02 July 2006 RFID: Keeping Track with Technology 0. Overview I. Learning situation/issues A. Grade level B. Mathematical topics C. Relevant PSSA standards D. Curriculum connection E. Steelton-Highspire curriculum connections F. Career considerations II. Underlying context A. Initial setting B. Detailed context C. Context Q&A III. Lesson ideas A. Open-ended project B. Free-response items C. Variety items IV. Particular Materials A. Mathematics tasks B. Mathematics handouts C. Technology guides D. Visual components E. Web in-sites F. Materials and equipment needed V. About this module A. Motivation C. Credits/disclaimer RFID module GE Math Excellence: Math in a “New Technology” Context project Page 1 of 53
Transcript
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RFID: Keeping Track with Technology
0. Overview I. Learning situation/issues
A. Grade levelB. Mathematical topicsC. Relevant PSSA standards D. Curriculum connectionE. Steelton-Highspire curriculum connectionsF. Career considerations
II. Underlying contextA. Initial settingB. Detailed contextC. Context Q&A
III. Lesson ideasA. Open-ended projectB. Free-response itemsC. Variety items
IV. Particular MaterialsA. Mathematics tasksB. Mathematics handoutsC. Technology guidesD. Visual componentsE. Web in-sitesF. Materials and equipment needed
V. About this moduleA. MotivationC. Credits/disclaimer
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Module Components
Overview
RFID: Keeping Track with Technology is a module for the GE Math Excellence: Math in a “New Technology” Context project. The module contains a collection of items that can be used with middle-school students to reinforce whole number operations, to connect rational number operations to whole number operations, to compare numbers, to apply geometric formulas, and to express patterns involving number calculations as a precursor to algebraic symbolism. For an explanation of how and why RFID became the topic of this module, please see the section on the motivation for this module.
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Learning situation/issuesThis section provides the basic background for this module and its use in middle school mathematics settings.
Grade levelThe primary intended audience for for the items in this module is middle school students of various abilities. The items target number and operations, data analysis, and algebra. As with many curriculum materials, this module may be used in its entirety or in part with many different learners.
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Mathematical topicsThis module addresses several topics of middle school mathematics. It relates particularly to number, data analysis, and algebra with less attention to geometry. The topics are described in the following sections in terms of Pennsylvania Mathematics Standards and Curriculum Connections for at least one school district.
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Relevant Pennsylvania Mathematics Standards
Each of the items in this set has been linked to Pennsylvania’s Assessment Anchors (Pennsylvania Department of Education, 2007 version). Clicking on the icon below should open a chart that lists all of the anchors and indicates how the items match the anchors. This module might seem to address some anchors or strands (e.g., Number and Operation) much more heavily than it addresses others. Other modules may have a different pattern of anchor emphasis.
In creating this chart, we used the following codes:
TARGET........The item would be particularly critical in addressing the anchor. The content of the item may be more demanding than the anchor, but students who adequately respond to the item certainly should be successful with respect to the anchor. For example, the anchors address conversions in limited ways (e.g., using only certain units [in/ft/yd, fl
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oz/cup/pint/quart/gallon, oz/Lb, sec/min/hours/days, metric units including milli/centi/kilo]). We included a broader range of conversions in the chart. Among other things, this approach honors the mathematics that is common and critical to success in engineering, science, and technology.
r......................The item involves ideas related to the anchor, but to a lesser extent than a TARGET item would. In some cases, the anchor content may be involved in only one of several valid responses to the item. We assume the totality of the solution and enactment is part of the classroom discussion. Some of the “r” items would be realized in discussion if not present in students’ initial work.
E.....................The anchor arises only in the “Enactment” aspect of the item.
We elaborated further for items involving calculators, reflecting our interest in the judicious use of calculators.
r......................The item would be one for which student learning or performance likely would be enriched by access to a calculator.
r-without.........The item could be used with students to make a point about a time in which using a calculator is not necessary, and may indeed be inefficient.
In the items, and in our school setting, we assumed access to a middle-school friendly graphing calculator, such as the TI-73.
There are some anchors that will not be marked as often as one might expect. For example, 7A.2.1 involves the order of operations. There are items for which students would attend to the order of operation in the process of using a formula for the area of a circle. We did not see this repetitious use of the order of operations in a familiar expression as a task that would contribute sufficiently to students’ development of deeper understanding of the anchor content. Similarly, we did not mark a computation anchor when the computation was critical in addressing another anchor. For example, we did not mark a computation anchor when the item targeted a conversion anchor. We chose to mark fewer anchors with recognition that suggesting an extremely large number of items, many of which had other key content, might not be useful in making decisions and selecting items for classroom use.
