物流自动化课后答案朱卫锋版
【作业参考】[福建师范大学]《物流自动化》在线作业一-00054
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福师《物流自动化》在线作业一-0005
物流系统中的托盘分为
A:平托盘
B:箱式托盘
C:柱式托盘
D:轮式托盘
答案：A,B,C,D
物流系统的监控的功能有
A:物流系统状态信息的采集
B:物流系统状态的监测
C:在计算机上用图形显示系统状态
D:异常情况的处理
答案：A,B,C,D
自动化仓库的机械设施包括
A:货架
B:货箱与托盘
C:运输系统
D:搬运设备
答案：A,B,C,D
自动化物流系统的一般包括
A:管理层
B:信息层
C:控制层
D:执行层
答案：A,C,D
自动化仓库主要组成设施包括
A:土建设施
B:监控设施
C:机械设施
D:电器设施
答案：A,C,D
入库原则有
A:均匀分布的原则
B:就近入/出库的原则
C:对货架受载有利的原则
D:先入先出的原则
答案：A,B,C
自动化仓库按照货架形式可分为
A:单元式货架仓库
1。
大工21春《物流自动化》线作业123满分答案
大工21春《物流自动化》线作业123满分答案1.AGV是指〔〕。
A.叉车B.货架C.自动导引小车D.物流监控系统该题正确选项是:C2.自动化仓库采用计算机管理应该包括如下功能：库存管理和〔〕。
A.信息化管理B.台账管理C.货位管理D.数据处理该题正确选项是:C3.以下〔〕不是自动化仓库的优点。
A.自动存取B.计算机控制C.高层货架存储D.提高库存周转期该题正确选项是:D4.〔〕托盘用于盛放各种液体或松散状物料。
A.箱式C.悬挂式D.架放式该题正确选项是:A5.重力式货架是〔〕的一种类型。
A.单元式货架B.移动式货架C.贯穿式货架D.垂直循环式货架该题正确选项是:C6.以下因素中，不属于影响装卸叉车使用因素的是〔〕。
A.单元负载单位B.储运作业单位C.货品布置D.物流条形码该题正确选项是:D7.RF是指〔〕。
A.无线射频终端系统B.无线数据采集器D.现代数据存储与处理技术该题正确选项是:A8.电子商务时代，现代物流业的必由之路是〔〕。
B.柔性化C.信息化D.全球化该题正确选项是:C9.自动控制的堆垛机必须具备自动认址功能，其中〔〕可靠性高，不易出错，但设备复杂，比拟昂贵，因此用的比拟少。
A.绝对数据认址系统B.相对数据认址系统C.非数字认址系统D.跟踪认址系统该题正确选项是:A10.把分布在不同地理位置的独立计算机，用传输介质、通信设备和网络操作系统等连接起来，实现资源共享的计算机系统是〔〕。
A.管理信息系统B.计算机网络C.专家系统D.决策支持系统该题正确选项是:B11.选择物流区域要注意的三个方面有〔〕。
A.政治经济条件B.效劳时效C.客户分布D.客户条件该题正确选项是:ABC12.物流设备选用与配置的原那么有〔〕。
A.系统化原那么B.适用性原那么C.经济性原那么D.超前性原那么该题正确选项是:ABCD13.现代物流信息处理技术主要有〔〕。
A.ERPB.条形码技术C.GPSD.EDIE.射频技术该题正确选项是:BCDE14.门式起重机按用途不同分为〔〕。
自动化课后答案ch10 material transport-3e-si
Chapter 10MATERIAL TRANSPORT SYSTEMSREVIEW QUESTIONS10.1Provide a definition of material handling.Answer: Material handling is defined by the Material Handling Industry of America (MHIA) as “the movement, storage, protection and control of materials throughout the manufacturing and distribution process including their consumption and disposal.”10.2How does material handling fit within the scope of logistics?Answer: Material handling is concerned with internal logistics – the movement and storage of material in afacility. By contrast, external logistics is concerned with the transportation of materials between facilities by rail, truck, seaway, air transport, and/or pipelines.10.3Name the four major categories of material handling equipment.Answer: As identified in the text, the four categories of material handling equipment are (1) material transport equipment, (2) storage systems, (3) unitizing equipment, and (4) identification and tracking systems.10.4What is included within the term unitizing equipment?Answer: As defined in the text, unitizing equipment refers to the containers used to hold individual itemsduring handling and the equipment used to load and package the containers.10.5What is the unit load principle?Answer: The unit load principle recommends that the unit load should be designed to be as large as is practical for the material handling system that will move or store it, subject to considerations of safety, convenience, and access to the materials making up the unit load.10.6What are the five categories of material transport equipment commonly used to move parts and materials inside afacility?Answer: The five categories identified in the text are (1) industrial trucks, manual and powered, (2) automated guided vehicles, (3) monorails and other rail guided vehicles, (4) conveyors, and (5) cranes and hoists.10.7Give some examples of industrial trucks used in material handling.Answer: Examples given in the text include nonpowered trucks such as dollies and pallet trucks, and powered trucks such as walkie trucks, forklift rider trucks, and towing tractors.10.8What is an automated guided vehicle system (AGVS)?Answer: As defined in the text, an automated guided vehicle system is a material handling system that uses independently operated, self-propelled vehicles guided along defined pathways. The vehicles are powered by on-board batteries.10.9Name three categories of automated guided vehicles.Answer: The three categories are (1) driverless trains, consisting of a towing vehicle that pulls one or more trailers, (2) pallet trucks, used to move palletized loads, and (3) unit load carriers, which move unit loads. 10.10What features distinguish self-guided vehicles from conventional AGVs?Answer: Conventional AGVs use either imbedded guide wires in the floor or paint strips on the floor surface as the guidance technology, whereas self-guided vehicles use a combination of dead reckoning, which refers to the capability of the vehicle to follow a given route by counting its own wheel rotations along a specified trajectory, and bea cons located throughout the facility that serve to verify the vehicle’s location in the facility.10.11What is forward sensing in AGVS terminology?Answer: Forward sensing involves the use of one or more sensors on each vehicle to detect the presence of other vehicles and obstacles ahead on the guide path. Sensor technologies include optical and ultrasonicdevices. When the