HW2-Solutions
Murata Power Solutions MP-HW80EVAL-01评估板用户指南说明书
MP-HW80EVAL-01MP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARY1.IntroductionMP-HW80EVAL-01 Evaluation Board User GuideThe MP-HW80EVAL-01 Evaluation Board is designed to assist with the evaluation of the IRH-W80, 250W, 10:1input range DC/DC converter module from Murata Power Solutions. The IRH-W80 series of isolated regulatedconverter modules, deliver an impressive 250W single output from a wide input range of 16V – 160Vdc, complyingwith the input battery voltage transient range of EN50155.The half brick module offers high efficiency levels up to 90%, with an input – output isolation voltage of 4242Vdc.The module features Overvoltage, Overcurrent, Short Circuit, Adjustable output voltage, Adjustable Current Limit,Positive or Negative Logic enable, Pulse output signal and Hold up function for an external capacitor. See Page 10for full schematic.Figure 1 – Evaluation BoardMP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARY2.Setup2.1 ConnectionsThis section describes the connector locations/pinouts on the evaluation board, to enable correct set up.Figure 2 - Connectors Position / Resistor Functions (Top View)H1 Input Voltage ConnectorH2 Signal Connector Resistor / Function Pin 1 VIN- Pin 1 PULSE R1 UVLOPin 2 VIN- Pin 2 VIN- R2 TRIM DOWNPin 3 VIN+ Pin 3 ON/OFF R3 TRIM UPPin 4 VIN+ Pin 4 PE R4 CHARGER5 SENSE (+) H3 Output Voltage Connector H4 External Hold Up Connector R6 SENSE (-)Pin 1 VOUT- Pin 1 VIN- R7 OCPPin 2 VOUT- Pin 2 CHOLD+ R1, R2, R3, R5, R6, R7 = SMD0805 Pin 3 VOUT+Pin 4 VOUT+MP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARY2.2 On Board Component Values2.2.3 OCP Resistor ValueR7 = OCP – Overcurrent Protection.By adding a resistor SMD0805 to position R7 on the PCB, it is possible to set the overcurrent threshold point.Leave unconnected if not used.Figure 3 - OCP Resistor SelectionR7 OCP Value 562Ω 665Ω 845Ω 1.13kΩ 1.69kΩ 3.32kΩ 6.65kΩ 10kΩ Open 24Vin 14.5A 16.2A 18.2A 20.2A 22.4A 24.3A 25A 25.5A 26A48Vin 15.5A 17.4A 19.4A 21.1A 23A 24.5A 25.4A 25.9A 26.5A72Vin 15.5A 17.4A 19.4A 21.1A 23A 24.5A 25.4A 25.9A 26.5A110Vin 11.6A 14A 17.1A 19.8A 22.1A 24.5A 25.4A 25.8A 27A2.2.4 Under Voltage Lockout. (UVLO)By adding a SMD0805 resistor to R1 position as per the table below, the converter will shut down if the inputvoltage drops below the threshold. The converter will automatically restart when the input voltages rises above theUVLO threshold. Leave unconnected if not used.Figure 4 - UVLO Resistor Value TableVin 24V 36V 48V 72V 96V 110VTurn Off 10V-12V 17.5V-19.5V 26V-28V 40V-43V 56V-60V 65V-70VTurn On 13V-15V 22V-24V 31V-34V 48V-51V 68V-72V 80V-84VResistor Value Open 27.4kΩ 13kΩ 6.8kΩ 4.3kΩ 3.57kΩ2.2.5 Output Voltage Trim ResistorsR3 and R2 – Trimming the Output Voltage.The trim pin of the converter allows the user to adjust the output either +10% or -20% by using SMD0805fixed value resistors.Figure 5 - Output Trim Resistor Values12V OutputOutput Voltage 9.6V 10.8V 11.4V 12.6V 13.2VR3 - Trim Up NA NA NA 188kΩ 97kΩR2 - Trim Down 4kΩ 8.9kΩ 18.7kΩ NA NAMP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARY2.2.6 Sense ConnectionsIf intended to utilize the output sense feature, please ensure that you short out, or add a zero ohm link in positionsR5, Sense (+) and R6, Sense (-). If it is proposed to use the sense function, then please leave R5 and R6 opencircuit, and connect output sense lines to the load. The sense connections can compensate up to 0.5V voltage dropof output leads.Figure 6 - Remote Sense Connection2.2.7 Hold Up CircuitThe BUS pin of the IRH250W80 module, is a voltage source output of 80Vdc to allow external capacitors to be connected in order to provide hold up power of the converter. The eval board houses a resistor (R4) to slowlycharge the capacitors up, and a feed forward diode (D1) for rapid discharge into the module during hold up mode.The capacitor value can be as per the below table to provide 10msec or 20msec of hold up. Connect the capacitorto H4 Pin 1 = VIN-, Pin 2 = CHOLD+. As per Figure 2.Figure 7 - Hold-Up Capacitor ValuesHold-up time 24Vin36Vin 48Vin 72Vin 96Vin 110Vin 10ms 2200uF 2200uF 2200uF 2200uF 1100uF 700uF20ms 4400uF 4400uF 4400uF 4400uF 2200uF 1400uF2.2.8 External FusingThe evaluation board does not have any fusing protection, the user must provide external fusing, circuit breaker protection as required.MP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARY2.2.9 Pulse OutThis pin outputs a 1kHz 50% duty cycle pulse voltage with 12V amplitude. It is designed to provide a bootstrapsignal for the input inrush current limit circuit and could also indicate operating status with a LED connected.Leave unconnected if not used.2.2.10 On/Off ControlConnect the On/Off pin (Pin 3, H2) to -VIN (Pin 2, H2) connection if “Negative” logic level is used in the part numberof the IRH-W80. Leave the On/Off pin open if “Positive” logic level is used.Note: A mechanical On/Off switch is also provided on the top side of the evaluation board.MP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARY3. EMI CircuitThe following schematic below meets EN55011 Class A.Figure 8 - EMI SchematicFigure 9 - EMI BOMReference Manufacturer MPN Type Specifications QtyMOV1 Epcos B72214S0141K101 Varistor 180V, 36J 1TVS1 Littel Fuse 1.5KE220A TVS diode 185V, 1.5KW 1C1 Faratronic C212E475K9AC000 Polyester capacitor 250V, 4.7uF 1C2, C3, C4 Murata GRM43DR72E474KW01L Capacitor MLCC 250V, 0.47uF 3C5, C6, C11, C12, C13, C14 Murata DE1E3RA102MA4BQ01F Safety ceramic cap 500V, 1000pF 6C9, C10 NCC EKXJ251EXX271ML40S E-cap 250V, 270uF 2C15, C16 Murata DE1E3RA472MA4BQ01F Safety ceramic cap 500V, 4700pF 2L1 Wurth 7448262013 CM choke 1.3mH, 20A 1L2, L3 Bourns 2300HT-220-V-RC1951 DM choke 22uH, 19A 2C7, C8 NA NOT USED NA NA 0C17, C18 NA NOT USED NA NA 0MP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARYFigure 10 - EMI ResultsLimit Line as per EN55011 level AVin = 110V, Line LMP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARY4. Circuit SchematicMP-HW80EVAL-01Analog DC-DC Evaluation BoardUser GuidePRELIMINARY5. Mechanical Drawings / Dimensions3: UNIT WEIGHT = 690 gramsMP-HW80EVAL-01 Analog DC-DC Evaluation Board User GuideDCAN-70_MP-HW80EVAL-01_A01.D08 Page 12 of 12PRELIMINARY6. Packaging InformationMurata Power Solutions, Inc.129 Flanders Rd., Westborough, MA 01581 USAISO 9001 and 14001 REGISTERED This product is subject to the following operating requirements and the Life and Safety Critical Application Sales Policy : Refer to: /requirements/Murata Power Solutions, Inc. makes no representation that the use of its products in the circuits described herein, or the use of other technical information contained herein, will not infringe upon existing or future patent rights. The descriptions contained herein do not imply the granting of licenses to make, use, or sell equipment constructed in accordance therewith. Specifications are subject to change w ithout notice. © 2021 Murata Power Solutions, Inc.MP-HW80EVAL-01。
腾讯NDA限制-腾讯NDA限制-Jacinto6Eco SoC电源解决方案-DRA72 TDA2
TPS22965 + TPS51200
8
New PDN Concept
0.40 = $3.25
(3 AVS @ 1‐2.5A, Dual 1.8/3.3V IO, DDR3L)
18 + 4 + 4.6 + 4.6 + 4 + 9 = 44
$4.42 128 $2.53 74
$7.32 190 $4.64 119
5. “PDN’s AVS Capability” is the achievable power if all AVS power rails are increased to 90% of capacity while other power rails remain at typical Use Case modelled values.
(3 AVS @ 1‐3.5A, Dual 1.8/3.3V IO, DDR3L) (similar to EVM PDN #0)
49 + 9 + 4 = 62
#8.2 – LP87524 + LP5912 + TLV713 + LP5907 +
9.92
1.26 + 0.19 + 0.07 + 0.10 + 0.10 +
2. PDN Support component (Rs, Cs & Ls) pricing from Mouser Distribution website using single 4k – 10k/reel qty costs as of May 2016. Both PDN Support & PDN Total Costs have been provided for relative comparison only, individual customer volume pricing may vary.