MS Excel Worksheet
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Curriculum connectionsThis module particularly fits with the study of large numbers and rational
numbers. It also fits with algebra topics and data analysis.Number and operation
Many of the items in this module require students to operate on large numbers (e.g., V8)1 or rational numbers (e.g., V10, V14). There are several items that involve percent (e.g., V3) or percent increase or percent decrease (e.g., V1, V13, V16). Students have opportunities to think about when an answer needs to be rounded (e.g., F2a) and when rounding an answer is not necessary (e.g., V4). Some of the items (e.g., P2) provide opportunities for students to think about number beyond computations. There is some exposure to the concept of inequality expressed in terms of on quantity being greater than, less than, at least or at most another quantity (e.g., F2b, V15).Algebra, pattern and function
There are opportunities for students to start to generalize operations on numbers or to use expressions with informal or formal variables (e.g., V7, V18). Some tasks (e.g., F1) ask students to create equations of lines in one or more of several ways they may have seen (e.g., two-point form of a line, using the regression feature of a calculator).Measurement
Students may have several experiences converting among different measures (e.g., V5, V8) as they think through what they are multiplying or dividing and how the units of measure appear in the calculations and the answers.Geometry
Students may have some opportunities to use geometric principles of area (e.g., P3). Geometry is not a central focus of this module.Data analysis
The projects offer students venues in which to design and implement data collection schemes. Throughout all items, students use data of several kinds. Some data are information explicitly given to students or information the students gather (e.g., P1). The information may involve measures and units they recognize (e.g., perhaps length and width of an auditorium in P1) or measures and units that are not familiar (e.g., V15). There is also an opportunity for students to extend a direct computation into a deeper understanding of average (e.g., V11).
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Steelton-Highspire curriculum matchThere are particularly fruitful uses of the materials in this module in the
Steelton-Highspire curriculum. The district uses the following textbooks: Grade 6 Altieri, M. B., Bezuk, N., Cole, P. B., Ferguson, B. W., Harrell, C.
P., & Lubcker, D. H. (2002). Mathematics. New York: McGraw Hill.
Grade 7 Charles, R. I., Branch-Boyd, J. C., Illingworth, M., Mills, D., & Reeves, A. (1999). Prentice Hall Mathematics: Course 2. Upper Saddle River, NJ: Pearson Prentice Hall.
1 To translate these codes to item in this module, please see Lesson Ideas.
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Grade 8 Charles, R. I., Branch-Boyd, J. C., Illingworth, M., Mills, D., & Reeves, A. (1999). Prentice Hall Mathematics: Course3. Upper Saddle River, NJ: Pearson Prentice Hall.
The specific connections for these grades and textbooks include the following:
P2 – Student EPC CodesV4 – Tired of PercentV5 – Reading DistanceV8 – UPC SongV10 – RFID Chip WidthV12 – Three Seaports RuleV14 – A Chip MeterV15 – Bits of Information
13 Percents (Clusters A & B) V1 – More SuppliersV3 – Seaport InspectionV12 – Three Seaports RuleV13 – In 6 to 8 YearsV15 – Bits of InformationV16 – Early RFID Users
14 Probability (Clusters A & B) P2 – Student EPC Codes
Prentice Hall
Mathe-matics: Course
2
1 Interpreting Data and Statistics (2, 3, 6, 8)
F1 – Expecting Cheaper RFID Tags
2 Applications of Decimals (1, 2, 3, 5, 6)
P1 - Auditorium TaggingF1 – Expecting Cheaper RFID TagsP3 – Detecting TagsV2 – Falling PricesV4 – Tired of PercentV5 – Reading Distance V8 – UPC SongV10 – RFID Chip WidthV13 – In 6 to 8 YearsV14 – A Chip MeterV15 – Bits of InformationV16 – Early RFID Users
P1 - Auditorium TaggingF1 – Expecting Cheaper RFID TagsV6 – UPC SlicesV7 – Check the DigitV9 – Second DigitsV11 – Average Military ShipmentV18 – Pallet Tags V19 – More Pallet TagsV20 – Tags for Another Warehouse
4 Fractions and Number Theory (9) P1 - Auditorium Tagging
5 Applications of Fractions (2, 5, 8) P1 - Auditorium TaggingV4 – Tired of Percent
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V5 – Reading DistanceV8 – UPC SongV10 – RFID Chip WidthV14 – A Chip MeterV15 – Bits of Information
6 Using Proportions and Percents (3, 5, 7, 9, 11)
P1 – Auditorium TaggingP3 – Detecting TagsV1 – More Suppliers V3 – Seaport InspectionV12 – Three Seaports RuleV13 – In 6 to 8 YearsV15 – Bits of InformationV16 – Early RFID Users
Career considerationsThere are number of mathematical applications in this module. Several
items require computing with different units and working with percent change, ideas that are important in the business world and help in everyday living. Facility in determining how to minimize cost or to weigh one product against another is useful in a variety of business and industry settings.