on-board sensor detects an obstacle in front of it, the vehicle is programmed to stop.10.12What are some of the differences between rail-guided vehicles and automated guided vehicles?Answer: Rail-guided vehicles ride on tracks on the floor or overhead, whereas AGVs ride on the building floor without rails. Rail-guided vehicles are guided by the tracks, whereas AGVs are guided by imbedded wires in the floor that emit a magnetic field, or by paint strips, or other means that does not rely on rails. Rail-guided vehicles obtain their electrical power from a “third rail”, whereas AGVs carry batteries as their electrical power source.10.13What is a conveyor?Answer: As defined in the text, a conveyor is a mechanical apparatus for moving items or bulk materials,usually inside a facility. Conveyors are used when material must be moved in relatively large quantitiesbetween specific locations over a fixed path, which may be in-the-floor, above-the-floor, or overhead.Conveyors divide into two basic categories: (1) powered and (2) non-powered.10.14Name some of the different types of conveyors used in industry.Answer: The conveyor types listed in the text include roller, skate wheel, belt, chain, in-floor towline, overhead trolley, and cart-on-track.10.15What is a recirculating conveyor?Answer: A recirculating conveyor is a closed-loop conveyor that allow parts or loads to remain on the return loop for one or more revolutions.10.16What is the difference between a hoist and a crane?Answer: A hoist is a mechanical device for lifting and lowering loads vertically, whereas a crane is a mechanical apparatus for horizontal movement of loads. A crane invariably includes one or more hoists. PROBLEMSAnalysis of Vehicle-based Systems10.1 A flexible manufacturing system is being planned. It has a ladder layout as pictured in Figure P10.1 and uses arail guided vehicle system to move parts between stations in the layout. All workparts are loaded into the system at station 1, moved to one of three processing stations (2, 3, or 4), and then brought back to station 1 forunloading. Once loaded onto its RGV, each workpart stays onboard the vehicle throughout its time in the FMS.Load and unload times at station 1 are each 1.0 min. Processing times at other stations are: 5.0 min at station 2,7.0 min at station 3, and 9.0 min at station 4. Hourly production of parts through the system is: 7 parts throughstation 2, 6 parts through station 3, and 5 parts through station 4. (a) Develop the from-to Chart for trips anddistances using the same format as Table 10.5. (b) Develop the network diagram for this data similar to Figure10.13. (c) Determine the number of rail guided vehicles that are needed to meet the requirements of the flexiblemanufacturing system, if vehicle speed = 60 m/min and the anticipated traffic factor = 0.85. Assume reliability = 100%.Solution: (a) First develop the distances from the FMS layout.Distance from 1 to 2: 10 + 10 + 10 = 30 mDistance from 2 to 1: 5 + 10 + 5 = 20 mDistance from 1 to 3: 10 + 20 + 10 = 40 mDistance from 3 to 1: 5 + 20 + 5 = 30 mDistance from 1 to 4: 10 + 30 + 10 = 50 mDistance from 4 to 1: 5 + 30 + 5 = 40 mFrom-To chart:(b) Network diagram:7/30(c) L d = 730206403055040765()()()+++++++=122018= 67.7 m L e = 0Average handling and processing time = 1.0 + 750670590765(.)(.)(.)+++++ 1.0 = 8.78 minT c = 8.78 + 67760.+60= 9.91 minR dv = 60085991(.).= 5.15 pc/hr per vehicle n c =765515++.= 18/5.15 = 3.5 →4 vehicles10.2In Example 10.2 in the text, suppose that the vehicles operate according to the following scheduling rules: (1)vehicles delivering raw workparts from station 1 to stations 2, 3, and 4 must return empty to station 5; and (2) vehicles picking up finished parts at stations 2, 3, and 4 for delivery to station 5 must travel empty from station 1.(a) Determine the empty travel distances associated with each delivery and develop a from-to Chart in the formatof Table 10.5 in the text. (b) Suppose the AGVs travel at a speed of 50 m/min, and the traffic factor = 0.90.Assume reliability = 100%. As determined in Example 10.2, the delivery distance L d = 103.8 m. Determine the value of L e for the layout based on your table. (c) How many automated guided vehicles will be required tooperate the system?Solution: (a) Enumeration of empty trips:Deliveries Associated empty trips Frequency Empty distance1 to2 2 to 1 9 1102 to 5 1 to 2 and 5 to 1 9 50 + 301 to 3 3 to 1 5 2003 to 5 1 to 3 and 5 to 1 3 120 + 301 to 4 4 to 1 6 1153 to4 1 to 3 and 4 to 1 2 120 + 1154 to5 1 to 4 and 5 to 1 8 205 + 30From-To chart:(b) L e = 95051208205911052008115203042()()()()()()()++++++=620042= 147.6 m(c) T c = 1.0 + 103840147640..+= 7.26 minR dv = 60090726(.).