hw2-consumer theory
Advanced Microeconomics Fall2014Homework#2Due date:Oct.28in class1.Let x1,x2,x3and x4be consumption bundles in the consumption set R n+.Use theaxioms and definitions in the lecture note to prove the following statements.(a)If both x1 x2and x2 x1are false,then x1∼x2.(b)If x1 x2and x2∼x3and x3 x4,then x1 x4.2.Consider the consumer’s problem:max x1,x2≥0x51x72s.t.p1x1+p2x2≤ywhere p1,p2and y are(positive)prices and income.(a)Is it possible for this problem to have a solution(x∗1,x∗2)with x∗1=0or x∗2=0?Why or why not?(b)Is it possible for this problem to have a solution(x∗1,x∗2)with p1x∗1+p2x∗2<y?Why or why not?(c)Derive the Marshallian demand functions x1(p1,p2,y)and x2(p1,p2,y),and theindirect utility function v(p1,p2,y).3.Suppose the utility function of a consumer is u(x1,x2)=min{a1x1,a2x2}where a1and a2are positive parameters.(a)Is the consumer’s preference continuous?Is it locally nonsatiated?Is it strictlymonotonic?Is it convex?Is it strictly convex?(b)Derive the Marshallian demand functions and indirect utility function.(c)Verify that the Marshallian demand functions and indirect utility function arehomogeneous of degree0.(d)Verify the Roy’s identity for this example.4.For CES utility function u(x1,x2)=(xρ1+xρ2)1/ρ,where0=ρ<1.(a)Find the Marshallian demand functions x i (p 1,p 2,y ),i =1,2;(b)Show that the limits of the Marshallian demand functions x i (p 1,p 2,y ),i =1,2that you found in part (a)exist when ρgoes to 0.Moreover,verify that the limits equals the Marshallian demand function for u (x 1,x 2)=√x 1x 2(Cobb-Douglas utility function).(c)Show that the limits of the Marshallian demand functions x i (p 1,p 2,y ),i =1,2that you found in part (a)exist when ρgoes to −∞.Moreover,verify that the limits equals the Marshallian demand function for u (x 1,x 2)=min (x 1,x 2).5.Find the Marshallian demand functions x i (p 1,p 2,y ),i =1,2with the following utility functions(a)u (x 1,x 2)=2√x 1+x 2.(b)u (x 1,x 2)=ln(1+x 1)+x 2.(Caution:the solution may be in the corner.)6.Suppose that there are n goods,and a consumer’s preference relation is represented by some utility function u (·).Suppose that the indirect utility function isv (p,y )=y ni =1p −a i i ,where a 1,...,a n >0are parameters with ni =1a i =1.Derive the Marshallian de-mand functions x i (p,y ),expenditure function e (p,u ),and Hicksian demand functions x h i (p,u ).7.LetM = 1ββ1 .What are the eigenvalues of M ?when is M positive definite?8.Suppose a consumer has the following quasi-linear utility function:u (x 1,x 2,x 3)=a 1x 1+a 2x 2+x 3−12(x 21+x 22+2βx 1x 2),where a 1>0,a 2>0,β∈(−1,1).Normalize the price of x 3to be 1,i.e.,p 3=1.(a)Derive the Marshall demand function x i (p 1,p 2,1,y ),i =1,2for this consumer.(You can assume the constants a 1,a 2are sufficiently large so that the solutions x i given by the FOCs are positive ).(b)Find the sign of∂x 1(p 1,p 2,1,y )∂p 2and relate it to the sign of β?How to interpret theparameter β?2。
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LTLX Swiss International Air LinesLY El Al Israel AirlinesLZMA Malev Hungarian AirlinesMB MNG AirlinesMD Air MadagascarME Middle East AirlinesMF Xiamen AirlinesMH Malaysia AirlinesMI Silk AirMJ Lineas Aereas Privadas Argentinas MK Air MauritiusMM Peach AviationMP MartinairMR Air MauritaniaMS EgyptairMT Thomas Cook AirlinesMU China Eastern AirlinesMV Armenian International Airways MX Mexicana de AviacionMZ Merpati Nusantara AirlinesN3Omskavia AirlineNB Sterling EuropeanNE Sky Europe AirlinesNG Lauda AirNH All Nippon AirwaysNI PortugaliaNK Spirit AirlinesNL Shaheen Air InternationalNQ Air JapanNS Hebei AirlinesNVNX Air MacauNW Northwest AirlinesNZ Air New ZealandO3SF AirlinesOA Olympic AirlinesOB Astrahan AirlinesOH ComairOK CSA Czech AirlinesOM Miat-Mongolian AirlinesOO SkyWestOQOS Austrian AirlinesOU Croatia AirlinesOV Estonian AirOW Executive AirlinesOX Orient Thai AirlinesOZ Asiana AirlinesPAP2UTairP7EastLinePC Fiji AirPDPE Air Europe SPAPG Bangkok AirwaysPK Pakistan International Airlines PN West AirPO PolarPR Philippine AirlinesPS Ukraine International AirlinesPX Air NiuginiPY Surinam AirwaysQ3PB AirQ540-Mile AirQF Qantas AirwaysQTQQ Alliance Airlines Reno AirQR Qatar AirwaysQS Travel Service AirlinesQS Smart WingsQV Lao AviationQWQX Horizon AirQZQZ Indonesia AirAsiaR2Orenburg AirlinesRA Royal Nepal AirlinesRB Syrian Arab AirlinesRC Atlantic AirwaysRE Aer ArannRF Florida West International Airways RG Varig BrasilRI Mandala AirlinesRJ Royal Jordanian AirlineRQ Kam AirRN Air HorizonsRN EuralairRO TaromS2Air SaharaS4SATA InternationalS7S7 AirlinesS9East African Safari AirSA South African AirwaysSC Shandong AirlinesSD Sudan AirwaysSG JetsgoSG SpicejetSH FlyMeSJ Freedom AirSK SAS Scandinavian Airlines SN Brussels AirlinesSQ Singapore AirlinesSR SwissAirSU Aeroflot Russian Airlines SV Saudi Arabian AirlinesSW Air NamibiaSY Sun Country AirlinesSZ Air China Southwest Company T4Hellas JetT5TurkmenistanT7Twin JetT9TransMerridian AirlinesTA Taca InternationalTE flyLALTG Thai AirwaysTH Transmile Air ServicesTH British Airways CitiExpress TL Trans Mediterranean Airways TL Airnorth RegionalTK Turkish AirlinesTN Air Tahiti NuiTP TAP AirTQ Tandem AeroTR Tiger AirwaysTS Air TransatTT Air LithuaniaTU Tunis AirTX Air CaraibesTZ ATAU2easyJetU2UFSU5USA 3000 AirlinesU6Ural AirlinesU8ArmaviaU9Tatarstan AirlinesUA United AirlinesUB Myanmar AirwaysUL SriLankan AirlinesUM Air ZimbabweUNUO HK ExpressUQ O'Connor AirlinesUS US AirwaysUU Air AustralUX Air Europa Lineas Aereas UY Cameroon AirlinesV7Air SenegalV9BALVA Virgin AustraliaVD Air LiberteVF ValuAirVH AeropostalVI Volga-Dnepr AirlinesVN Vietnam AirlinesVR Cape Verde AirlinesVS Virgin Atlantic Airlines VV AerosvitVY Formosa AirlinesVY Vueling AirlinesVZ MyTravel AirwaysW2Canadian Western Airlines W3Flyhy Cargo AirlinesW5Mahan AirW6WizzairW8Cargojet Airways Ltd.W9Sky Service BvbaW9Eastwind Airlines Inc. 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hygropalm hp32说明书
English1 The HP32 is a highly accurate handheld instrument compatible with all HC2 and HC2A probes to perform measurement and logging of relative humidity, temperature and calculated psychro-metric parameters.1.1 • HygroPalm HP32• Soft Case• Micro-USB Service Cable (AC0006)• Belt Buckle Set • Lanyard • Certificate• HW4-LITE license key 22.1 PROBEThe HP32 is compatible with all HC2 and HC2A probes. Simply connect it to the probe input on top.A calibration of the HP32 after changing the probe is not necessary.range of conventional mounting accessories. For example, camera tripods or, the belt buckle delivered with the device.2.3 MAGNETICThe magnet integrated at the back of the HP32 enables easy installation on metallic surfaces such as, for example, ventilation shafts.ered cable to charge it.The device has a stand-by and an auto switch off feature. After 5 minutes without user interac-tion, the display is dimmed. Moreover, after a total of 10 minutes without user interaction, the device is switched off to save battery.These features do not affect data logging operations. While data logging is active, although the screen is dimmed or off, data points are still logged and saved.3.3 DISPLAY AND HOME SCREEN4 GENERAL SETTINGS 4.1 LANGUAGETo change the display language, go into:Main menu [OK] ✂Settings ✂ LanguageAvailable languages: English, Deutsch, Français, Italiano 4.2 DATE/TIMETo set up the date and time, go into:Main menu [OK] ✂Settings ✂ Date / TimeThe following settings can also be set beside the date and time.• Date format: DIN, US, ISO • Time format: 12 h, 24 hThe date and time must be set correctly for proper operation of the device.4.3 UNITSSet units used and displayed by the HP32 into:Main menu [OK] ✂ Settings ✂ Units Your options are as follow:• Distance: m, ft • Temperature: °C, °F • Pressure: hPa, mmHg • Weight: kg, lbs[POWER]Switches the HP32 on or off.Probe connector Tripod socket Reset button Integrated magnet Lanyard fixation USB port[DOWN]Cursor down[LEFT] [RIGHT]Action buttons, their action are indicated by the on-screen labels.[UP]• Cursor up• Parameter selection (home screen)Status Bar• Time and date • Hold indicator • Log indicator • Battery status Measured valuesAction Buttons labels[OK]• Opens the main menu / submenus • Confirm entries 4.4 Beside relative humidity and temperature, the HP32 can display 2 of the following 10 psychro-metric parameters:• Dew / Frost point (Dp/Fp)• Wet bulb temperature (Tw)• Enthalpy (H)• Vapor concentration (Dv)• Specific humidity (Q)however, still possible.You must name storage locations before use. Simply use the [Rename] action button when selecting a storage location to assign it a name.6 VIEWING SAVED DATA 6.1 SPOT MEASUREMENTSReview the spot measurements by selecting their storage location within:Main Menu [OK] ✂ Memory – Capture6.2 LOGGED DATAReview the logged data by selecting its storage location within:Main Menu [OK] ✂ Memory – LogSwitch between graphic displays and MIN / MAX / AVG values of the 4 parameters with [UP] and [DOWN].7 ALARMSet up upper, lower limits as well as hysteresis for every displayed parameter in:Main Menu [OK] ✂ AlarmsBroken alarm limits are shown in red on the home screen.You can also enable an acoustic signal for broken alarms in:Main Menu [OK] ✂ Settings ✂Buzzerlicense key.To download the HW4 software and learn more about its capabilities, please visit our webpage: /hw4.10 Belt Buckle SetProbe HC2A-S (not included)HP32Lanyard。
新思科技光学解决方案客户技术支持使用指南说明书
新思科技光学解决方案客户技术支持使用指南（2022.12版）目录一 、简介1.服务对象2.平台使用注意事项3.关于软件版本4.客户支持服务内容二、 SolvNetPlus 技术资源整合平台1.SolvNetPlus 注册及登入方法2.下载软件安装程序、授权档案3.Learning Center 线上学习平台4.咨询技术问题三、常见问题一、 简介感谢您使用新思科技光学产品（软件和设备）。
为了提供更好的服务，我们为软件 （CODE V、LightTools、LucidShape 产品系列、RSoft 产品系列）在租赁或维护期内的用户，以及购买硬件测量设备（Mini-DiffV2、Mini-Diff VPro、REFLET、TIS Pro）的用户提供技术支持和服务。
我们使用 SolvNetPlus 技术资源整合平台提供软件与授权文件下载、线上课程学习（Learning Center）以及线上技术支持。
1.服务对象我们对以下用户提供技术支持和服务●软件维护期内用户●软件租赁期内用户●硬件测量设备用户2.平台使用注意事项我们在技术支持过程中会涉及光学专业术语，因此我们预设您对光学术语已经有所了解。
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维克斯45VPF系列-21设计-高压高性能梭形泵部件手册说明书
45VPF Series – 21 DesignHigh Pressure, High PerformanceVane PumpParts ManualAX439856912957en-000101High Pressure, High Performance Vane Pump45VPF Series – 21 DesignService DataViton is a registered trademark of E.I. DuPont Co.Parts prefixed with symbols areavailable only in kits (see back page).NOTE: Lubricate all parts and seals with a thin coat of oil at assembly.45VPF–***–***–*–*–01Model Code45VPF–***–***–*–*–0245VPF–***–***–*–*–0345VPF–***–***–*–*–0545VPF–***–***–*–*–06883356883357883358883359860370Key 928543928547Shaft Included in Cartridge Kit (see table)Included in Flange Kit (see table)928542Model CodeTypical sectional viewTo reverse Cartridge rotation remove Retaining Ring and Bearing Carrier S/A. Install Bearing Carrier S/A and Retaining Ring in opposite end ofCartridge and reinstall Cartridge facing in opposite direction. For R.H. rotation the end of the Cartridge with screw heads showing will be installed toward the cover end. For L.H. rotation the end of the Cartridge with screw heads showing will be installed toward the shaft end.02–142427 STD Seal Kit, Single Shaft Seal or02–142429 Viton Seal Kit, Single Shaft seal.02–142428 STD Seal Kit, Double Shaft Seal or02–142430 Viton Seal Kit, Double Shaft Seal.Shaft Seal OptionC –Primary and secondary seals:both with spring loaded sealing member facing inward.B –Primary and secondary seals:primary seal with spring loaded sealing member facing inward and secondary seal with spring loaded sealing member facing outward.A –Primary seal only: sealing member facing inward.Danfoss Power Solutions is a global manufacturer and supplier of high-quality hydraulic and electric components. We specialize in providing state-of-the-art technology and solutions marine sector. Building on our extensive applications expertise, we work closely with you to ensure exceptional performance for a broad range of applications. We help you and other customers around the world speed up system development, reduce costs and bring vehicles and vessels to market faster.Danfoss Power Solutions – your strongest partner in mobile hydraulics and mobile Go to for further product information.outstanding performance. And with an extensive network of Global Service Partners, we also provide you with comprehensive global service for all of our components.Local address:DanfossPower Solutions GmbH & Co. OHG Krokamp 35D-24539 Neumünster, Germany Phone: +49 4321 871 0DanfossPower Solutions ApS Nordborgvej 81DK-6430 Nordborg, Denmark Phone: +45 7488 2222DanfossPower Solutions (US) Company 2800 East 13th Street Ames, IA 50010, USA Phone: +1 515 239 6000DanfossPower Solutions Trading (Shanghai) Co., Ltd.Building #22, No. 1000 Jin Hai Rd Jin Qiao, Pudong New District Shanghai, China 201206Phone: +86 21 2080 6201Danfoss can accept no responsibility for possible errors in catalogues, brochures and other printed material. Danfoss reserves the right to alter its products without notice. This also applies to productsagreed.All trademarks in this material are property of the respective companies. Danfoss and the Danfoss logotype are trademarks of Danfoss A/S. All rights reserved.© Danfoss | September 2022•Cartridge valves •DCV directional control valves•Electric converters •Electric machines •Electric motors •Gear motors •Gear pumps •Hydraulic integrated circuits (HICs)•Hydrostatic motors •Hydrostatic pumps •Orbital motors •PLUS+1® controllers •PLUS+1® displays •PLUS+1® joysticks and pedals•PLUS+1® operator interfaces•PLUS+1® sensors •PLUS+1® software •PLUS+1® software services,support and training •Position controls and sensors•PVG proportional valves •Steering components and systems •TelematicsHydro-GearDaikin-Sauer-Danfoss。
hw2
【输出样例】
14
【限制】
1 ≤ N ≤ 10^5，最后解的范围为[1, 2^31)。
【提示】
树的直径
2012 年春季学期
第 3 / 10 页
清华大学计算机系(c)版权所有
a2,1 a3,1 ... an-2,1 an-1,1
a3,2 ... an-2,2 an-1,2
... ... ...