RFID falls under the broad category of automatic identification technologies. The science behind the most common way to read passive RFID engages principles of inductive coupling. Simply put, the coiled antenna of the reader creates a magnetic field with the coiled antenna of the tag. The tag draws energy from this field and uses it to send back waves to the readers. These waves are turned into digital information - the tag's Electronic Product Code. Knowing about radio wave technology is useful in many careers that draw on mathematics and science. For example, automatic identification technology has roots in warfare and military operations.2
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2 The answer to the first question under “Context Q&A” provides historical background.
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Underlying context
Initial settingRFID stands for Radio Frequency Identification. RFID is part of a system
that allows companies and organizations to track objects and people. Instances of RFID use appear on the Pennsylvania Turnpike, in airports, in warehouses, and in many other familiar places.
An RFID system consists of three components: tag(s), reader, and communication software. RFID tags, consisting of a chip and an antenna, are placed on objects or people. The tag is capable of transmitting information about the object or person. The second piece of equipment in the system is a reader. The reader picks up the radio waves from the tag and translates them into a digital signal. Communication software is needed to change the digital signal into information that other programs and people can understand.
RFID tags are important to all of us. In particular, these tags could replace the UPC codes that we see on the products we buy in stores. Understanding how the tags work, knowing their benefits and their limitations, and looking into the costs of the tags helps us to understand what we may soon see in our everyday life.
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Detailed contextThe brief introduction above likely is enough for students to know as they
begin working on the tasks in this module. Additional information about the RFID context may be useful. Among the topics here are details about the parts of the system, examples of RFID use, background about automatic identification systems, comments about the benefits of RFID, and its relationship to bar codes.
Parts of the system. RFID is a generic term used to describe a system that transmits the identity of an object or person wirelessly via radio waves. An RFID system includes three main parts: a tag, a reader, and communication software. One RFID system is capable of collecting hundreds of signals per second. The signals can be stored in a networked database that has the capability of checking whether any of the signals are repeated signals from the same tag or if any of the signals were not read correctly.
Tags. The Radio Frequency Identification Tags are primarily made up of an EERPROM (electrically erasable programmable read-only memory) chip and a coil. The coil serves as an antenna. There are different types of tags for different uses. Figure 1 is a photograph of a tag taken from a shirt purchased at the GAP. The tag is not much larger than a quarter.
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Figure 1. RFID tag removed from shirt purchased at the GAP
Readers. A reader is needed to retrieve data stored on an RFID tag. A typical reader is a device that has one or more antennas that emit radio waves and receive signals back from the tags. The reader then passes the information in digital form to a computer system.
Communication software. The communication software translates the digital messages into messages that are usable by software or people using a workstation connected to a network. The data may be meshed with another database that allows people to find more information about the person or product.
What information is involved. Minimally, a unique identification number would be stored on a tag. A database would use this unique number to connect other information to the object. This other information could include basic product data, tracking history, and processing instructions. The tag also might contain configuration instructions, what time the item traveled through a certain zone and temperature or other data provided by sensors.
Passive and active tags. Active RFID tags have a battery, which is used to run the microchip's circuitry and to broadcast a signal to a reader. Since they have a battery power source, active tags have the largest coverage area. These types of tags are good in systems where a continuous scan is required, like the ones in security systems and toll collection system such as EZ Pass on the Pennsylvania Turnpike.
Passive tags have no battery. Instead, they draw power from the reader, which sends out electromagnetic waves that induce a current in the tag's antenna. These tags are less expensive than other types of tags, but they compromise on the coverage area, which is limited to 10 feet or less. Passive tags are useful in the retail industry where stationary objects are placed on shelves and scanners can be placed in close proximity to the objects. There is a third type of tag, which is not used in the items in this module; semi-passive tags use a battery to run the chip's circuitry, but communicate by drawing power from the reader.