= 7.44 del/hr per vehicle n c = 42/7.44 = 5.65 → 6 vehicles10.3In Example 10.2 in the text, suppose that the vehicles operate according to the following scheduling rule in orderto minimize the distances the vehicles travel empty: vehicles delivering raw workparts from station 1 to stations 2, 3, and 4 must pick up finished parts at these respective stations for delivery to station 5. (a) Determine the empty travel distances associated with each delivery and develop a from-to Chart in the format of Table 10.5 in the text. (b) Suppose the AGVs travel at a speed of 40 m/min, and the traffic factor = 0.90. Assume reliability = 100%. As determined in Example 10.2, the delivery distance L d = 103.8 m. Determine the value of L e for the layout based on your table. (c) How many automated guided vehicles will be required to operate the system?Solution: (a) Enumeration of empty trips:Deliveries Associated empty trips Frequency Empty distance1 to2 none2 to 5 5 to 1 9 301 to 3 none3 to 5 5 to 1 3 301 to 4 none3 to4 none4 to5 5 to 1 8 30From-To chart:(b) L e = 203042()= 14.3 m(c) T c = 1.0 + 10384014340..+= 3.95 minR dv = 60090395(.).= 13.66 del/hr per vehicle n c = 42/13.66 = 3.07 → 4 vehicles10.4 A planned fleet of forklift trucks has an average travel distance per delivery = 500 ft loaded and an averageempty travel distance = 350 ft. The fleet must make a total of 60 deliveries per hour. Load and unload times are each 0.5 min and the speed of the vehicles = 300 ft/min. The traffic factor for the system = 0.85. Availability =0.95, and worker efficiency = 90%. Determine (a) ideal cycle time per delivery, (b) the resulting average numberof deliveries per hour that a forklift truck can make, and (c) how many trucks are required to accomplish the 60 deliveries per hour.Solution: (a) T c = 0.5 + 500/300 + 0.5 + 350/300 = 3.83 min/delivery(b) Ideally, R dv =60383.= 15.66 deliveries/hr per truckAccounting for traffic factor, availability, and worker efficiency, R dv = 15.66(0.85)(0.95)(0.90) = 11.39 deliveries/hr per truck (c) n c = 60/11.39 = 5.27 → 6 forklift trucks10.5An automated guided vehicle system has an average travel distance per delivery = 200 m and an average emptytravel distance = 150 m. Load and unload times are each 24 s and the speed of the AGV = 1 m/s. Traffic factor =0.9. How many vehicles are needed to satisfy a delivery requirement of 30 deliveries/hour? Assume thatavailability = 0.95.Solution: T c = 24 + 200/1 + 24 + 150/1 = 398 s = 6.63 minR dv = 60090095663(.)(.).= 7.73 deliveries/hr per vehiclen c = 30/7.73 = 3.88 →4 vehicles10.6Four forklift trucks are used to deliver pallet loads of parts between work cells in a factory. Average traveldistance loaded is 350 ft and the travel distance empty is estimated to be the same. The trucks are driven at an average speed of 4 miles/hr when loaded and 5 miles/hr when empty. Terminal time per delivery averages 1.0 min (load = 0.5 min and unload = 0.5 min). If the traffic factor is assumed to be 0.90, availability = 100%, and worker efficiency = 0.95, what is the maximum hourly delivery rate of the four trucks?Solution: When loaded, v c = (4 miles/hr)5280160ftmilexhrmin.⎛⎝⎫⎭⎪= 352 ft/minWhen empty, v c = (5 miles/hr)5280160ftmilexhrmin.⎛⎝⎫⎭⎪= 440 ft/minT c = 1.0 + 350350352440+= 2.79 min/deliveryR dv = 60(1.0(0.90)(0.95)2.79= 18.39 deliveries/hr per vehicleWith four trucks, R d = 4(18.39) = 73.6 deliveries/hr.10.7An AGVS has an average loaded travel distance per delivery = 400 ft. The average empty travel distance is notknown. Required number of deliveries per hour = 60. Load and unload times are each 0.6 min and the AGV speed = 125 ft/min. Anticipated traffic factor = 0.85 and availability = 0.95. Develop an equation that relates the number of vehicles required to operate the system as a function of the average empty travel distance L e.Solution: T c = 0.6 + 4000.6125125eL++= 4.4 +125eLAT = 60(0.95)(0.85)(1.0) = 48.45 min/hr per vehicleWL = 50(4.4 + L e125) = 220 + 0.4 L en c =2200.448.45eLWLAT+=n c = 4.54 + 0.00825 L e10.8 A rail-guided vehicle system is being planned as part of an assembly cell. The system consists of two parallellines, as in Figure P10.8. In operation, a base part is loaded at station 1 and delivered to either station 2 or 4,where components are added to the base part. The RGV then goes to either station 3 or 5, respectively, where further assembly of components is accomplished. From stations 3 or 5, the product moves to station 6 forremoval from the system. Vehicles remain with the products as they move through the station sequence; thus, there is no loading and unloading of parts at stations 2, 3, 4, and 5. After unloading parts at station 6, the vehicles then travel empty back to station 1 for reloading. The hourly moves (parts/hr) and distances (ft) are listed in the table below. RGV speed = 100 ft/min. Assembly cycle times at stations 2 and 3 = 4.0 min each, and at stations 4 and 5 = 6.0 min each. Load and unload times at stations 1 and 6 respectively are each 0.75 min. Traffic factor =1.0 and availability = 1.0. How many vehicles are required to operate the system?To: 1 2 3 4 5 6From: 1 0/0 14L/200 0/NA 9L/150 0/NA 0/NA2 0/NA 0/0 14L/50 0/NA 0/NA 0/NA3 0/NA 0/NA 0/0 0/NA 0/NA 14L/504 0/NA 0/NA 0/NA 0/0 9L/50 0/NA5 0/NA 0/NA 0/NA 0/NA 0/0 9L/1006 23E/400 0/NA 0/NA 0/NA 0/NA 0/0 Solution: Assembly through stations 2 and 3:T c = 0.75 + 20010040501004050100075+++++...