an-2,n-3 an-1,n-3
an-1,n-2
其中 n = |V|为顶点总数，顶点从 0 到 n-1 编号； ai,j 表示顶点 i 和 j 之间有无联边，有则为 1，没有为 0。
【输入样例】
6 2 2 5 1 1 2 2 2 2 3 1 2 100 200 150 100 300 200
【输出样例】
200.00
【限制】
1 ≤ N < 10,000,000 1 ≤ X < 1,000 股市的营业期不超过 2^31 - 1 天
【提示】
在队列基础上扩充操作接口
2012 年春季学期
【输入】
第一行是正整数 N，表示村庄的总数，各村庄从 1 到 N 编号。 接下来的 N-1 行各有以空格分隔的三个整数，表示由一条贸易通道联通的两个村庄（的编号） ， 以及 Donkey 在走这段路的时候会吃掉的草料的量。
【输出】
仅有一行，含一个整数，即 Shrek 所需背包的最小体积。
【输入样例】
【输入】
第一行含一个正整数 n，表示初始在墙上的 01 串总数。接下来的 n 行依次给出第 1~n 个 01 串。
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DC Arc Fault Has Been Found the Major Cause of Fire in PV System
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Section_13_Exercises_Solutionspdf
Section 1.3 Exercises (Solutions)1.109, 1.110, 1.111, 1.114*, 1.115, 1.119*, 1.122, 1.125, 1.127*, 1.128*, 1.131*, 1.133*, 1.135*, 1.137*, 1.139*, 1.145*, 1.146 - 148.1.109Sketch some normal curves.(a)Sketch a normal curve that has mean 10 and standard deviation 3.(b)On the same x axis, sketch a normal curve that has mean 20 and standard deviation 3.(c)How does the normal curve change when the mean is varied but the standard deviationstays the same.Solution(a) It is easiest to draw the curve first, and then mark the scale on the axis. (b) Draw acopy of the first curve, with the peak over 20. (c) The curve has the same shape, but is translated left or right.The curve has the same shape, but is translated left or right.1.110The effect of changing the standard deviation.(a)Sketch a normal curve that has mean 10 and standard deviation 3.(b)On the same x axis, sketch a normal curve that has mean 10 and standard deviation 1.(c)How does the normal curve change when the standard deviation is varied but the meanstays the same.Solution(a) As in the previous exercise, draw the curve first, and then mark the scale on the axis.(b) In order to have a standard deviation of 1, the curve should be 1/3 as wide, and threetimes taller. (c) The curve is centered at the same place (the mean), but its height andwidth change. Specifically, increasing the standard deviation makes the curve wider and shorter; decreasing the standard deviation makes the curve narrower and taller.1.111Know your density. Sketch density curves that might describe distributions with the following shapes:(a)Symmetric, but with two peaks (that is, two strong clusters of observations).(b)Single peak and skewed to the left.Solution1.111. Sketches will vary.1.114Total scores. Here are the total scores of 10 students in an introductory statistics course. Previous experience with this course suggests that these scores should come from a distribution that is approximately normal with mean 70 and standard deviation 10.(a)Using these values for µ and σ, standardize the first exam scores of these 10 students.(b)If the grading policy is to give grades of A to the top 15% of scores based on the normaldistribution with mean 70 and standard deviation 10, what is the cut-off for an A in terms of a standardized score?(c)Which students earned a grade of A on the final exam for this course?Solution1.114. (a) For example, (68−70)/10 = −0.2. The complete list is given below.(b) The cut-off for an A is the 85th percentile for the N(0, 1) distribution. From Table A,this is approximately 1.04; software gives 1.0364. (c) The top two students (with scores of 92 and 98) received A’s.68 54 92 75 73 98 64 55 80 70-0.2 -1.6 2.2 0.5 0.3 2.8 -0.6 -1.5 1 01.115Assign more grades. Refer to the previous exercise. The grading policy says the cut-offs for the other grades correspond to the following: bottom 5% receive F, next 10% receive D, next 40% receive C, and next 30% receive B. These cut-offs are based on the N(70, 10) distribution.(a)Give the cut-offs for the grades in this course in terms of standardized scores.(b)Give the cut-offs in terms of actual total scores.(c)Do you think that this method of assigning grades is a good one? Give reasons for youranswer.Solution1.115. (a) We need the 5th, 15th, 55th, and 85th percentiles for a N(0, 1) distribution.These are given in the table below. (b) To convert to actual scores, take the standard-score cut-off z and compute 10z + 70. (c) Opinions will vary.Note: The cut-off for an A given in the previous solution is the lowest score that gets an A—that is, the point where one’s grade drops from an A to a B. These cut-offs are thepoints where one’s grade jumps up. In practice, this is only an issue for a score that falls exactly on the border between two grades.1.116A uniform distribution. If you ask a computer to generate “random numbers” between 0 and 1, you will get observations from a uniform distribution. Figure 1.34 graphs the density curve for a uniform distribution. Use areas under this density curve to answer the following questions.(a)Why is the total area under this curve equal to 1?(b)What proportion of the observations lie below 0.35?(c)What proportion of the observations lie between 0.35 and 0.65?Figure 1.34 The density curve of a uniform distribution, for Exercise 1.116.1.118Find the mean, the median, and the quartiles. What are the mean and the median of the uniform distribution in Figure 1.34? What are the quartiles?Solutions (1.116 - 1.118)1.119Three density curves.Figure 1.35 displays three density curves, each with three points marked on it. At which of these points on each curve do the mean and the median fall?Figure 1.35 Three density curves, for Exercise 1.119.Solution1.119. (a) Mean is C, median is B (the right skew pulls the mean to the right). (b) MeanA, median A. (c) Mean A, median B (the left skew pulls the mean to the left).1.120Length of pregnancies. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Draw a density curve for this distribution on which the mean and standard deviation are correctly located.Solution1.122Pregnancies and the 68–95–99.7 rule. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Use the 68–95–99.7 rule to answer the following questions.(a)Between what values do the lengths of the middle 95% of all pregnancies fall?(b)How short are the shortest 2.5% of all pregnancies? How long do the longest 2.5% last?Solution1.122. See the sketch of the curve in the solution to Exercise 1.120. (a) The middle 95%fall within two standard deviations of the mean: 266 ± 2(16), or 234 to 298 days. (b) The shortest 2.5% of pregnancies are shorter than 234 days (more than two standarddeviations below the mean).1.123Horse pregnancies are longer. Bigger animals tend to carry their young longer before birth. The length of horse pregnancies from conception to birth varies according to a roughly Normal distribution with mean 336 days and standard deviation 3 days. Use the 68–95–99.7 rule to answer the following questions.(a)Almost all (99.7%) horse pregnancies fall in what range of lengths?(b)What percent of horse pregnancies are longer than 339 days?Solution1.123. (a) 99.7% of horse pregnancies fall within three standard deviations of the mean:336 ± 3(3), or 327 to 325 days. (b) About 16% are longer than 339 days since 339days or more corresponds to at least one standard deviation above the mean.1.125Acidity of rainwater. The Normal quantile plot in Figure 1.32 (page 66) shows that the acidity (pH) measurements for rainwater samples in Exercise 1.36 are approximately Normal. How well do these scores satisfy the 68–95–99.7 rule? To find out, calculate the mean andstandard deviation s of the observations. Then calculate the percent of the 105 measurements that fall between − s and + s and compare your result with 68%. Do the same for the intervalscovering two and three standard deviations on either side of the mean. (The 68–95–99.7 rule is exact for any theoretical Normal distribution. It will hold only approximately for actual data.) Solution1.125. The mean and standard deviation are x = 5.4256 and s = 0.5379. About 67.62%(71/105 = 0.6476) of the pH measurements are in the range x ± s = 4.89 to 5.96. About95.24% (100/105) are in the range x ± 2s = 4.35 to 6.50. All (100%) are in the rangex ± 3s = 3.81 to 7.04.1.126Find some proportions. Using either Table A or your calculator or software, find the proportion of observations from a standard Normal distribution that satisfies each of the following statements. In each case, sketch a standard Normal curve and shade the area under the curve that is the answer to the question.(a)Z > 1.65(b)Z < 1.65(c)Z > −0.76(d)−0.76 < Z < 1.65Solution1.126. Using values from Table A: (a) Z > 1.65: 0.0495. (b) Z < 1.65: 0.9505.(c) Z > −0.76: 0.7764. (d) −0.76 < Z <1.65: 0.9505 − 0.2236 = 0.7269.1.127Find more proportions. Using either Table A or your calculator or software, find the proportion of observations from a standard Normal distribution for each of the following events. In each case, sketch a standard Normal curve and shade the area representing the proportion.(a)Z ≤ −1.8(b)Z ≥ −1.8(c)Z > 1.6(d)−1.8 < Z < 1.6Solution1.127. Using values from Table A: (a) Z ≤ −1.8: 0.0359. (b) Z ≥ −1.8: 0.9641.(c) Z > 1.6: 0.0548. (d) −1.8 < Z < 1.6: 0.9452 − 0.0359 = 0.9093.1.128Find some values of z. Find the value z of a standard Normal variable Z that satisfies each of the following conditions. (If you use Table A, report the value of z that comes closest to satisfying the condition.) In each case, sketch a standard Normal curve with your value of z marked on the axis.(a)22% of the observations fall below z.(b)40% of the observations fall above z.Solution1.128. (a) 22% of the observations fall below −0.7722. (This is the 22nd percentile of thestandard Normal distribution.) (b) 40% of the observations fall above 0.2533 (the 60thpercentile of the standard Normal distribution).1.129Find more values of z. The variable Z has a standard Normal distribution.