Distance at which a tag can be read. The stated distances at which a tag can be read vary across the tasks in this module and in the resources referenced. One reader may be able to read different types of tags at different distances. As the previous paragraphs suggest, the differences in distances might be related to the use of passive tags as opposed to active tags. In general, cheap tags cannot be read at as great a distance as more costly tags. A usual inventory tag can be read
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by a typical system at 5 meters to 50 meters, as noted in Item P3. There are other times when the mathematical goals for students are better met by assuming (without contradicting RFID principles) that a particular reader is used and a specific distance is given, as in Item V5.
Examples of RFID use. RFID tags could be used to monitor shipments from suppliers to customers. When Supplier A ships a pallet full of soft drinks, the tags on the cases and pallet are scanned as the shipment leaves the warehouse. Software is used to automatically let Customer B know the shipment has left the warehouse. Customer B can look up data associated with the serial numbers on the shipment and learn what is in the shipment, when it will arrive, and so on. When Customer B receives the shipment, a reader scans the tags automatically, and a message can be immediately sent to Supplier A to let it know the shipment arrived.
RFID tags also can be used to move goods within a business. Placing RFID readers on shelves would monitor how many products are being sold. The readers would be capable of signaling the backroom when the stock on the shelves gets low and requesting for the shelves to be re-stocked. When inventory in the backroom gets low, readers there could signal the warehouse to send more products. When inventory in the warehouse gets low, readers would signal the manufacturer to send more products.
Electronic Product Codes. The Electronic Product Code (EPC) is a sequence of numbers and letters consisting of a header and three sets of data, as shown in Figure 23. The header identifies the EPC's version number. Starting an EPC with a version number allows for different lengths or types of EPCs to be used in the future. The second part of the EPC identifies the EPC Manager or the manufacturer of the product that is attached to the EPC. The third part of the EPC, called object class, describes the exact type of product. The fourth part is the serial number that is unique to the particular item on which the EPC appears. The serial number is useful when it is necessary to locate a specific product for a particular reason, such as locating a perishable item that is nearing its expiration date.
Figure 2. Four parts of an EPC number
Automatic identification technologies. Automatic identification technologies include bar code systems, optical character readers, and some biometric technologies (e.g., retinal scans). Automatic identification technologies provide an alternative to manually entering data. These technologies reduce the
3 Facts and pictures of electronic product codes appeared at (www.smartcode.corp.com.images/EPC/EPC_type1.gif).
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amount of time needed to input data. They also improve data accuracy. For example, in our daily lives, this means that grocery store workers can enter the prices of all the items we purchase and make fewer mistakes in the process.
Some automatic identification technologies, such as bar code systems, require a person to manually scan a label or tag across a reader to capture the information. RFID is designed to enable readers to capture data on tags and transmit it to a computer system without needing to involve a person.
Cost as a major issue. Until recently, use of RFID was limited by the costs of RFID tags and systems. The cost often exceeded one dollar per tag (as compared to less than one cent to put a bar code on a product). For some applications, such as tracking parts for just-in-time manufacturing, companies could justify the cost of tags. This type of company needs to know what it has where immediately and the money the company saved by being able to make and move things quickly offset the cost of the RFID system. When RFID was used to track assets or reusable containers within a company’s own four walls, the costs are less of a problem because tags can be reused. In other cases, a company cannot reuse an RFID tag. For example, a tag on something we would buy likely would be thrown out with the package. A tag used for products sold to the general public (such as the tag in Figure 1) would have to be reasonably disposable. Some companies are looking to develop ways to recycle tags on corrugated cardboard boxes.
Advantages of RFID. The previous paragraphs suggested several good things about RFID. The following chart summarizes the advantages of this new form of technology.
Advantages of RFID Unlike some other identification systems, there is no need for
physical contact between the tag and the reader. Tags can be used repeatedly. Tags can withstand extreme conditions and temperatures. Maintenance costs are low. Tags come in a variety of types, sizes, and materials. Information can be put on or read from tags in questionable (e.g.,
dirty) conditions. There is an extremely low chance of errors in RFID systems.
Comparison of RFID and bar codes. A big reason why we have this module is that RFID may replace bar codes (as in UPC labels) on the products we buy. The following chart shows how these two types of technology compare.
Characteristic Bar Codes RFIDReading Capability Optical-line of sight is
needed.No line of site is needed.
Reading Speed One code can be read in each scan.
Multiple tags can be read at once.