+400100= 16.5 minR dv = 60/16.5 = 3.636 deliveries/hr per vehicle n c = 14/3.636 = 3.85 vehicles.Assembly through stations 2 and 3:T c = 0.75 + 150100605010060100100075400100++++++...= 20.5 minR dv = 60/20.5 = 2.927 deliveries/hr per vehiclen c = 9/2.927 = 3.07 vehicles.Total vehicles n c = 3.85 + 3.07 = 6.92 →7 vehicles10.9An AGVS will be used to satisfy material flows indicated in the from-to Chart in the table below, which showsdeliveries per hour between stations (above the slash) and distances in meters between stations (below the slash).Moves indicated by "L" are trips in which the vehicle is loaded, while "E" indicates moves in which the vehicle is empty. It is assumed that availability = 0.90, traffic factor = 0.85, and efficiency = 1.0. Speed of an AGV = 1.1 m/s. If load handling time per delivery cycle = 1.0 min, determine the number of vehicles needed to satisfy the indicated deliveries per hour? Assume that availability = 0.90.To: 1 2 3 4From: 1 0/0 9L/90 7L/120 5L/752 5E/90 0/0 0/NA 4L/803 7E/120 0/NA 0/0 0/NA4 9E/75 0/NA 0/NA 0/0Solution: v c = 1.1 m/s(60 s/min) = 66 m/minRoute 1 → 2 → 1: T c = 1.0 + (90 + 90)/66 = 3.73 min, 5 deliveries.Route 1 → 3 → 1: T c = 1.0 + (120 + 120)/66 = 4.64 min, 7 deliveries.Route 1 → 4 → 1: T c = 1.0 + (75 + 75)/66 = 3.27 min, 5 deliveries.Route 2 → 4 → 1*: T c = 1.0 + (80 + 75)/66 = 3.35 min, 4 deliveries.Route 1 → 2*: T c = 1.0 + 90/66 = 2.36 min, 4 deliveries.* Assumes vehicles on route 1 → 2 are used to make deliveries on route 2 → 4 → 1.Average T c = 5(3.73)7(4.64)5(3.27)4(3.35)4(2.36)25++++= 3.613 min/delivery cycleR dv = 60(0.85)3.613= 14.12 deliveries/hr per vehicleIncluding effect of availability factor, R dv = 14.12(0.90) = 12.7 deliveries/hr per vehiclen c = 25/12.7 = 1.97 →2 vehicles10.10An automated guided vehicle system is being proposed to deliver parts between 40 workstations in a factory.Loads must be moved from each station about once every hour; thus, the delivery rate = 40 loads per hour.Average travel distance loaded is estimated to be 250 ft and travel distance empty is estimated to be 300 ft.Vehicles move at a speed = 200 ft/min. Total handling time per delivery = 1.5 min (load = 0.75 min and unload =0.75 min). Traffic factor F t becomes increasingly significant as the number of vehicles n c increases; this can bemodeled as:F t = 1.0 - 0.05(n c-1) for n c = Integer > 0Determine the minimum number of vehicles needed in the factory to meet the flow rate requirement. Assume that availability = 1.0 and worker efficiency = 1.0.Solution: T c = 1.5 + 250300200+= 4.25 min/cycleR dv = 60(1.00.05(1))4.25cn--=60(1.050.05)4.25cn-= 14.824 - 0.706 n c deliveries/hr per vehiclen c =4014.8240.706cn-n c (14.824 - 0.706 n c) = 4014.824 n c - 0.706 n c2 = 400.706 n c2 - 14.824 n c + 40 = 0Use quadratic equation to find roots:n c = --±-(.).(.)()(.)14824148244070640207062= 17.82 or 3.18 → Use n c = 4 vehiclesCheck: F t = 1.0 - 0.05(4 - 1) = 1.0 - 0.15 = 0.85R dv = 60085425(.).= 12 deliveries/hr per vehicle, n c = 40/12 = 3.33 → Use n = 4 vehicles10.11An automated guided vehicle system is being planned for a warehouse complex. The AGVS will be a driverlesstrain system, and each train will consist of the towing vehicle plus four carts. Speed of the trains will be 160ft/min. Only the pulled carts carry loads. The average loaded travel distance per delivery cycle is 2000 ft and empty travel distance is the same. Anticipated travel factor = 0.95. Assume reliability = 1.0. The load handling time per train per delivery is expected to be 10 min. If the requirements on the AGVS are 25 cart loads per hour, determine the number of trains required.Solution: T c = 10 + 20002000160+= 35.0 min/delivery cycleR dv = 60095350(.).= 1.629 deliveries/hr per vehicleR f = 254cartloads hrcartloads train//= 6.25 trainloads/hr = 6.25 deliveries/hrn c = 6.25/1.629 = 3.84 → Use n c = 4 AGV trains.10.12The from-to Chart in the table below indicates the number of loads moved per 4-hour period (above the slash)and the distances in ft (below the slash) between departments in a particular factory. Fork lift trucks are used to transport materials between departments. They move at an average speed = 350 ft/min (loaded) and 400 ft/min (empty). Load handling time per delivery is 1.5 min, and anticipated traffic factor = 0.9. Use an availabilityfactor = 95% and worker efficiency = 100%. Determine the number of trucks required under each of thefollowing assumptions: (a) the trucks never travel empty; and (b) the trucks travel empty a distance equal to their loaded distance.To Dept. A B C D EFrom Dept A - 62/500 51/450 45/350 0B 0 - 0 22/400 0C 0 0 - 0 76/200D 0 0 0 - 65/150E 0 0 0 0 -Solution: (a) L d = 625005145045350224007620065150625145227665()()()()()()++++++++++= 322.27 ftUsing L e = 0, T c = 1.5 + 322.270350400+= 2.42 min/delivery cycleR dv = 60(0.95)(0.90)(1.0)2.42= 21.19 deliveries/hr per vehicleR f = 321/4 = 80.25 deliveries/hrn c = 80.25/21.19 = 3.79 → Use n c = 4 vehicles.