(a)Find the number z that has cumulative proportion 0.65.(b)Find the number z such that the event Z > z has proportion 0.45.Solution1.129. (a) z = 0.3853 has cumulative proportion 0.65 (that is, 0.3853 is the 65thpercentile of the standard Normal distribution). (b) If z = 0.1257, then Z > z hasproportion 0.45 (0.1257 is the 55th percentile).1.130Find some values of z. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test. The scale of scores is set separately for each age group and is approximately Normal with mean 100 and standard deviation 15. People with WAIS scores below 70 are considered mentally retarded when, for example, applying for Social Security disability benefits. What percent of adults are retarded by this criterion?Solution1.130. 70 is two standard deviations below the mean (that is, it has standard score z =−2), so about 2.5% (half of the outer 5%) of adults would have WAIS scores below 70.1.131High IQ scores. The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test. The scale of scores is set separately for each age group and is approximately Normal with mean 100 and standard deviation 15. The organization MENSA, which calls itself “the high IQ society,” requires a WAIS score of 130 or higher for membership. What percent of adults would qualify for membership?There are two major tests of readiness for college, the ACT and the SAT. ACT scores are reported on a scale from 1 to 36. The distribution of ACT scores are approximately Normal with mean µ = 21.5 and standard deviation σ = 5.4. SAT scores are reported on a scale from 600 to 2400. The SAT scores are approximately Normal with mean µ = 1509 and standard deviation σ= 321. Exercises 1.132 to 1.141 are based on this information.Solution1.131. 130 is two standard deviations above the mean (that is, it has standardscore z = 2), so about 2.5% of adults would score at least 130.1.132Compare an SAT score with an ACT score. Tonya scores 1820 on the SAT. Jermaine scores 29 on the ACT. Assuming that both tests measure the same thing, who has the higher score? Report the z-scores for both students.Solution1.132. Tonya’s score standardizes to z = (1820−1509)/321 = 0.9688, while Jermaine’sscore corresponds to z = (29−21.5)/5.4 = 1.3889. Jermaine’s score is higher.1.133Make another comparison. Jacob scores 16 on the ACT. Emily scores 1020 on the SAT. Assuming that both tests measure the same thing, who has the higher score? Report the z-scores for both students.Solution1.133. Jacob’s score standardizes to z = (16−21.5)/5.4 = −1.0185, while Emily’s scorecorresponds to z = (1020−1509)/321 = −1.5234. Jacob’s score is higher.1.134Find the ACT equivalent. Jose scores 2080 on the SAT. Assuming that both tests measure the same thing, what score on the ACT is equivalent to Jose’s SAT score?Solution1.134. Jose’s score standardizes to z = (2080−1509)/321 = 1.7788, so an equivalentACT score is 21.5 + 1.7788 × 5.4 = 31.1. (Of course, ACT scores are reported as whole numbers, so this would presumably be a score of 31.)1.135Find the SAT equivalent. Maria scores 30 on the ACT. Assuming that both tests measure the same thing, what score on the SAT is equivalent to Maria’s ACT score?Solution1.135. Maria’s score standardizes to z = (30−21.5)/5.4 =1.5741, so an equivalent SATscore is 1509 + 1.5741 × 321 = 2014.1.136Find an SAT percentile. Reports on a student’s ACT or SAT usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than this one. Maria scores 2090 on the SAT. What is her percentile?Solution1.136. Maria’s score standardizes to z = (2090−1509)/321 = 1.81, for which Table Agives 0.9649. Her score is the 96.5 percentile.1.137Find an ACT percentile. Reports on a student’s ACT or SAT usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as a percent: the percent of all scores that were lower than this one. Jacob scores 19 on the ACT. What is his percentile?Solution1.137. Jacob’s score standardizes to z = (19−21.5)/5.4 = −0.4630, for which Table Agives 0.3228. His score is the 32.3 percentile.1.138How high is the top 10%? What SAT scores make up the top 10% of all scores?Solution1.138. 1920 and above: The top 10% corresponds to a standard score of z = 1.2816,which in turn corresponds to a score of 1509 + 1.2816 × 321 = 1920 on the SAT..1.139How low is the bottom 20%? What SAT scores make up the bottom 20% of all scores?Solution1.139. 1239 and below: The bottom 20% corresponds to a standard score ofz = −0.8416, which in turn corresponds to a score of 1509 − 0.8416 × 321 =1239 on the SAT.1.140Find the ACT quartiles. The quartiles of any distribution are the values with cumulative proportions 0.25 and 0.75. What are the quartiles of the distribution of ACT scores?Solution1.140. The quartiles of a Normal distribution are ±0.6745 standard deviations from themean, so for ACT scores, they are 21.5 ± 0.6745 × 5.4 = 17.9 to 25.1.1.141Find the SAT quintiles. The quintiles of any distribution are the values with cumulative proportions 0.20, 0.40, 0.60, and 0.80. What are the quintiles of the distribution of SAT scores?Solution1.141. The quintiles of the SAT score distribution are 1509 − 0.8416 × 321 = 1239,1509−0.2533×321 = 1428, 1509+0.2533×321 = 1590, and 1509+0.8416×321 = 1779. 1.142Do you have enough “good cholesterol?” High-density lipoprotein (HDL) is sometimes called the “good cholesterol” because low values are associated with a higher risk of heart disease. According to the American Heart Association, people over the age of 20 years should have at least 40 mg/dL of HDL cholesterol.36 U.S. women aged 20 and over have a mean HDL of 55 mg/dL with a standard deviation of 15.5 mg/dL. Assume that the distribution is Normal.(a)What percent of women have low values of HDL (40 mg/dL or less)?(b)HDL levels of 60 mg/dL are believed to protect people from heart disease. What percentof women have protective levels of HDL?(c)Women with more than 40 mg/dL but less than 60 mg/dL of HDL are in the intermediaterange, neither very good or very bad. What proportion are in this category?Solution1.142. For a Normal distribution with mean 55 mg/dl and standard deviation 15.5 mg/dl:(a) 40 mg/dl standardizes to z = (40−55)/15.5 = −0.9677. Using Table A, 16.60% ofwomen fall below this level (software: 16.66%).(b) 60 mg/dl standardizes to z = (60−55)/15.5 = 0.3226. Using Table A, 37.45(c) Subtract the answers from (a) and (b) from 100%: Table A gives 45.95% (software:45.99%), so about 46% of women fall in the intermediate range.1.143Men and HDL cholesterol. HDL cholesterol levels for men have a mean of 46 mg/dL with a standard deviation of 13.6. Answer the questions given in the previous exercise for the population of men.Solution1.143. For a Normal distribution with mean 46 mg/dl and standard deviation 13.6 mg/dl:(a) 40 mg/dl standardizes to z = (40−46)/13.6 = −0.4412. Using Table A, 33% of men fallbelow this level (software: 32.95%).(b) 60 mg/dl standardizes to z = (60−46)/13.6 = 1.0294. Using Table A, 15.15(c) Subtract the answers from (a) and (b) from 100%: Table A gives 51.85% (software:51.88%), so about 52% of men fall in the intermediate range.1.145Length of pregnancies. The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days.(a)What percent of pregnancies last less than 240 days (that’s about 8 months)?(b)What percent of pregnancies last between 240 and 270 days (roughly between 8 monthsand 9 months)?(c)How long do the longest 20% of pregnancies last?Solution1.145. (a) About 5.2%: x < 240 corresponds to z < −1.625. Table A gives 5.16% for−1.63 and 5.26% for −1.62. Software (or averaging the two table values) gives 5.21%.(b) About 54.7%: 240 < x < 270 corresponds to −1.625 < z < 0.25. The area to theleft of 0.25 is 0.5987; subtracting the answer from part (a) leaves about 54.7%.(c) About 279 days or longer: Searching Table A for 0.80 leads to z > 0.84, whichcorresponds to x > 266 + 0.84(16) = 279.44. (Using the software value z > 0.8416 gives x > 279.47.)1.146Quartiles for Normal distributions. The quartiles of any distribution are the values with cumulative proportions 0.25 and 0.75.(a)What are the quartiles of the standard Normal distribution?(b)Using your numerical values from (a), write an equation that gives the quartiles of theN(µ, σ) distribution in terms of µ and σ.(c)The length of human pregnancies from conception to birth varies according to adistribution that is approximately Normal with mean 266 days and standard deviation 16 days. Apply your result from (b): what are the quartiles of the distribution of lengths of human pregnancies?Solution1.146. (a) The quartiles for a standard Normal distribution are ±0.6745.(b) For a N(µ, σ) distribution, Q1 = µ − 0.6745σ and Q3 = µ + 0.6745σ.(c) For human pregnancies,Q1 = 266 − 0.6745 × 16 .=255.2 and Q3 = 266 + 0.67455 ×16 =276.8 days.1.147IQR for Normal distributions. Continue your work from the previous exercise. The interquartile range IQR is the distance between the first and third quartiles of a distribution.(a)What is the value of the IQR for the standard Normal distribution?(b)There is a constant c such that IQR = cσ for any Normal distribution N(µ, σ). What is thevalue of c?Solution1.147. (a) As the quartiles for a standard Normal distribution are ±0.6745,we have IQR = 1.3490.(b) c = 1.3490: For a N(µ, σ) distribution, the quartiles are Q1 = µ − 0.6745σ andQ3 = µ + 0.6745σ.1.148Outliers for Normal distributions. Continue your work from the previous two exercises. The percent of the observations that are suspected outliers according to the 1.5 × IQR rule is the same for any Normal distribution. What is this percent?Solution1.148. In the previous two exercises, we found that for a N(µ, σ) distribution,Q1 = µ − 0.6745σ, Q3 = µ + 0.6745σ, and IQR = 1.3490σ.Therefore, 1.5 × IQR = 2.0235σ, and the suspected outliers are belowQ1 − 1.5 × IQR = µ − 2.698σ, and above Q3 + 1.5 × IQR = µ + 2.698σ.The percentage outside of this range is 2 × 0.0035 = 0.70%.1.149Deciles of Normal distributions. The deciles of any distribution are the 10th, 20th,…, 90th percentiles. The first and last deciles are the 10th and 90th percentiles, respectively.(a)What are the first and last deciles of the standard Normal distribution?(b)The weights of 9-ounce potato chip bags are approximately Normal with mean 9.12ounces and standard deviation 0.15 ounce. What are the first and last deciles of thisdistribution?Solution1.149. (a) The first and last deciles for a standard Normal distribution are ±1.2816.(b) For a N(9.12, 0.15) distribution, the first and last deciles are µ − 1.2816σ= 8.93 andµ + 1.2816σ = 9.31 ounces.。
