponents can be written directly for many processor design problems! Study Set 01 — Course Introduction slide 3

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& Memory Design EEL [ . - UCF Example: DDR4 Cost Calculation 3801 Just how inexpensive ... is DRAM | - (Dynamic Random Access Memory) these days? i et S M i » DRAM Given: In the year 2000, main memory cost $100 for a 64MB module. __P"/'g;’;:"_;s Currently, DDR4 DRAM costs about $200 for a 16GB module. Year 2015... 16GB DDR4 288-pin module “main Note: $100 in year 2000 equates to roughly $137 dollars today. . T memory” used by a processor Partial Credit 1: What was the cost of 4GB memory in the year 20007? _ Solution: [4GB/64MB]*[$100] = (272)*(2"30)/(2"6)*(2720)*$100 = (2732)/(2"26) = (2"6)*$100 = $6400 Partial DRAM is different than non-volatile memory: Credit 2: What was the cost per bit in 2000, in terms of today’s dollars? Solution: ($100 / 64M Byte) * [(1 Byte )/(8 bits)] = [{$100 * ($137/$100)} / (2426) ] * [1/(273)] = [$100 * 1.37 ]/ (2729) = $137 /[ 1024*1024*1024 / 2] = $(2.552e-7) per bit Study Set 01 — Course Introduction slide 4

Considerations in Memory Design Example: DDR4 Module Selection eeL 3801 Given: Choose memory module X or Y to add 16GB per processor board in a large server farm. Larger Image: / Partial Credit 1: Solution: Partial Credit 2: Solution: Partial Credit 3: Solution: Partial Credit 4: Study Set 01 - Module-X @ orly $177.75 72 Module-Y @ Mfr List $249-60 w4 Hewlett Packard HP HP 8GB 1RX4 PC4- Save $71.25 (29%) W Hewlett Packard HP HP 16GB 2RX4 PC4- 2133P-R Kit -y 2133P-R Kit Manufacturer Part# 726718-B21 AbD T e .. Manufacturer Part# 726719-B21 Provantage Code: CMPMOEM Provantage Code: CMPMOEN UPC Code: 887758376614 [ 354INSTOCK | UPC Code: 887758376621 » Product Type: RAM Module Larger Image: / » Product Tvoe: RAM Modul » Memory Size: 8 GB [O[0F7 (apoTowshLisT) b Memory Size: 16GB 1(0]0)7% » Memory Technology: DDR4 SDRAM SATISFACTION » Memory Technology: DDR4 SDRAM SATISFACTION » Memory Speed: 2133 MHz GUARANTEED! » Memory Speed: 2133 MHz | GUARANTEED! Which DDR4 module above offers the least cost per bit? Mod-X:$177.75/8GB=%22.22/GB; Mod-Y:$283.40/16GB=%$17.71/GB —->Module Y costs less for 16GB Which DDR4 module above offers a higher maximum data transfer rate”? Both have Memory Speed of 2133 MHz so speed is the same (details later) If average power dissipation of Mod-X and Mod-Y is 5W and 8W, respectively, then which design will minimize the cooling cost in the server farm? For 16GB then Mod-X dissipates (5W/module)*(2 modules) = 10W For 16GB then Mod-Y dissipates (8W/module)*(1 module) = 8W Less power dissipation = less cooling cost, thus Module-Y design preferable. Given all of the above then which design do you recommend? Ans: Module Y since preferable cost and cooling, at identical speed, and occupies 1 less memory slot. Course Infroduction Only $28340 Mfr List $399-60 Save $115.60 (29%) ADD TO CART (ADD TO WISH LlST) slide 5

| @ Comparing Processor Characteristics EEL ver Example: classic IBM Mainframe vs today’s iPhone 3801 Given: |BM’'s System 370 Mainframe had a single 32-bit processor with a 12.5MHz clock for single- clock instructions. It has a memory capacity sufficient to store 4M distinct word-sized values and the system cost was $5M. The Apple A8 processor used in iPhones has a dual-core 64-bit processor with a 1.4GHz clock for single clock instructions. When configured with memory sufficient to store 16G distinct word- sized values, the iphone costs $800. Sought: What is the word-width of the System 370 in units of Bytes? Solution: 4 Bytes, because the word width is given as “32-bit processor” and there are 8 bits per Byte. Sought: Which system has higher data processing throughput of 32-bit values? And by what performance ratio? And by what cost performance ratio even if inflation is ignored? Solution: IBM 370: (1 core)*(12.5M instructions/sec/core)= (12.5E6) instructions/sec iPhone: (2 core)*(1.4G instructions/sec/core)*(64-bit/32-bit) = (5.6E9) instructions/sec Perf Ratio: iPhone as higher throughput by (5.6E9 instr/sec)/(12.5E6 instr/sec)=448-fold higher Cost Perf Ratio: iPhone delivers (448-fold)*($5,000,000/$800) = 2,800,000-fold increase per dollar Sought: What is the memory capacity of the iPhone as given, expressed as precise value in Bytes? Solution: The iPhone’s word size is 64-bits each = 8 Bytes/word. The memory capacity of the iPhone is 16G words = (2*4)*(2*30) words. Thus, [2*3 Bytes/word ] * [(2*4)*(2”30) words]= 2*37 Bytes = (277) GBytes = 128 GB = 128%1024*1024*1024 = 137,438,953,472 Bytes Reminder: Memory capacities (including quantities of values) are always powers of two; clock rates, instructions, and dollars are powers of ten as are all non-storage-related quantities. Only storage-related quantities are powers of 2. Study Set 01 — Course Introduction slide 6

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© Computer Abstractions geL vor Units and Exponent Rule 2801 Problems: Partial Credit 1: For a color display using 8 bits for each of the primary colors (red, green, blue) 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? Partial Credit 2: If a computer has a main memory of 2 GB, how many frames could it store, assuming the memory contains no other information? Partial Credit 3: If a computer connected to a 1 gigabit Ethernet network needs to send a 256 Kbytes file, how long it would take? Solutions: Perspective on problems: the purpose of assighing these problems is to reinforce that each Byte is defined as 8 bits and to practice using the Exponent Rule to compute quickly for quantities using powers of two. Solution 1: A Byte is always 8 bits. Thus, (1 Byte/8 bits)(8 bits/color)(3 color/pixel)(1280*800) = 3,072,000 Bytes Solution 2: When working with memory capacities, use the Exponent Rule shown on Lecture Slide 1.6 to compute quick he using powers of two. A memory capacity of 2GB = 2*(2730) Bytes = (2721)*(2/30) Bytes =2/(30+1) Bytes = 27231 Bytes. Thus, (2231) Byte) * (1 frame/3072000 Byte). = 699.05 frames. Since an image requires an integral number of frames then the number of complete frames which can reside in memory is 699 Solution 3: Using precise numbers, a 1Gbit network has throughput of (2230 bits)/second. Meanwhile, a file of size 256KB contains (2/8) KBytes = (228)*(2710) Bytes= 2*(8+10) Bytes = (2218)Bytes. Thus, (218 Bytes)(8 bits/1Byte)(second/2/30 bits) = (2218 Bytes)(2”*3 bits/1 Byte)(second/2”30bits) =(27(18+3))/(2”~30) second = 27(21-30) second = 27(-9) second = 1/(2”9) second = 1/512 second = 0.001953125 second = 1.95msec Study Set 01 — Course Introduction slide 7