(b) L d = 322.27 ft, same as before.Using L e = L d = 322.26 ft, T c = 1.5 + 322.27322.27350400+= 3.23 min/delivery cycleR dv = 60(0.95)(0.90)(1.0)3.23= 15.9 deliveries/hr per vehiclen c = 80.25/15.9 = 5.05 → Use n c = 6 vehicles.10.13 A warehouse consists of five aisles of racks (racks on both sides of each aisle) and a loading dock. The racksystem is four levels high. Forklift trucks are used to transport loads between the loading dock and the storage compartments of the rack system in each aisle. The trucks move at an average speed = 140 m/min (loaded) and 180 m/min (empty). Load handling time (loading plus unloading) per delivery totals 1.0 min per storage/retrieval delivery on average, and the anticipated traffic factor = 0.90. Worker efficiency = 100% and vehicle reliability (availability) = 96%. The average distance between the loading dock and the centers of aisles 1 through 5 are 200 m, 300 m, 400 m, 500 m, and 600 m, respectively. These values are to be used to compute travel times. Therequired rate of storage/retrieval deliveries is 100 per hour, distributed evenly among the five aisles, and thetrucks perform either storage or retrieval deliveries, but not both in one delivery cycle. Determine the number of forklift trucks required to achieve the 100 deliveries per hour.Solution: L d = (200 + 300 + 400 + 500 + 600)/5 = 400 m, L e = L d = 400 mT c = 1.0 + 400/140 + 400/180 = 6.08 minR dv = 60/6.08 = 9.87 del/hr per truckn c = 100/(9.87 x 0.90 x 0.96) = 11.73 rounded to 12 trucks10.14Suppose the warehouse in the preceding problem were organized according to a class-based dedicated storagestrategy based on activity level of the pallet loads in storage, so that aisles 1 and 2 accounted for 70% of thedeliveries (class A) and aisles 3, 4, and 5 accounted for the remaining 30% (class B). Assume that deliveries in class A are evenly divided between aisles 1 and 2, and that deliveries in class B are evenly divided between aisles 3, 4, and 5.How many trucks would be required to achieve 100 storage/retrieval deliveries per hour?Solution: Class A L d = (200 + 300)/2 = 250 m, L e = L d = 250 mClass B L d = (400 + 500 + 600)/3 = 500 m, L e = L d = 500 mWeighted average L d = L e = 0.70(250) + 0.30(500) = 325 mT c = 1.0 + 325/140 + 325/180 = 5.13 minR dv = 60/5.13 = 11.69 del/hr per truckn c = 100/(11.69 x 0.90 x 0.96) = 9.9 rounded to 10 trucks10.15Major appliances are assembled on a production line at the rate of 55 per hour. The products are moved alongthe line on work pallets (one product per pallet). At the final workstation the finished products are removed from the pallets. The pallets are then removed from the line and delivered back to the front of the line for reuse.Automated guided vehicles are used to transport the pallets to the front of the line, a distance of 900 ft. Return trip distance (empty) to the end of the line is also 900 ft. Each AGV carries four pallets and travels at a speed of 200 ft/min (either loaded or empty). The pallets form queues at each end of the line, so that neither theproduction line nor the AGVs are ever starved for pallets. Time required to load each pallet onto an AGV = 15 sec; time to release a loaded AGV and move an empty AGV into position for loading at the end of the line = 12 sec. The same times apply for pallet handling and release/positioning at the unload station located at the front of the production line. Assume the traffic factor is 1.0 since the route is a simple loop. How many vehicles areneeded to operate the AGV system?Solution: T L = T u = 12 sec + 4(15 sec) = 72 sec = 1.2 minT c =9001.2200++9001.2200+= 11.4 min/delivery cycle.R dv = 60/11.4 = 5.26 delivery cycles/hr per vehicleEach delivery means 4 pallets, so R dv = 4(5.26) = 21.05 pallets/hr per vehiclen c = 55/21.05 = 2.613 → Use n c = 3 vehicles.10.16For the production line in the previous problem, assume that a single AGV train consisting of a tractor andmultiple trailers are used to make deliveries rather than separate vehicles. Time required to load a pallet onto a trailer = 15 sec; and the time to release a loaded train and move an empty train into position for loading at the end of the production line = 30 sec. The same times apply for pallet handling and release/positioning at theunload station located at the front of the