抽象代数Chapter2习题答案
HW 2 Solutions, (October 5th, 2011)
All problems from Chapter 2 of Artin’s Algebra. 1.3 If map r : N → N was a right inverse for the shift map s, then the composition sr would send 1 to 1. However, the number 1 is not in the image of s so such a right inverse is impossible. For any n ∈ N, define the map ℓn by ℓn (i) = i − 1 if i ≥ 2 and ℓn (1) = n. Then the composition ℓn s is the identity on N. Thus we have exhibited an infinite number of left inverses. 2.1 Group multiplication table for S3 = {e, (12), (13), (23), (123), (132)} with first row (resp. column) corree (12) (23) (13) (123) (132) (12) e (123) (132) (23) (13) (23) (132) e (123) (13) (12) sponding to left (resp. right-) multiplication by identity e: . (13) (123) (132) e (12) (23) (123) (13) (12) (23) (132) e (132) (23) (13) (12) e (123) 2.2 Let S ′ be the subset of S consisting of invertible elements. We must show that the associative law of composition, ◦, on S restricts to a law of composition on S ′ . In other words, we need to show closure: if s1 and s2 are invertible (i.e. in S ′ ), then s1 ◦ s2 is also invertible (i.e. in S ′ ). But this is clearly true since 1 −1 ′ s− 2 ◦ s1 is the inverse of s1 ◦ s2 . This law of composition on S is associative since it is associative on S . To complete the proof that subset S ′ is a group, we need to check that identity and inverses are in S ′ , and these follow quickly. 2.4 a) Yes, GLn (R) is a subgroup of GLn (C), clearly the product of two invertible matrices with real entries is an invertible matrix with real entries (implies closure ). The identity matrix has real entries, and the inverse of a matrix with real entries also has real entries. b) Yes, {−1, 1} is a subgroup of R× . (Similar technique as in part (a).) c) No, the inverse of a positive integer (under addition) is not a positive integer. d) Yes, {positive reals} is a subgroup of R× . (Similar technique as in part (a).) e) No, matrix a 0 is not invertible. 0 0 1
Twain翻译第一章二章
需要的一致性通过引入扫描仪、数码相机、和其他图像采集设备,userseagerly发现的值将图像纳入他们的文件和其他工作。
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他们需要创建用户界面和构建在设备控制设备为各种各样的形象。
一旦他们的应用程序准备支持给定的设备,他们面对令人沮丧的现实,设备继续beupgraded新的功能和特性。
应用程序开发人员发现themselvescontinually修改他们的产品保持电流。
开发人员的图像采集设备和软件应用程序承认需要一个标准的图像设备和应用程序之间的通信。
一个标准将有利于两组的用户以及他们的产品。
将允许设备供应商的产品被更多的应用程序和应用程序供应商couldaccess访问数据从没有关心哪种类型的设备,这些设备或特定设备,它提供.TWAIN开发,因为这需要一致性和简化。
吐温的元素吐温定义了一个标准的软件协议和API(应用程序编程接口)forcommunication软件应用程序和图像采集设备之间数据的来源。
吐温的三要素是:•应用软件,应用程序必须使用吐温被修改。
•源代码管理软件,该软件管理应用和源之间的交互。
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•源软件,这个软件控制图像采集设备,writtenby设备开发人员遵守吐温规范。
传统devicedrivers现在包含在源代码软件,不需要byapplications运来。
使用吐温的好处Benefits of Using TWAIN为应用程序开发人员•允许您应用程序的用户提供一个简单的方法将图像从任何兼容的光栅设备无需离开您的应用程序。
•节省时间和金钱。
如果你目前为扫描仪提供底层设备驱动程序,等等,你不再需要编写、支持或船这些驱动程序。
TWAIN-compliant形象采集设备将提供源码软件模块,消除对你的需要创建和船舶设备驱动程序。
奇石乐发动机燃烧分析仪说明书
压阻式压力传感器
压阻原理以1954年首次发现的半导体 效应为基础，即在机械应力作用下， 半导体的阻值会发生变化。与传统的 应变仪测量相比，这一发现开拓了全 新的应用领域。从那时起，其他相近 的技术开始出现，如金属薄膜技术和 陶瓷厚膜技术。
开始： – 奇石乐 – 传感器型号 – 版本 – 传感器版本号 – 传感器序号
初始校准： – 工作温度范围 – 初始校准日期 – 初始校准值
再校准： – 参考温度 – 再校准日期 – 再校准值
结尾： – 注释
传感器
密封
连接电缆
传感器、连接电缆和 TEDS 元件
PiezoSmart®适配器
压阻式低压
压电式缸压
PiezoStar®-10 多年来奇石乐在厂内培养高灵 敏和高温稳定的晶体。
2
目录
从晶体到传感器需要 200 道工序
4
奇石乐压力传感器 - 设计的多样化和创新
6
PiezoSmart® 意味着传感器的自动识别
8
技术先驱 - 引导发动机测试技术发展
压电式缸压
PiezoSmart® 传感器
优点概述
● 电荷和电压放大器参数的自动设 置
● 精确压力指示数据测量简便 ● 简化设置过程缩短试验台的测量
准备时间 ● 通过分析系统数据自动交换，简
化测量分析和记录 ● 传感器自动识别技术极大地提高
多通道校准系统的自动化程度
更多关于升级的信息， 请参考第 18 页
诺瓦科技多画面拼接处理器VS2用户手册
在主界面下，按下旋钮，进入主菜单操作界面，按旋钮进入子菜单，按 ESC 返回上一层。
西安诺瓦电子科技有限公司
屏体设置
设置两个输出的拼接方式以及各自的大小。 -6-
多画面拼接处理器 VS2 用户手册
屏体结构
屏体 1 配置 屏体 2 配置
即：仅一路输出、水平拼接以及垂直拼接。 屏体 1 水平宽度、屏体 1 垂直高度 屏体 2 水平宽度、屏体 2 垂直高度
司 源插头。 公 远离易燃易爆危险物品。 有限 禁止液体、金属碎片浸入机器内部，以免引起安全事故。 西安诺瓦电子科技 插座应当装在设备的附近，而且应当便于触及到。
多画面拼接处理器 VS2 用户手册
目录
第一章
功能简介 ......................................................................................... 1 前面板示意图 .............................................................................................. 1 后面板示意图 .............................................................................................. 2 电气参数 ..................................................................................................... 3
输入源
司 输入-A 限公 输入-B 技有 输入-C 科 输入-D 子 输出接口 电 预监 DVI 瓦 DVI1 诺 DVI2 安控制接口 西ETHERNET
HW2Solution
Homework Assignment 2InstructionsThis assignment is due to be turned in at the start of class on Wednesday February 11. You can work on this assignment by yourself or in a team of two. If you choose to work in a team of two, you should turn in only one assignment per team (with both member’s student IDs on the assignment).The assignment consists of two parts. Part A contains discussion questions. These are NOT to be turned in, they are intended to solidify your understanding of the theoretical concepts that we have discussed. I suggest writing down your answers and comparing them with the solutions that I will upload once the assignments have been turned in. Part B contains assignment problems that must be turned in. Remember that these assignments are graded on a √- ,√, √+ system as detailed in the syllabus. You are free to turn in printouts or a clear, legible, handwritten assignment.Part A: Discussion questions (not to be turned in).1.Explain carefully why liquidity preference theory is consistent with the observation thatthe term structure of interest rates tends to be upward sloping more often than it is downward sloping.If long-term rates were simply a reflection of expected future short-term rates, we would expect the term structure to be downward sloping as often as it is upward sloping. (This is based on the assumption that half of the time investors expect rates to increase and half of the time investors expect rates to decrease). Liquidity preference theory argues that long term rates are high relative to expected future short-term rates. This means that the term structure should be upward sloping more often than it is downward sloping.2.What is meant by (a) an investment asset and (b) a consumption asset? Why is thedistinction between investment and consumption assets important in the determination of forward and futures prices?An investment asset is an asset held for investment by a significant number of people or companies. A consumption asset is an asset that is nearly always held to be consumed(either directly or in some sort of manufacturing process). The forward/futures price can be determined from the spot price for an investment asset. In the case of a consumptionasset all that can be determined is an upper bound for the forward/futures price.3.What is the cost of carry for (a) a non-dividend-paying stock, (b) a stock index, (c) acommodity with storage costs, and (d) a foreign currency?a)the risk-free rate, b) the excess of the risk-free rate over the dividend yield c) therisk-free rate plus the storage cost, d) the excess of the domestic risk-free rate over theforeign risk-freerate.4. Do we expect the futures price of a stock index to be greater than or less than the expected future value of the index? Explain your answer.The futures price of a stock index is always less than the expected future value of the index. This follows from Section 5.14 and the fact that the index has positive systematic risk. For an alternative argument, let μ be the expected return required by investors on the index so that ()0()q T T E S S e -=μ. Because r >μ and ()00r q T F S e -=, it follows that 0()T E S F >.Part B: Homework problems (to be turned in)1. The cash prices of six-month and one-year Treasury bills are 94.0 and 89.0. A 1.5-year bond that will pay coupons of $4 every six months currently sells for $94.84. A two-year bond that will pay coupons of $5 every six months currently sells for $97.12. Calculate the six-month, one-year, 1.5-year, and two-year zero rates.The 6-month Treasury bill provides a return of 6946383%/=. in six months. This is 2638312766%⨯.=. per annum with semiannual compounding or2ln(106383)1238%.=. per annum with continuous compounding. The 12-month rate is 118912360%/=. with annual compounding or ln(11236)1165%.=. with continuous compounding. For the 112year bond we must have 012380501165115441049484R e e e -.⨯.-.⨯-.++=.where R is the 112year zero rate. It follows that 15153763561049484084150115R R e e R -.-..+.+=.=.=.or 11.5%. For the 2-year bond we must have012380501165101151525551059712R e e e e -.⨯.-.⨯-.⨯.-+++=. where R is the 2-year zero rate. It follows that 2079770113R e R -=.