Durability Codes tend to be Tags withstand harsh
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damaged in harsh processes.
environments.
Amount of Information 20 to 2K characters 15 characters to 4k bitsFlexibility of Information Limited Read/write tags can be
updated. Security RFID tags have
manufacturer-installed ID codes that make counterfeiting difficult.
Costs Typical bar code costs less than $0.01 to use.
RFID tags cost as little as $0.35 to $0.50.
Standards Standards for use are widely accepted.
There is a lack of global standardization.
References. Portions of the information in this section came from the following references:
RFID Handbook, ”Fundamental and applications in Contactless Smart Cards and Identification”, Klaus Finkenzeller.
RFID Presentation by John Martinez of RFID, Inc. for Council of Logistics Management Group of the Lehigh Valley in March 2004.
Intermec White Paper “Supply Chain RFID: How It Works and Why It Pays”, http://home.intermec.com/eprise/main/Intermec/Content/Products/Products_Articles?Category=RFID&Family=RFID4&Product=RFID4_01&ArticleType=White%20Paper
Lauren R. Hartman, Systech International, Wal-Mart’s on schedule with RFID revolution, http://www.packagingdigest.com/articles/200407/36.php
SkyeTek RFID Demonstration Kits, http://www.skyetek.com/products.htmlIntersoft Corporation, RFID Demokit-1,
http://www.intersoft-us.com/demokit.htmUnderstanding RFID system frequency choices, September 13, 2004,
Context Q&ATo reflect the changes in the context, this module will continue to evolve
and this Context Q&A section will be the place to gather more information about the context as the module is used in mathematics teaching and learning.
Q1: What was the motivation for RFID? What area of engineering gave birth to RFID?A: It is generally said that the roots of radio frequency identification technology can be traced back to World War II. The Germans, Japanese, Americans and British were all using radar—which had been discovered in 1935 by Scottish physicist Sir Robert Alexander Watson-Watt to track the approaching planes while they were still miles away. The problem
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associated with radar for a military group was the fact that there was no way to distinguish between the enemy planes coming towards the site and its own planes returning from missions. The Germans were the first people who identified that if the pilots rolled their planes while on the way back to the site, the reflected radio signal changed, which alerted the people at the base that it was their own plane coming in. This was the first passive RFID.The British government, with the help of Watson Watt, developed the first active transmitter, which could be fixed on each British plane. When it received signals from radar stations on the ground, it began broadcasting a signal back that identified the aircraft as friendly. RFID works on this same basic concept. A signal is sent to a transponder, which wakes up and either reflects back a signal (passive system) or broadcasts a signal (active system).
This led to much research in Europe Japan and the U.S. for RF (Radio frequency) communication in 1950s and 60s. The U.S. government was also working on RFID systems. At the request of the Agricultural Department, Los Alamos National laboratory developed a passive RFID tag to track cows. The problem was that cows were being given hormones and medicines when they were ill, but it was hard to make sure each cow got the right dosage and was not given two doses accidentally. Los Alamos came up with a passive RFID system that used 125 kHz radio waves. A transponder encapsulated in glass could be injected under the cow’s skin. It drew energy from the reader and simply reflected back a modulated signal to the reader using a technique known as backscatter. This system is still used in cows around the world today. Low-frequency transponders are also put in cards and used to control the access to buildings.
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Lesson ideasThere are three types of task in this module (and in every Math in a “New Technology” Context module). Open-ended projects are major activities that engage students in answering non-trivial questions about the real-world contexts. These activities could be treated as projects that take several days or class meetings. Free-response items are individual problems or sets of problems. They require students to make decisions, calculate values, and do other things to construct their own answers. For all or at least most of these open-ended items, there are specific suggestions (Enactment) about how the items may play out in the classroom. Variety items are stand-alone problems that could be used in many different ways, including review and practice. These items may be used to revisit previously learned materials while students are working either on a project or on several open-ended items related to RFIDs. The variety items also may be incorporated as applications of mathematics into lessons in a standard mathematics curriculum and used separately from the projects and free-response items. However, some of the variety items may provide insights into things to consider in the projects or free-response answers.
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Open-ended projects
P1. Auditorium Tagging
Your school’s auditorium is a warehouse. Each seat in the audience represents an object on which an RFID tag will be placed. How many readers would you need so that all of the tags can be read at any time? Where would you place the reader in your auditorium? Why does this placement make sense?