production line. If each trailer is capable of carrying four pallets, how many trailers should be included in the train?Solution: Let n p = number of pallets per trainT L = T u = 30 sec + n p (15 sec) = 0.5 + 0.25 n p (min)T c = 1.0 + 0.5n p + 900200+900200= 10.0 + 0.5n pR dv =60100.5pn+delivery cycles/hr per train.Each delivery cycle means n p pallets are delivered; thus, R dv =60100.5ppnn+pallets/hr per train.n c =55/60(100.5)pppallets hrnn⎛⎫⎪+⎝⎭= 1 train (given that a single AGV train is used to satisfy delivery requirements)55 =60 100.5ppnn+55(10 + 0.5n p) = 60 n p550 + 27.5 n p = 60 n p60 n p - 27.5 n p = 32.5 n p = 550 n p = 16.9 round up to 17 pallets/trainWith 4 pallets per trailer, the train must have 17/4 = 4.25 rounded up to 5 trailers.10.17An AGVS will be implemented to deliver loads between four workstations: A, B, C, and D. The hourly flowrates (load s/hr) and distances (m) within the system are given in the table below (travel loaded denoted by “L”and travel empty denoted by “E”). Load and unload times are each 0.45 min, and travel speed of each vehicle is1.4 m/sec. A total of 43 loads enter the system at station A, and 30 loads exit the system at station A. In addition,six loads exit the system from workstation B each hour and seven loads exit the system from station D. This is why there are a total of 13 empty trips made by the vehicles within the AGVS. How many vehicles are required to satisfy these delivery requirements, assuming the traffic factor is 0.85 and the reliability (availability) is 95%?Hourly rate (loads/hr) Distances (m)From ABCDSolution: L d =()()()()()()30221215101815030802265121501580109518++++++++++L d = 11,800/107 = 110.28 mL e =()()1071507956+= 1620/107 = 15.14 mT c = 0.45 + 110.28/(1.4 x 60) + 0.45 + 15.14/(1.4 x 60) = 2.393 minWL = 107(2.393) = 256.04 min of work per hourAT = 60(0.85)(0.95) = 48.45 min/hr per vehiclen c = 256.04/48.45 = 5.28 rounded to 6 vehiclesAnalysis of Conveyor Systems10.18An overhead trolley conveyor is configured as a continuous closed loop. The delivery loop has a length of 150 mand the return loop = 130 m. All parts loaded at the load station are unloaded at the unload station. Each hook on the conveyor can hold one part and the hooks are separated by 4 m. Conveyor speed = 1.75 m/s. Determine (a)maximum number of parts in the conveyor system, (b) parts flow rate; and (c) maximum loading and unloading times that are compatible with the operation of the conveyor system? Solution : (a) Number of parts on the conveyor =1504/mm part= 37.5 parts (average)(b) R f = p c cn v s =(1/)(1.75/)(4/)part carrier m s m carrier = 0.4375 parts/s = 1575 parts/hr(c) T L = T u =1fR = 1/0.4375 = 2.29 sec 10.19 A 300 ft long roller conveyor, which operates at a velocity = 80 ft/min, is used to move pallets between load andunload stations. Each pallet carries 12 parts. Cycle time to load a pallet is 15 sec and one worker at the loadstation is able to load pallets at the rate of 4 per min. It takes 12 sec to unload at the unload station. Determine (a) center-to-center distance between pallets, (b) the number of pallets on the conveyor at one time, and (c) hourly flow rate of parts. (d) By how much must conveyor speed be increased in order to increase flow rate to 3000 parts/hour.Solution : (a) s c = T L v c = (15/60 min)(80 ft/min) = 20 ft/carrier (b) Number pallets on conveyor =d c L s = 30020ft ft carrier/ = 15 carriers (c) R f = p c cn v s =(/)(/min.)/128020parts carrier ft ft carrier= 48 parts/min = 2880 parts/hr(d) Increasing v c would have not effect towards increasing R f in this problem. The loading rate, set by T L = 15 sec, is what limits the flow rate in this system.10.20 A roller conveyor moves tote pans in one direction at 150 ft/min between a load station and an unload station, adistance of 200 ft. With one worker, the time to load parts into a tote pan at the load station is 3 sec per part.Each tote pan holds 8 parts. In addition, it takes 9 sec to load a tote pan onto the conveyor. Determine (a) spacing between tote pan centers flowing in the conveyor system and (b) flow rate of parts on the conveyor system. (c) Consider the effect of the unit load principle. Suppose the tote pans were smaller and could hold only one part rather than 8. Determine the flow rate in this case if it takes 7 sec to load a tote pan onto the conveyor (instead of 9 sec for the larger tote pan), and it takes the same 3 sec to load the part into the tote pan.Solution : (a) T L = 9 + 3(8) = 33 sec = 0.55 min/tote pan s p = (150 ft/min)(0.55 min/tote pan) = 82.5 ft/tote pan (b) R f =p c cn v s =(/)(/min.)./8150825pc totepan ft ft totepan= 14.55 pc/min = 872.7 pc/hr .(c) T L = 7 + 3(1) = 10 sec = 0.167 min/tote pans p = (150 ft/min)(0.167 min/tote pan) = 25 ft/tote panR f = (/)(/min.)