=.or 11.3%.2. Problem 4.9. What rate of interest with continuous compounding is equivalent to 15% per annum with monthly compounding?The rate of interest is R where: 12015112R e .⎛⎫=+ ⎪⎝⎭i.e., 01512ln 112R .⎛⎫=+ ⎪⎝⎭01491=.The rate of interest is therefore 14.91% per annum.3. Problem4.12. A three-year bond provides a coupon of 8% semiannually and has a cash price of 104. What is the bond’s yield?Ans: The bond pays $4 in 6, 12, 18, 24, and 30 months, and $104 in 36 months. The bondyield is the value of y that solves05101520253044444104104y y y y y y e e e e e e -.-.-.-.-.-.+++++= Using the Solver or Goal Seek tool in Excel 006407y =. or 6.407%.4. Problem 4.14. Suppose that zero interest rates with continuous compounding are as follows:Calculate forward interest rates for the second, third, fourth, and fifth years.For year4 2,$1*exp(2+f)=exp(2*3)f=6-2 = 4%Year 2: 4.0%Similarly, for year 3, 2*3+f = 3.7*3. Therefore f=5.1%Year 3: 5.1%Year 4: 5.7%Year 5: 5.7%5. A stock is expected to pay a dividend of $1 per share in two months and in five months. The stock price is $50, and the risk-free rate of interest is 8% per annum with continuous compounding for all maturities. An investor has just taken a short position in a six-month forward contract on the stock.a. What are the forward price and the initial value of the forward contract?b. Three months later, the price of the stock is $48 and the risk-free rate of interest is still 8% per annum. What are the forward price and the value of the short position in the forward contract?a) The present value, I , of the income from the security is given by:0082120085121119540I e e -.⨯/-.⨯/=⨯+⨯=. From equation (5.2) the forward price, 0F , is given by:008050(5019540)5001F e .⨯.=-.=.or $50.01. The initial value of the forward contract is (by design) zero. The fact that the forward price is very close to the spot price should come as no surprise. When the compounding frequency is ignored the dividend yield on the stock equals the risk-free rate of interest.b) In three months:00821209868I e -.⨯/==. The delivery price, K , is 50.01. From equation (5.6) the value of the short forward contract, f , is given by008312(48098685001)201f e -.⨯/=--.-.=.and the forward price is008312(4809868)4796e .⨯/-.=.6. The current price of silver is $30 per ounce. The storage costs are $0.48 per ounce per year payable quarterly in advance. Assuming that interest rates are 10% per annum for all maturities, calculate the futures price of silver for delivery in nine months.The present value of the storage costs for nine months are0.12 + 0.12e −0.10×0.25 + 0.12e −0.10×0.5 = 0.351or $0.351. The futures price is from equation (5.11) given by 0F whereF 0 = (30 + 0.351)e 0.1×0.75= 32.72i.e., it is $32.72 per ounce.7. In early 2012, the spot exchange rate between the Swiss Franc and U.S. dollar was 1.0404 ($ per franc). Interest rates in the U.S. and Switzerland were 0.25% and 0% per annumrespectively, with continuous compounding. The three-month forward exchange rate was1.0300 ($ per franc). What arbitrage strategy was possible? How does youranswer change if the exchange rate is 1.0500 ($ per franc).The true forward price should be:F =Se(r-rf)T= 1.0404 * exp((0.25% -0)*3/12)= 1.04105With the three month forward at 1.03 $ per franc , rather than the 1.04105 $ per franc it should be, I want to convert $s to Francs 3 months forward. Therefore, I borrow 1 franc, convert it spot to get 1.0404 dollars and invest them at the risk free rate. I also buy a 3 month swiss francs forward. My US investment yields:1.0404*exp(0.25%*0.25)=1.04105. I convert that using the forward contract, to yield1.04105/1.03 = 1.01073 francs. I thus make a risk free profit of 0.01073 per francinvested..With three month forward rate as 1.05, my strategy is: Borrow 1 dollar, convert it spot to francs, (I get 1/1.0400 francs) in Switzerland, sell francs forward. Use the forward to convert it back to dollars. At the end of three months, I need to pay out exp(0.25%*0.25) = 1.00063 from my US borrowing. I have 1.0500*1/1.0404 =1.00927. I thereby make a profit of 1.00927-1.00063 =0.00864 for every dollar invested.8.The current spot price of oil is $67.25 per barrel, the 6 month futures price is$69.75/barrel, and the risk-free rate is 4.2% (annualized and continuously compounded).What is the implied storage cost of oil expressed as an annualized continuously compounded rate? Whose storage cost do you think this reflects (Hint: Could you store oil at this rate?)F0 = S0e(r+s)T69.75=67.25e(0.042+s)0.5s = 2* (Ln(69.75/67.25)-0.021)= 3.1%This is the storage cost for the firm with the lowest marginal storage cost in the market. If for example this was not true, and there exists someone with lower marginal storage costs than 3.1% per year,they would execute the strategy of buying spot, and storing and be able to make a risk free arbitrage profit. They would do so until either (1) their marginal storage cost rose, or (2) the forward price fell to reflect their cost.。
Mellanox WinOF-2 产品知识库说明书
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All Rights Reserved.Mellanox®, Mellanox logo, Accelio®, BridgeX®, CloudX logo, CompustorX®, Connect -IB®, ConnectX®, CoolBox®, CORE-Direct®, EZchip®, EZchip logo, EZappliance®, EZdesign®, EZdriver®, EZsystem®, GPUDirect®, InfiniHost®, InfiniBridge®, InfiniScale®, Kotura®, Kotura logo, Mellanox CloudRack®, Mellanox CloudXMellanox®, Mellanox Federal Systems®, Mellanox HostDirect®, Mellanox Multi-Host®, Mellanox Open Ethernet®, Mellanox OpenCloud®, Mellanox OpenCloud Logo®, Mellanox PeerDirect®, Mellanox ScalableHPC®, Mellanox StorageX®, Mellanox TuneX®, Mellanox Connect Accelerate Outperform logo , Mellanox Virtual Modular Switch®, MetroDX®, MetroX®, MLNX-OS®, NP-1c®, NP-2®, NP-3®, NPS®, Open Ethernet logo , PhyX®, PlatformX®, PSIPHY®, SiPhy®, StoreX®, SwitchX®, Tilera®, Tilera logo, TestX®, TuneX®, The Generation of Open Ethernet logo , UFM®, Unbreakable Link®, Virtual Protocol Interconnect®, Voltaire® and Voltaire logo are registered trademarks of Mellanox Technologies , Ltd.All other trademarks are property of their respective owners .For the most updated list of Mellanox trademarks, visit /page/trademarksNOTE:THIS HARDWARE, SOFTWARE OR TEST SUITE PRODUCT (“PRODUCT(S)”) AND ITS RELATED DOCUMENTATION ARE PROVIDED BY MELLANOX TECHNOLOGIES “AS -IS” WITH ALL FAULTS OF ANY KIND AND SOLELY FOR THE PURPOSE OF AIDING THE CUSTOMER IN TESTING APPLICATIONS THAT USE THE PRODUCTS IN DESIGNATED SOLUTIONS. THE CUSTOMER'S MANUFACTURING TEST ENVIRONMENT HAS NOT MET THE STANDARDS SET BY MELLANOX TECHNOLOGIES TO FULLY QUALIFY THE PRODUCT(S) AND/OR THE SYSTEM USING IT . THEREFORE, MELLANOX TECHNOLOGIES CANNOT AND DOES NOT GUARANTEE OR WARRANT THAT THE PRODUCTS WILL OPERATE WITH THE HIGHEST QUALITY. ANY EXPRESS OR IMPLIED WARRANTIES , INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY , FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT ARE DISCLAIMED. IN NO EVENT SHALL MELLANOX BE LIABLE TO CUSTOMER OR ANY THIRD PARTIES FOR ANY DIRECT, INDIRECT, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES OF ANY KIND (INCLUDING, BUT NOT LIMITED TO, PAYMENT FOR PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY FROM THE USE OF THE PRODUCT(S) AND RELATED DOCUMENTATION EVEN IF ADVISED OF THE POSSIBILITY OF SUCHDAMAGE.Doc #: N/A 2Mellanox TechnologiesRev 1.03Mellanox Technologies Table of ContentsTable of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3List Of Tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4Chapter 1 Archived Known Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5Rev 1.04Mellanox Technologies List Of TablesTable 1:Archived Known Issues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .5Archived Known Issues Rev 1.05Mellanox Technologies 1Archived Known IssuesThe following table describes archived known issues in the former releases on WinOF-2.Table 1 - Archived Known Issues (Sheet 1 of 5)Internal Ref.Issue1121688Description: Mellanox WinOF-2 Device Diagnostics and Mellanox WinOF-2 PCI Device Diag-nostics counters in Perfmon are available only on the adapter that was initialized first, which is thecounters’ owner.In case both adapters are viewed in Perfmon:1. The owner will receive correct values, while the second adapter will only receive zeros.2. If the owner goes down, the ownership will move to the second adapter.Workaround : N/AKeywords: WinOF-2 Device Diagnostics, Mellanox WinOF-2 PCI Device Diagnostics, Perfmon,ownershipDetected in version: 1.70.0.1961251Description: On a dual port device, when the first port is IPoIB and the second port is Ethernet, thedevice will not start over iSCSI and Windows 2012 in the Ethernet port.Workaround : N/AKeywords: IPoIB port, Ethernet port, ISCSI, Windows 2012Detected in version: 1.701000390Description: Configuring a different RoCE mode between VF and PF is not supportedWorkaround : N/AKeywords: RoCE, VF, PFDetected in version: 1.70917856Description: All folders under %PROGRAMFILES%Mellanox\MLNX_WINOF2 are deleted incase of upgrade failure. The old driver remains, and no connectivity loss should occur. The issueoccurs only when the upgrade fails after the old version was removed.Workaround : Following the upgrade failure, reinstall the new package folder.Keywords: Installation, setup, missing foldersDetected in version: 1.70915981Description: On IPoIB adapters, the network adapter and task manager network counters count allport traffic when only non-RDMA traffic should be shown.Workaround : N/AKeywords: IPoIB, counters, RDMADetected