/115025pc totepan ft ft totepan= 6.0 pc/min = 360 pc/hr .10.21 A closed loop overhead conveyor must be designed to deliver parts from one load station to one unload station.The specified flow rate of parts that must be delivered between the two stations is 600 parts per hour. Theconveyor has carriers spaced at a center-to-center distance that is to be determined. Each carrier holds one part. Forward and return loops will each be 90 m long. Conveyor speed = 0.5 m/s. Times to load and unload parts at the respective stations are each = 12 s. Is the system feasible and if so, what is the appropriate number of carriers and spacing between carriers that will achieve the specified flow rate?Solution : Relationships to use in this problem:(1) R f = p c cn v s = c c v s = 0.5c s for v c = 0.5 m/s and n p = 1。
大工18春《物流自动化》在线作业3
------------------------------------------------------------------------------------------------------------------------------ (单选题) 1: MRP展开的依据是()。
A: 主生产计划B: 粗能力计划C: 产品结构信息D: 能力需求计划正确答案:(单选题) 2: 在生产物流中,()是主要的控制对象。
A: 人B: 机器C: 物流过程D: 物流环节正确答案:(单选题) 3: 仓库内应划分的区域包括()。
A: 生产作业区、行政生活区、辅助生产区B: 储存区、装卸作业区、待验区C: 验收区、储存区、出库区D: 装车区、出货区、保管区正确答案:(单选题) 4: 确定仓库面积的主要依据是()。
A: 最高物资储备量B: 物资性质C: 物资质量D: 物资体积正确答案:(单选题) 5: 合计拣货作业前,应累计订单中每一商品项目的总量,再按这一总量进行检取的方式为()分批。
A: 定时B: 智慧型C: 固定订单量D: 总合计量正确答案:(单选题) 6: 分拣时订单进行分批作业后,需要进行()作业。
A: 工作分区B: 订单分类C: 订单分割D: 订单不分割正确答案:(单选题) 7: 下列不需要人工控制的设备是()。
A: 普通叉车B: 高架叉车C: 拣选叉车D: AGV叉车正确答案:------------------------------------------------------------------------------------------------------------------------------ (单选题) 8: 下面只有一种设备不能进行集装箱的堆垛作业,这种设备是()。
A: 底盘车B: 叉车C: 轮胎门式起重机D: 跨运车正确答案:(单选题) 9: 配送中心的库存应该()位置管理。
大连理工大学智慧树知到“物流管理”《物流自动化》网课测试题答案5
大连理工大学智慧树知到“物流管理”《物流自动化》网课测试题答案（图片大小可自由调整）第1卷一.综合考核(共15题)1.企业信息资源从狭义的定义包括()。
A.信息B.信息系统C.资金流D.物流2.操作系统是()的接口。
A.主机和外设B.用户和计算机C.系统软件和应用软件D.高级语言和机器语言3.下列()不是自动化仓库的优点。
A.自动存取B.计算机控制C.高层货架存储D.提高库存周转期4.实践证明，基于()管理的物流战略联盟能给企业带来实质性的成本节约。
A.生产链B.供应链C.销售链D.客户链5.光敏电阻的成本高于光电池。
()A.正确B.错误6.物流系统中的托盘分为()。
A.平托盘B.箱式托盘C.柱式托盘D.网箱托盘7.物流自动化系统中的自动控制技术主要是对相关机械机构运动的控制。
()A.错误B.正确8.()托盘用于盛放各种液体或松散状物料。
A.箱式B.插孔式C.悬挂式D.架放式9.以下不属于RFID优势的是()。
A.安全性高B.扫描速度快C.成本低D.无屏障阅读10.自动化仓库的机械设备包括()。
A.货架B.货箱与托盘C.运输设施D.搬运设施11.以下EDI通信方式中，数据安全性高的是()。
A.InternetB.点对点C.传真D.Intranet12.叉车的基本作业功能分为()。
A.水平搬运B.堆垛/取货C.装货/卸货D.拣选13.语句表是PLC编程语言的一种。
()A.错误B.正确14.GIS是一种以()研究和决策服务为服务目标的计算机技术系统。
A.地理B.天文C.生物D.智能15.条形码在配送管理上的应用可以分为两类，即类别管理和单品管理。
()A.正确B.错误第2卷一.综合考核(共15题)1.以下物流运作环节中，条码技术未广泛应用的是()。
A.运输B.零售C.仓储配送D.生产制造2.RFID识别速度低于条码技术。
()A.正确B.错误3.把分布在不同地理位置的独立计算机，用传输介质、通信设备和网络操作系统等连接起来，实现资源共享的计算机系统是()。
智慧物流复习题与参考答案
智慧物流复习题与参考答案一、单选题（共48题，每题1分，共48分）1.下列关于成熟电商的说法，错误的是 ( )A、一般情况下，会固定一家快递网点作为自己的主要发件果道B、了解不同品牌的快递在不同地区的网络覆盖时效、价格上的黄异C、会把所有其他品牌作为备用网点正确答案：C2.可以预知有些项目可交付成果需要返工，却不知道返工的工作量是多少，为应对这些未知数量的返工工作，可以预留（）。
A、管理储备B、组织储备C、应急储备D、控制储备正确答案：C3.请使用SWOT分析法进行战略选择，某物流公司拥有雄厚的资金基础，但传统的主营业务面临复杂的竞争导致盈利能力不断下降，应采取的战略是（）。
A、W-O战略B、S-T战略C、S-O战略D、W-T战略正确答案：B4.服务水平是影响顾客满意度的重要指标，下列服务水平的等级中顾客满意度最高的是（）。
A、专业顾问B、有问必答C、长期伙伴D、保持沟通正确答案：C5.评价表格法指的是设计多个指标对多个参评单位进行评价的方法，又称（）。
A、相对指标法B、综合评价方法C、平均指数法D、平衡积分法正确答案：A6.本月包装A产品400件，实际使用工时1000小时，支付工资7000元；包装单位产品的人工标准成本是15元/件，每件产品的标准工时是3小时，试计算直接人工成本差异是（）。
A、1000元B、1200C、1800D、1650正确答案：A7.快件的处理为( ）小时。
A、4B、2C、1D、3正确答案：B8.2017年10月，尼康中国相机工厂正式停产,是什么原因导致的呢?尼康方面相关人士表示,小型数码相机市场正在出现极大的萎缩,这主要是由于智能手机迅速发展对其业务造成了冲击。
这属于供应链“五力分析”中的（）。
A、新进入者威胁B、替代品威胁C、上游供应商的威胁D、行业内部竞争正确答案：B9.在网点管理数据的实际应用中,往往以监测为基础。
A、业务量指标B、质量指标C、时效指标D、成本效益指标正确答案：A10.多式联运经营人在统一责任制下对货物承担的运输责任是（）。
物流自动化技术与应用智慧树知到答案章节测试2023年山东农业工程学院
第一章测试1.（）是当前最主要的控制装置，是物流系统自动化的核心设备。
A:电子数据设备B:可编程控制器C:射频标签识别系统D:监控管理计算机答案:B2.以下关于EDI系统的描述，不准确的是（）。
A:EDI系统的运用，可实现远程通信，取代了计算机、数据库B:EDI系统可实现数据交换和信息共享C:运输领域可采用EDI系统实现货运单证的电子数据传输D:仓储领域利用EDI系统可加速货物的提取、周转，提高仓储利用率答案:A3.物流自动化系统工作的前提是（），通过收集和记录流通实物的相关数据信息，如货物信息、工作状态信息等，才能实现实物流动的自动化控制。
A:可编程控制器B:信息管理系统C:外接执行设备D:自动标识与数据采集系统答案:D4.物流自动化系统的核心是（），它具有机电一体化系统的特征，接收PLC控制器的指令，完成自动化作业任务。
A:监控管理系统B:前端执行系统C:信息采集系统D:网络通信系统答案:B5.以下物流自动化系统中的先进技术，哪些属于自动标识与信息采集技术的范畴？（）A:GPS技术B:GIS技术C:RFID技术D:条形码技术答案:ABCD第二章测试1.以下不符合可编程控制器特征描述的一项是（）。
A:点对点连接方式下，接线工作较复杂B:为特定工作系统设计，专用性强C:采用扫描工作方式，抗干扰能力强D:编程简单，面向过程控制，使用方便答案:B2.当PLC执行用户程序时，不可能同时执行多个操作，而是分别单独执行，由于CPU的运算处理速度极快，所以从结果呈现上来看，似乎是同时执行的，而实际是采取的（）。
A:封闭式操作B:分时操作C:集中输出D:集中采样答案:B3.PLC运行过程中需要生成或调用中间结果数据和组态数据，这类数据存放在（）中，采用随机存取存储器存放。
A:过程数据库B:工作数据存储器C:系统程序存储器D:用户程序存储器答案:B4.PLC执行程序的一个扫描周期必经阶段包括：（）。
A:输出刷新B:输入采样C:任务分解D:程序执行答案:ABD5.PLC中的存储器包含可读/写操作的随机存储器和只读存储器，主要的作用是存储：（）。
物流自动化技术与应用智慧树知到期末考试章节课后题库2024年山东农业工程学院
物流自动化技术与应用智慧树知到期末考试答案章节题库2024年山东农业工程学院1.自由路径导引方式下，AGV小车采用激光导航方式进行导引。
（）答案:对2.可编程控制器的灵活性一方面体现在：系统程序可以在应用过程中随时修改，利用编程器生成新的程序指令。
（）答案:对3.在中心转运型物料搬运路线中，物料由起点至终点以最短的距离移动。
答案:错4.借助于GPS技术的绝对定位测量，可以测出山峰高度、海面任意两点间的距离等。
（）答案:错5.现场总线的互可操作性，使得不同厂家性能类似的设备，可以通过统一的应用层协议进行通信连接。
（）答案:对6.球坐标型机器人由回转、旋转、平移三个自由度构成，具有较大的动作范围。
（）答案:对7.自动分拣系统对商品外包装要求高，自动分拣机只适于分拣底部平坦且具有刚性的包装规则的商品。
（）答案:对8.自动分拣技术水平高、资金回笼快，只要有建设基金，企业都应该改建自动分拣线，才能在市场竞争中处于优势。
（）答案:对9.中心转运型物料搬运路线下，物料在预定路线上移动，与来自不同地点的其他物料一起运到同一终点，适合布置不规则或搬运距离较长的物料。
（）答案:错10.工业机器人的运动控制是对机器人整体进行的位置移动控制，通过控制关节运动来实现。