in version: 1.60939227Description: When upgrading from WinOF-2 1.50 to 1.60 or later, the MAC address used for theIPoIB interface changes.Workaround : N/AKeywords: IPoIB, MAC addressDetected in version: 1.60Rev 1.06Mellanox Technologies 928999Description: When installing a new driver, driver version queries via WMI may not be up-to-date,due to information caching in the WMI service.Workaround : Stop and restart the WMI service, running the following CMD commands:net stop winmgmtnet start winmgmtKeywords: Driver version queries, WMI, information cachingDetected in version: 1.60964973Description: SRQ limit event is not supported in the ND and NDK RDMA programming inter-faces.Workaround : N/AKeywords: SRQ, ND, NDK RDMADetected in version: 1.60900928Description: The packet sniffer is currently not supported in InfiniBand mode.Workaround : N/AKeywords: InfiniBand, packet snifferDetected in version: 1.50-Description: The installation process or restart of the driver does not close any RoCE userspace applications running in the background, and may cause a bug check as a result of a stuckcmdWorkaround : It is recommended to close all running RoCE user space applications prior toupgrading the driver.Keywords: Installation/Upgrade-Description: Installation/upgrade fails due to PNP failure to copy the driver files to the driverstore, and the following text is printed in the event logs: Fault bucket, type 0Event Name: PnPDriverImportError Response: Not availableAttached Files: C:\Users\<user>\AppData\Local\Temp\DMI151A.tmp.log.xmlC:\Program Files\Mellanox\ MLNX_WinOF2\Drivers\”Current OS”\mlx5.infWorkaround : Reboot the machine and reinstall.Keywords: Installation/Upgrade-Description: Installing both WinOF for ConnectX-3 and ConnectX-3 Pro, and WinOF-2 forConnectX-4 is supported only from WinOF version 5.00 and above.Workaround : N/AKeywords: Installation/UpgradeTable 1 - Archived Known Issues (Sheet 2 of 5)Internal Ref.IssueArchived Known Issues Rev 1.07Mellanox Technologies 654674Description: When trying to uninstall the mlx5 driver manually (by using pnputil/DPINST orDIFX API), additional hardware scan will be required before viewing the device in the DeviceManager or before reinstalling.This used to happen due to a bug in the NetCfgx.dll.(Microsoft case ID is: 115020112345121).Workaround : Rescan the hardware after performing the uninstallation.Keywords: Installation/Upgrade650489Description: While installing the driver on Windows Server 2012, and if SR-IOV mode isdisabled in the BIOS and enabled in the firmware, the server might reboot, and the BIOS willhang while loading.Workaround : To work in SR-IOV mode, enable SR-IOV in BIOS. Otherwise, disable SR-IOV in the firmware using mlxconfig.For further information on how to enable/disbale SR-IOV, please refer to the “Single Root I/OVirtualization (SR-IOV)” section in the User Manual.Keywords: Installation/Upgrade-Description: On machines configured with NVGRE encapsulation with the encapsulation taskoffload enabled, incoming VXLAN traffic on the interface may be reported with wrong check-sum status.Workaround : N/AKeywords: Virtualization-Description: Running Ntttcp without the “-a X” flag (X >1) in a NIC configured with10GbE may cause low bandwidth in TCP single streamWorkaround : Run Ntttcp with “-a 8” for best performanceKeywords: Performance-Description: RDMA read on single QP 100GbE RoCE cannot achieve more than 50Gb/s.Workaround :•Use more than one QP•Use Jumbo packets (4K)Keywords: Performance576556Description: “TCP RSC Average Packet Size” counter under network adapter does not countcorrectly. This is a known operating system issue.Workaround : N/AKeywords: GeneralWorkaroundTable 1 - Archived Known Issues (Sheet 3 of 5)Internal Ref.IssueRev 1.08Mellanox Technologies 683840Description: In Windows Server 2016, the following RDMA counters for VPorts that werecreated to use RDMA capability will show statistics for all the VPorts connected to the sameinterface and not for a specific VPort:•RDMA Inbound Bytes/sec•RDMA Inbound Frames/sec•RDMA Outbound Bytes/sec•RDMA Outbound Frames/secWorkaround : N/AKeywords: RDMA786035Description: Running applications on top of MS MPI may result in failure.Workaround : N/AKeywords: RDMA877750Description: Occasionally, the adapter card shows an error of duplicate IPv4 address whendisabling and enabling the adapter through Device Manager.Workaround : Reboot the machine instead of disabling and enabling it in the Device Manager.Keywords: Device Manager, IPv4825154Description: Mellanox WinOF-2 Device Diagnostic and PCI Device Diagnostic countersreported in PerfMon are per device. The counters that are reported per adapter under these setsshow the counters for all the devices and not only for the specific adapter.Workaround : N/AKeywords: perfmon, counters894614Description: The nd_*_bw and nd_*_lat tools do not work on Windows server 2012, Win-dows 8.1 and Windows Server 2012 R2. An error message appears, notifying that the api-ms-win-crt.dll is missing.Workaround : Windows update kb2999226 must be downloaded and installed, in order toobtain the universal c run time dlls. To download the update, go to https://support.micro-/en-us/kb/2999226.To verify that the update is installed, run the following powershell command:get-hotfix -id kb2999226Keywords: nd tools, Windows 2012, Windows 2012 R, missing dll, Universal C Runtime899853Description: Uninstallation of the driver does not reset all network adapter configurations tothe default values.Workaround : Upon completion of the uninstallation process, run the following powershellcommand for each network adapter, while replacing <AdapterName> with the name of the rel-evant network adapter:Reset-NetAdapterAdvancedProperty -Name "<AdapterName>" -DisplayName"*"Keywords: Uninstallation, network adapter configurations, Windows Server 2016, Windows10Table 1 - Archived Known Issues (Sheet 4 of 5)Internal Ref.IssueArchived Known Issues Rev 1.09Mellanox Technologies 778631Description: IB utils are currently not a part of the WinOF-2 packageWorkaround : Fabric diagnostic can be done from a managed switch or a different node in the fabric with IB utils support.Keywords: InfiniBand, IB utils, fabric diagnosticTable 1 - Archived Known Issues (Sheet 5 of 5)Internal Ref.Issue。
FusionModule2000智能微模块数据中心安装指南说明书
FusionModule2000 智能微模块数据中心V200R002安装指南(2200mm高机柜,单排密封通道)文档版本04发布日期2017-06-15版权所有© 华为技术有限公司2017。
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华为技术有限公司地址：深圳市龙岗区坂田华为总部办公楼邮编：518129网址：前言概述本文档主要介绍FusionModule2000智能微模块数据中心的安装方法。
读者对象本文档主要适用于以下工程师：●硬件安装工程师●调测工程师●维护工程师●技术支持工程师符号约定在本文中可能出现下列标志，它们所代表的含义如下。
修订记录文档版本04（2017-06-15）第一次正式发布。
目录前言 (ii)1 安全注意事项 (1)1.1 通用安全注意事项 (1)1.2 电气安全 (2)1.3 电池 (3)1.4 机械安全 (5)1.5 其它 (6)2 安装准备 (8)2.1 搬运注意事项 (8)2.2 开箱验货检查 (9)2.3 安装环境检查 (10)2.4 场地要求 (11)2.5 文档准备 (12)2.6 工具仪表准备 (13)2.7 施工人员需具备的技能和条件 (17)3 安装场景 (18)4 硬件安装 (20)4.1 划线定位机柜或底座位置 (20)4.2 安装底座 (22)4.2.1 安装可调尺寸底座 (22)4.2.1.1 支架规格 (22)4.2.1.2 支架安装注意事项 (23)4.2.1.3 安装机柜底部支架（可选） (23)4.2.1.3.1 拼装支架 (23)4.2.1.3.2 定位支架 (26)4.2.1.3.3 钻孔 (28)4.2.1.3.4 安装膨胀螺栓 (29)4.2.1.3.5 安装支架 (30)4.2.2 安装固定尺寸底座 (32)4.2.2.1 摆放300mm宽固定底座 (32)4.2.2.2 摆放600mm宽固定底座 (32)4.2.2.3 摆放600mm宽活动底座 (33)4.2.2.4 摆放800mm宽固定底座 (35)4.2.2.5 固定底座 (36)4.3 空调安装前操作 (39)4.3.1 系统连接和物料准备 (39)4.3.2 检查气压 (46)4.3.3 安装水管转接头 (48)4.4 安装机柜 (51)4.4.1 拆除IT柜栈板 (51)4.4.2 拆除配电柜栈板 (52)4.4.3 拆除NetCol5000-C 空调栈板 (54)4.4.4 摆放机柜 (56)4.4.5 固定机柜 (56)4.4.5.1 固定IT柜 (56)4.4.5.2 固定配电柜 (57)4.4.5.3 固定NetCol5000-C (58)4.4.6 安装通道密封件 (59)4.4.6.1 安装UPS5000-E可调尾框 (59)4.4.6.2 安装UPS可调顶框 (66)4.4.6.3 安装300mm宽空调可调顶框 (70)4.4.6.4 安装配电柜顶框 (75)4.4.7 机柜并柜 (76)4.4.8 安装机柜配件 (78)4.4.8.1 取出PDU2000工业连接器及线缆 (78)4.4.8.2 安装分接地铜排（可选） (79)4.5 安装密封通道 (80)4.5.1 安装流程 (80)4.5.2 安装围板 (81)4.5.3 天窗介绍 (83)4.5.4 固定天窗 (90)4.5.5 安装翻转天窗电磁锁 (91)4.5.6 安装双开旋转门（支持门禁安装） (92)4.5.6.1 安装通道门下安装件 (94)4.5.6.2 安装门框 (95)4.5.6.3 安装门板 (98)4.5.6.4 安装闭门器 (102)4.5.6.5 安装门立柱侧板并拆卸辅助拉杆 (106)4.6 安装走线槽 (107)4.6.1 安装600mm宽走线槽 (107)4.6.2 安装300mm宽走线槽 (112)4.6.3 安装800mm宽走线槽 (114)4.6.4 安装走线槽端板 (114)4.7 安装门禁系统（可选） (115)4.7.1 安装双门电磁锁 (116)4.7.2 安装S3-RC34-A21B05门禁读卡器 (124)4.7.3 安装开关 (128)4.8 安装通道级照明LED灯（可选） (133)4.9 安装监控设备 (136)4.9.1 网络柜设备布局 (136)4.9.2 安装服务器 (138)4.9.3 安装交换机 (140)4.9.4 安装ECC采集器 (142)4.9.5 安装WCON-14Z天窗控制器 (142)4.9.6 安装VCN500（可选） (143)4.9.7 安装门禁控制器（可选） (145)4.9.8 安装LCON-24V照明控制箱 (147)4.9.9 安装UF-LTS16-1P静态切换开关（可选） (148)4.9.10 安装独立部署AI/DI模块（可选） (148)4.9.11 安装控制天窗上的监控部件 (149)4.9.11.1 安装摄像机 (150)4.9.11.2 安装SDLH-94R感烟探测器 (154)4.9.11.3 安装红外传感器 (155)4.9.12 安装温度传感器 (156)4.9.13 安装声光告警器 (157)4.9.14 安装WLDS900水浸传感器（用于智能微模块） (158)4.9.14.1 连接WLDS9000水浸传感器线缆 (158)4.9.14.2 敷设水浸检测绳（上走水管场景） (160)4.9.14.3 敷设水浸检测绳（下走水管场景） (160)4.10 系统走线 (163)4.10.1 系统走线总则 (163)4.10.2 单排密封场景走线路径 (167)4.10.3 监控部件安装位置 (168)4.10.4 旋转门监控部件走线路径 (169)4.10.5 控制天窗监控部件走线路径 (170)4.10.6 通道级照明接线 (171)4.10.7 天窗电磁锁走线路径 (174)4.10.8 电源线走线 (175)4.10.9 连接模块接地线 (177)4.10.10 供配电系统接线 (180)4.10.10.1 电池包接线及安装 (180)4.10.10.2 蓄电池接线 (185)4.10.10.3 电池架安装接线 (190)4.10.10.4 供配电接线 (190)4.10.11 温控系统接线 (191)4.10.12 机房管理系统接线 (191)4.10.12.1 制作监控线缆 (191)4.10.12.2 监控线走线 (193)4.10.12.3 连接服务器、交换机、采集器间线缆 (193)4.10.12.4 连接服务器监控线缆 (194)4.10.12.5 连接交换机监控线缆 (195)4.10.12.6 连接ECC500采集器监控线缆 (195)4.10.12.7 连接GS-06门磁监控线缆 (196)4.10.12.8 连接摄像机线缆 (197)4.10.12.9 连接SDLH-94R感烟探测器监控线缆 (198)4.10.12.10 连接WLDS900水浸传感器线缆 (198)4.10.12.11 连接声光告警器监控线缆 (199)4.10.12.12 连接A8802T门禁控制系统监控线缆 (199)4.10.12.13 连接温度传感器线缆 (203)4.10.12.14 连接短信Modem线缆 (204)4.10.12.15 连接PDU2000监控线缆 (205)4.10.12.16 连接MS899电子门锁监控线缆 (206)4.10.12.17 连接WCON-14Z天窗控制器监控线缆 (207)4.10.12.18 连接LCON-24V照明控制箱监控线缆 (208)4.10.12.19 连接VCN500监控线缆（可选） (210)4.10.12.20 连接STS静态转换开关电源线 (211)4.10.12.21 连接NetCol5000-C 30kW空调监控线缆 (211)4.10.12.22 连接NetCol8000-C 50kW/70kW空调监控线缆 (213)4.10.12.23 连接配电柜监控线缆 (214)4.10.12.24 连接UPS2000-G监控线缆 (215)4.10.12.25 连接UPS5000-E监控线缆 (215)4.10.12.26 连接PD510电量仪监控线缆 (215)4.10.12.27 连接PDU8000监控线缆 (217)4.11 安装底座饰板（可选） (217)5 安装检查 (219)6 上电调测 (220)6.1 上电调测顺序 (220)6.2 调测供配电系统 (220)6.3 调测温控系统 (221)6.4 配置机房管理系统 (222)A 附录 (224)A.1 安装除湿机 (224)A.2 制作监控线缆 (229)A.3 绕柱场景 (231)A.3.1 绕柱安装场景 (231)A.3.2 绕柱安装流程 (233)A.3.3 安装绕柱密封件 (233)A.3.3.1 安装围板及天窗连接板 (233)A.3.3.2 安装可调天窗 (235)A.3.3.2.1 安装托架 (235)A.3.3.2.2 安装侧边连接式活动托架 (237)A.3.3.3 侧边连接固定横梁 (239)A.3.3.4 安装可调天窗顶部PC板 (239)A.3.3.5 安装侧面压板 (241)A.3.3.6 安装线槽支撑板 (242)A.4 连接气体灭火控制器主机与ATS联动线缆（可选） (243)A.5 连接机房消防系统与门禁控制器联动线缆（可选） (245)A.6 线缆清单 (246)A.7 ECC采集器接口介绍 (252)A.8 天窗控制器接口介绍 (256)B 常用螺钉力矩 (258)C 安装检查 (260)D 缩略语 (263)1 安全注意事项1.1 通用安全注意事项在安装、操作、维护智能微模块设备时，本文介绍的所应遵守的部分安全注意事项可指导选择测量设备和测试设备。
【尚择优选】HW-IT建设与发展
自行开发
HWIT建设历程
•协同办公自动化系统 •MRPII •全国广域网 •人力资源管理系统
1998
1992
重点实施
•IT战略规划 •业务变革
•IPD •ISC •财务四统一 •企业级数据中心 •数据仓库 •WEB应用平台
全面建设
•经过十年的建设和发展，从最初的临时满足客户需求逐渐转为提高公司的整体工作效率，朝着为全球市场、客户、 员工、供应链提供服务这个方向发展。已经初步建立了统一的信息化发展建设体系，公司信息化应用水平迅速提高 ，信息化已经成为公司业务拓展和运做效率提升的重要E
ISCBPE
BIE
HR FPE
Fin FPE
R&D FPE Mktg FPE
Sales FPE TS FPE
注:实线为行政汇报关系，虚线为业务汇报关系。
IT价值和使命 IT建设历程 IT体系 IT建设展望 总结
目录
•自行开发 管理信息系统， 局部业务 自动化
优化的供应链
电子商务实现外部供应链
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安装指南：Blade Series插座与Interlock Box
Interlock Box Installation InstructionsPart #: 0290-00815, 0290-00815-HW,0290-00820, 0290-00820HWHow to Install the Interlock BoxStep 1 - TURN OFF ALL POWER AND VERIFYWITH MULTIMETER.Step 2 - Mount Interlock Box within 3 feet of theDocking Drawer location by fastening it with the 4 Interlock Box Mounting Screws (part #: 3490-00028) through the mounting holes in the mounting bracket.Interlock Box is applicable with the following models:NOTE: It may be easier to mount the Interlock Box in the cabinet and mount the limit switch to the rear mounting bracket before installingthe Blade Series outlet.Part #: 0024-00227Step 3 - Locate the two screw holes on the outlet’srear mounting bracket. Install Interlock Switch Cable by fastening bracket (part #: 0024-00227), as shown using the 2 provided Interlock Switch Mounting Screws (part #: 3940-00021). Tighten securely.Step 4 - Install the Blade Series outlet in thePull the drawer all the way open and slide the limit switch back and forth until the limit switch roller is closed. You will hear an audible click when this occurs. When the limit switch is in the correct position, securely tighten the limit switch mounting screws.Step 5 - Plug in the Interlock Switch Cable to themating connector on the Interlock Box.Step 6 - Turn on the power to the system. Testthe function of the interlock by moving the drawer in and out. The Blade Series outlet should de-energize when the drawer is 1-2” from the fully open position. INSTALLATION IS NOT COMPLETE UNTIL THE UNIT HAS BEEN TESTED. You may find it is necessary to further adjust the position of the limit switch to de-energize the outlet at the correct drawer position to achieve proper operation.BLADE TM PART #:1514-1x02012-1x0BLADE DUO TM PART #:1514-2xx 2012-2xx *********************** (530) 205-3625 12893 Alcosta Blvd, Suite M,San Ramon, CA 94583AG110620Interlock Switch Cablecomes pre-assembled with the following brackets:Interlock Box Mounting Screws:Part #: 3940-00028 (4)Interlock Switch Mounting Screws:Part #: 3940-00021 (2)These instructions are for the corded Interlock Box options. For additional instructions on hardwiring, call us at (925) 233-5598.0290-00820HWInterlock Switch CableInterlock Switch Mounting ScrewsLoosen the 2Limit Switch screws to adjust when the outlet de-energizesInterlockBox Mounting ScrewsInterlock Box that de-energizes the outlet when the drawer starts to close.Outlets with all USB ports are certified for use in Canada without an Interlock Box.。
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PART A: Minimum Inventory Production Plan (i.e., one that allows arbitrary hiring and firing) A B C D E F G H I J KForecast Net Cummulative Minimum # # Units Cummulative Inv. above Actual Demand Demand Net Demand of Workers Produced Production Min required InventoryC/32 E x 32 I - DJuly 1250 1050 1050 33 2 0 1056 1056 6 306Aug 1100 1100 2150 35 0 2 1120 2176 26 326Sep 950 950 3100 30 5 0 960 3136 36 336Oct 900 900 4000 29 1 0 928 4064 64 364Nov 1000 1000 5000 32 0 3 1024 5088 88 388Dec 1150 1250 6250 40 0 8 1280 6368 118 5188 13 338 2238A B C D E F G H I J KForecast Net Cummulative Minimum # # Units Cummulative Inv. above Actual Demand Demand Net Demand of Workers Produced Production Min required InventoryC/32 E x 32 I - DJuly 1250 1050 1050 33 2 0 1056 1056 6 306Aug 1100 1100 2150 35 0 2 1120 2176 26 326Sep 950 950 3100 29 6 0 928 3104 4 304Oct 900 900 4000 28 1 0 896 4000 0 300Nov 1000 1000 5000 32 0 4 1024 5024 24 324Dec 1150 1250 6250 39 0 7 1248 6272 22 4229 13 82 1982A B C D E G H I J Forecast Net Cummulative Cum. Units Monthly Cummulative Inv. Above Actual Demand Demand Net Demand produced Production Production Min required Inventory per worker 32 x 34 H - DJuly 1250 1050 1050 32 1088 1088 38 338Aug 1100 1100 2150 64 1088 2176 26 326Sep 950 950 3100 96 1088 3264 164 464Oct 900 900 4000 128 1088 4352 352 652Nov 1000 1000 5000 160 1088 5440 440 740Dec 1150 1250 6250 192 1088 6528 278 6781298 3198PROBLEM # 1Month Fire Hire Minimum inventory required per month = 300Ending Inventory = 400Employee-hours required per hour = 5 hoursWorking days per months = 20Each employee works 8-hour-dayWorkforce at the end of June = 35K CALCULATION K (Units produced by one worker in one day)Initial Inventory = 500TOTALK = 8/5 = 1.6# Units Produced by 1 worker in a month = 1.6 x 20 = 32In this problem there is sufficient inventory on hand to reduce the workforce by one worker in the months of Sep, Oct and Dec.That decision will allow to reduce the total inventory above minimum required from 338 units to 82.Month FMonth Fire Hire TOTALD/EPART B: Production Plan that meets demand but does not hire or fire workers during the six-month periodTOTAL323233Because there are 35 workers employed at the end of June, the constant workforce plan require fire 1 worker at thebeginning of July. No further hiring or firing of workers are required. The inventory levels that results from a constantworkforce of 34 workers appear in the following table.33Ratio 3334PROBLEM # 2Constant Workforce = 137. Therefore, hire 51 workers at the beginning of April.The inventory levels that results from a constant workforce of 137 shown in the table below.In 26 days 100 workers produced 12000 cases of beer Holding Cost = $0.75 per case per month Initial Inventory = 4500Ending Inventory = 3000Minimum inventory required per month = 1000PART A: Minimum constant workforce planCOST = (107) x (300) + (111) x (125) + (8802) x (0.75) = 52576.50PART B: Zero Inventory PlanPART B: Zero Inventory PlanK CALCULATION K (Units produced by one worker in one day)K = 12000 / (26 x 100) = 4.6153Hiring Cost = $125 per worker Firing Cost = $300 per worker Workforce at the end of March = 86COST = (51) x (125) + (31976) x (0.75) = 30357COST = (106) x (300) + (110) x (125) + (8285) x (0.75) = 51763.75In this problem there is sufficient inventory on hand to reduce the workforce by one worker in the months of June and August.PART C & DLINDO INPUTMINIMIZE125H1+125H2+125H3+125H4+125H5+125H6+300F1+300F2+300F3+300F4 +300F5+300F6+0.75I1+0.75I2+0.75I3+0.75I4+0.75I5+0.75I6SUBJECT TOW1-H1+F1=86W2-W1-H2+F2=0W3-W2-H3+F3=0W4-W3-H4+F4=0W5-W4-H5+F5=0W6-W5-H6+F6=0P1-50.7692W1=0P2-101.5384W2=0P3-92.3077W3=0P4-106.1538W4=0P5-73.8462W5=0P6-92.3077W6=0P1-I1=4000P2+I1-I2=9300P3+I2-I3=12200P4+I3-I4=17600P5+I4-I5=14000P6+I5-I6=6300I1>1000I2>1000I3>1000I4>1000I5>1000I6>3000ENDGIN W1GIN W2GIN W3GIN W4GIN W5GIN W6LINDO OUTPUTOBJECTIVE FUNCTION VALUE24445.35VARIABLE VALUE REDUCED COSTW199.000000228.461395W299.000000380.768982W3158.000000276.923096W4158.000000363.846039W5158.000000410.769287W6158.000000-230.769226H113.000000 0.000000H20.0000000.000000H359.0000000.000000H40.0000000.000000H50.000000125.000000H60.000000425.000000F10.000000425.000000F20.000000425.000000F30.000000425.000000F40.000000425.000000F50.000000300.000000F60.0000000.000000I11026.1507570.000000I21778.4522710.000000I34163.0688480.000000I43335.3696290.000000I51003.0689700.000000I69287.6855470.000000P15026.1508790.000000P210052.3017580.000000P314584.6171880.000000P416772.3007810.000000P511667.6992190.000000P614584.6171880.000000I NTERPRETATION OF THE SOLUTIONTotal Cost = $24445.35Total # of workers in April and May = 99Total # of workers in June, July, August and September = 158Total # of workers hired in April = 13Total # of workers hired in June = 59Total # of workers hired in May, July, August and September = 0Total # of workers fired in April, May, June, July, August and September = 0Inventory at the end of April = 1026 cases of beer.Inventory at the end of May = 1778 cases of beer.Inventory at the end of June = 4163 cases of beer.Inventory at the end of July = 3335 cases of beer.Inventory at the end of August = 1003 cases of beer.Inventory at the end of September = 9287 cases of beer.Total # of cases of beer produced in April = 5026Total # of cases of beer produced in May = 10052Total # of cases of beer produced in June = 14584Total # of cases of beer produced in July = 16772Total # of cases of beer produced in August = 11667Total # of cases of beer produced in September = 14584P LAN FOR P ART B WOULD BE THE SAME AS FOR P ART D. (N O FIRING)。