（）答案:错11.以下关于物料搬运系统的搬运原则，哪些是合理的？（）答案:重力化原则###集装化原则###空间最大利用原则###中转衔接原则12.自动分拣系统的构成要素除基本的动力输送线、电子计算机以外，还包含以下哪些项目？（）答案:分拣控制决策信息###信息读取装置###货物分拣标志或信息###可编程控制器13.全球定位系统技术可向全球用户提供实时、连续、高精度的位置、速度和时间信息，系统构成包括：（）。
答案:空间部分###用户接收设备###地面监控部分14.自动分拣系统中，输送线在节点元素处实施通过策略的目的有哪些？（）。
答案:绩效目的###通畅目的###时间目的###最大能力目的15.RFID系统的体系结构包含（）。
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1.1 简述物流自动化系统的结构及其特点。
答：结构：1--前提，信息采集系统（条形码扫描、标签识别、图像识别、语音识别、传感器）2--核心，前端执行系统（自动输送系统、运输设备、搬运机器人、自动引导车系统、自动存储和分发系统、自动分拣系统）3--基础，信息管理系统，4--指令，控制系统，5--通信，计算机网络（局域网、广域网）6--管理目标，人特点：系统化、集成化、自动化、智能化、网络化、信息化1.2略1.3物流自动化先进技术有哪些？答：1、自动标志与数据收集（1）条形码；（2）射频标签识别系统；（3）销售时点信息系统；2、EDI系统及其应用3、GPS、GIS原理与应用4、PLC编程技术5、现场总线控制系统技术6、监控组态软件开发技术7、机器人、AGV、分拣机构2.1 可编程控制器由哪几部分组成？它有什么作用？PLC的特点？答：CPU：按系统程序赋予的功能，指挥PLC有条不紊的进行工作。
存储器：存储系统程序，用户程序及数据。
输入输出接口:用于数据传输和控制。
通信接口:用于与外部设备连接。
扩展接口：用于连接扩展单元的接口。
电源：提供电力。
特点：1，可靠性强，抗干扰性强；2，编程简单，使用方便；3，功能完善，通用性强；4，设计安装简单，维护方便；5，体积小，重量轻，能耗低。
2.2 PLC采用何种扫描方式？工作过程分为哪五个阶段？请简要阐述。
答：扫描方式：分时操作。
阶段：1.内部处理：检查CPU等内部硬件，对监视定时器复位以及其他工作；2.通信服务：有其他智能装置实现通信；3输入采样：按顺序对所有输入端的状态进行采样，并存入相应寄存器。
4.程序执行：对用户程序扫描执行，并将结果存入相应的寄存器；5.输出刷新：将寄存器中的输出状态，转入输出锁存器，输出驱动外部负载。
2.3梯形图和继电器控制电路图有什么区别？答：继电器控制属于普通开关控制电路，梯形图控制电路属于微电脑控制。
前者控制精度较高，后者可靠性较高。
3.1简述现场总线的定义及特点。
答：定义：现场总线是指安装在制造或过程区域的现场装置之间，以及现场装置与控制室内的自动控制装置之间的数字式、串行和多点通信的数据总线。
特点：布局简单、开放性、实时性、可靠性、适应性、分散控制。
3.2略3.3、现场总线主要考虑哪些网络性能？答：（1）在选择现场总线时，用户对性能方面考虑有如下几方面：①传输速率；②时间同步准确度；③执行同步准确度；④媒体访问控制装置；⑤发布/预订接收能力；⑥报文分段能力；⑦可寻址的最大网络段数；（2）与实时性要求相关的技术指标有如下几个方面：①媒体访问控制机制；②媒体访问控制的时间准确程度；③系统有无公共时间源及后背时间源；④通信模式；⑤编码效率；⑥通行效率；⑦传输速率；4.0 RFID系统的构成？答：电子标签，天线，射频模块，控制模块，应用系统。
4.1简述条形码识读的基本工作原理，并说明条形码识读设备的分类及各类设备的工作方式和性能？答：条形码符号的识读是通过不同颜色对红色光源的反射率不同，进而分辨条、空的边界和宽窄来实现的。
识读设备的分类：光笔条形码扫描器、手持式条形码扫描器、台式条形码自动扫描器、卡式条形码阅读器、便携式数据采集器4.2简述RFID技术的特点，并说明RFID系统的基本工作原理。
答：特点：读取方便快捷；识别速度快；数据容量大；实验寿命长；可重复使用；安全性和可靠性高；动态实时通信。
工作原理：（1）读写器经过发射天线向外发射无线电载波信号。
（2）当电子标签进入发射天线的工作区域时，电子标签被激活，将自身信息经天线发射出去。
（3）系统的接收天线接收到电子标签发出的载波信号，经天线的调节器传给读写器。
读写器对接收到底信号进行解调解码，送后台计算机控制器。
（4）计算机控制器根据逻辑运算判断电子标签的合法性，针对不同的设定做出相应的处理和控制，发出指令信号控制执行机构的动作。
（5）执行机构按计算机的指令动作。
（6）计算机网络将各个监控点连接起来，构成总信息平台。
4.3 RFID系统中工作频率对电子标签的工作有什么影响？答：工作频率是射频识别系统最基本的技术参数之一。
工作频率的选择在很大程度上决定了射频识别系统的应用范围、技术可行性及系统成本的高低。
从本质上说，射频识别系统是无线电传播系统，必须占据一定的无线通信信道。
在无线通信信道中，射频信号只能以电磁耦台或电磁波传播的形式表现出来。
因此，射频识别系统的工作性能必然会受到电磁波空间传输特性的影响。
不同射频频率的电磁波存在较大的差异。
低频电磁波具有很强的穿透能力，能够穿透水、金属、动物等导体材料，但是传播距离比较近。
另外，由于频率比较低，可以利用的频带窄，数据传输速率较低，信噪比较低，容易受到干扰。
高频系统的发射功率较小，设备比较简单，成本也比较低。
高频电磁波的数据传输速率较高，没有低频的信噪比限制，但是高频电磁波的穿透能力较差，高频电磁波对障碍物的敏感性较强，很容易被水等导体媒质所吸收。
4.4比较条形码技术和RFID技术的区别与联系。
自动识别技术信息载体信息量读/写性读取方式保密性干扰能力寿命成本条形码纸、塑料薄膜小只读CCD或激光束扫描差差较短最低RFID EEPROM 大读/写无线通信最好很好最长较高4.5 简述GPS定位技术的特点及其定位原理？答：特点：1全球范围内连续覆盖、2实时导航、定位精度高、数据内容多3静态定位观测效率高4抗干扰能力强、保密性好5功能多、应用广定位原理;利用空间分布的卫星，以及卫星与地面点的距离交会得出地面点位置。
简言之，GPS定位原理是一种空间的距离交会原理4.6简述地理信息系统GIS分类及其构成？答：按内容分为三类 1专题地理信息系统2区域地理信息系统3地理信息系统工具。
构成：系统硬件、系统软件、空间数据、系统的用户。
5.1简述监控组态软件的发展趋势。
答：1通过OPC统一数据采集方式2用脚本语言扩充组态系统功能3组态环境的可扩展性4组态软件的开发性5由单一的人机界面朝数据处理剂方向发展，管理的数据量越来越大6基于Internet的远程监控7组态软件的控制功能日益强大5.3简述监控组态软件的内部通信方式答：内部通信机制 1 请求式--客户周期性向服务器发出请求，然后服务器为客户准备数据，传送给客户、2 订阅是--客户向服务器注册所关心的数据，服务器记录每个用户关心的数据。
当数据变化时通知相应的用户。
3 广播室--当数据发生变化时，服务器向所有客户发出通知。
5.4介绍监控组态软件与外部程序的通信机制答：1 OPC方式---这是一种基于微软的OLE自动化技术开发而成的，主要用于工业控制领域的设备通信接口标准，是目前主流的通信方式。
2.API方式---客户方直接通过编程方式调用设备提供商的API函数实现与设备间的通信。
3ODBC 方式----主要用于访问存放在关系数据库的历史数据和事件记录数据。
4DDE方式----是一种动态数据交换机制。
5.5 详细描述监控组态软件的主要功能。
答：1、过程报警监控组态软件具有过程报警的功能，这是过程数据库的基本功能。
报警是对测量值的范围、变化速度的预警。
报警包括限值报警、变化率报警、偏差报警、常报警等。
更复杂的报警可以通过对测量值进行数学运算，然后对运算结果进行报警检查而产生。
报警后，操作员可以通过报警画面对报警进行“确认”，“确认”就是告诉系统我知道发生事了，以免系统再次提醒。
报警信息、报警确认信息、报警恢复(报警消除)等信息都可以被系统自动记录下来。
报警按照重要程度可分为多个优先级，如低级、高级、紧急。
报警发生时系统可以通过发出声响，甚至可以发送短信或电子邮件，至于使用哪种方式，用户可以自由指定。
其中有多种方式通知用户，如弹出报警窗。
2、历史存储监控组态软件可以对实时数据进行历史存储。
历史数据对于生产状况分析、实现先进／优化控制，以及生产事故的分析非常重要，所以重要的工艺数据都会进行长期保存。
为了节省存储空间，对保存的数据使用压缩保存。
常用的保存方式有周期性保存和变化保存(数据只有变化到一定程度才保存)。
3、脚本语言监控组态软件除了固定功能外，如点内部处理(能够完成几种固定形式的功能，量程转换、报警检查、PID运算)、动画连接(能够完成几种固定形式的动画，如颜色、位置改变)等，监控软件还允许用户定义特殊的逻辑、运算，这是通过监控软件的内置编程语言实现的。
监控组态软件的脚本语言与一般编程语言类似．在脚本中提供多种运算操作(如赋值、数学运算、逻辑运算)、控制语句(如条件判断、循环、分支)及内置函数等。
脚本的触发方式有多种：一次性执行(如进入窗口时执行)、周期执行和事件触发执行(如数据改变时执行、按键触发)。
脚本也能产生多种输出动作，如向过程数据库写数据、发送短信、调用窗口、产生声响等。
6.1分拣的基本方法有哪些？1.单一分拣2.批量分拣 3 单一分拣与批量分拣的组合6.2试述分拣作业所消耗时间的组成？1.形成拣货指令的订单信息处理过程所需的时间2.行走或货物运动的时间3.准确找到储位并确认所拣货物及其数量所需的时间4.拣取完毕，将货物分类集中的时间6.3 简述摘果式分拣作业法和播种式分拣作业法各自的优缺点，以及适用范围。
（一）摘果式分拣作业法(1)摘果式分拣作业法的优点主要有：作业方法单纯；订单处理前置时间短(收到订单到拣选时间间隔短)；作业人员责任明确，派工容易、公平；拣货后不必再进行分拣作业，适用于数量大、品种少的订单的处理。
(2)摘果式分拣作业法的缺点主要有：商品品种多时，拣货行走路线过长，拣取效率低；拣取区域大时，搬运系统设计困难；少量、多批次拣取时，会造成拣货路径重复费时，效率降低(3)摘果式分拣作业法的适用范围如下：①出货量少、频率小的物品拣选；②种类多，数量少，但识别条件多(颜色＼尺寸等)；③体积小，单价高；④涉及批号管制，且每批数量不定；⑤用户不稳定，波动较大，不能建立相对稳定的用户分货货位；⑥用户之间共同需求差异较大；⑦用户需求种类不多；⑧用户配送时间要求不一；⑨传统仓库改造为配送中心，或新建的配送中心初期运营。
（二）播种式分拣作业法(1)播种式分拣作业法的优点主要有：适合订单数量大(多)的系统；可以缩短拣取时的行走搬运距离，增加单位时间的拣取量；对于少量、多批次的配送十分有效。
(2)播种式分拣作业法的缺点主要有：由于必须等订单达到一定数量时才做一次处理，因此订单处理前置时间长。
(3)播种式分拣作业法的适用范围如下：①用户稳定，而且用户数量很多；②用户需求具有很强的共同性，品种差异小，只是数量有一定差异；③用户需求种类有限；④用户配送时间要求没有严格限制；6.4什么叫分拣配货作业？有哪几种方式？分拣配货作业是指按客户的要求将商品从储存区分拣出来，配货后送入指定发货区的物流活动。