Follow-up question for class discussion of different methods: For each method for placing the readers, what percent of the area of the auditorium do the readers not cover?Answer:Answers will vary depending on expectations for the project as well as the students’ creation of a scaled map of the auditorium versus a layout of the auditorium seating arrangements obtained through the principal or building staff personal. One strategy is to consider the RFID reader’s range to be in a circle with the reader in the center, the radius of the circle will be determined by the quality of the reader. For example, a reader that has a range of 5 meters might be at the center of a circle with a radius of 5 meters. Strategies could involve fitting the appropriate size and number of circles with a reader at each of the centers. See the following figure below for an example of a possible layout of a rectangular auditorium.
Enactment: There are some things about RFID tags and readers that students need
to know at the start of the project. Among these facts might be the components of a RFID system, the function of readers and tags, the difference between active and passive tags, and the respective distances at which these tags can be read.
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This project could be carried out in several ways. For example, the information about the auditorium could be supplied by the teacher or gathered by the students. In the latter case, after the students are introduced to the task and understand the requirements, the students could take a trip to the auditorium. Before the trip, it may worthwhile for students to discuss materials they should take on the trip (e.g. measuring tools, paper, pencils, clipboards, etc.) and information they will need to gather (e.g. dimensions of the auditorium, seat arrangements, total number of seats, pillars or obstacles that may cause interference, etc.) After taking the trip to the auditorium and gathering necessary information, the students should create a scaled layout of the auditorium including the dimensions and the arrangement and number of seats.
The next step of the project could involve students working individually or in small groups to develop a plan for the best locations for the readers. Using the knowledge that the active tags have a larger range of 5 to 50 meters and the passive tags have a range of 10 feet or less, students should create two plans for the readers’ placements (one for passive tags and one for active tags).
Students should justify the placements of the readers using written explanations as well as mathematical explanations.
Upon completion of the project, a class discussion should ensue to address the strengths and weaknesses of each of the plans while addressing issues of accuracy, originality, thoroughness, and justification.
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P2. Student EPC Codes
Create an EPC for each student in your school. What version number would you use? How many possible EPCs could you create?Answer: To create the code, students need a version number. Students should reason about how long the “serial number” needs to be to allow each student in the school to be “named” by a number. This part of the tasks involves estimating and rounding as well as knowing some facts about the size of the school. Enactment: Several points of interests on the creation of the EPC codes. The
“version number” is the first part of the EPC code and it consists of only two digits or letters. The second part of the EPC code identifies the “EPC manager” which is most likely the manufacturer of a product. There are a total of seven digits or letters allotted to this part of the code. An example of how a student may assign this part of the code is to make this a school wide code thus each of the students in the school would have the same EPC manager. The third group of digits or letters represents the “object class” which refers to the exact type
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of product, or, in this case, to a particular student. This has a total of six digits or letters. One possible example that a student may create for this part of the EPC code could be the student’s month of birth (from 01-12) followed by day of birth (01-31) followed by the year of birth (90-05) followed by a number 1, 2,3… depending on the number students that have the same date of birth (e.g. 1008931, 1008932 indicates that two students were born on October 8, 1993). The last group of numbers on the EPC code is the “serial number” of a code. This has nine total slots or letters. One possible solution for this part of the code would be to use the student’s ID number if the school uses such identification numbers.
It is important to have a discussion about the lasting power of the numbers that are chosen by the students. For example if the codes involve students’ homerooms or grade levels, what happens when they change homerooms or advance to the next grade? If they have to be assigned a new number, what impact does that have on the usefulness of the numbering system? This discussion should enhance the importance of the assigning an EPC code that is useful, practical, and requires the least amount of work and upkeep, which in turn will be cost effective.
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P3. Detecting Tags
An RFID reader usually can detect a tag that is 5 meters to 50 meters away from the reader. Suppose that for inventory purposes, a store wants to have enough readers in the right places throughout the store. Lowe’s is building a new store in State College. That store will cover 138,134 square feet. Home Depot is building a new 102,000-square foot store in the same town. Does one of these stores need to have more readers than the other store? If so, how many? Explain your thinking.Answer:4
A first glance at the relative square footage might suggest that the Lowe’s store would need to have more readers than the Home Depot store. However, the dimensions of the store matter and it does matter how the readers are located. There is nothing said about the shapes of the stores. It seems natural to think of the footprint of the stores as rectangles.Way : The students could design the buildings to be squares. Since the RFID readers have a range of 5 meters to 50 meters, it helps to convert the original stoer measurements in square feet to measurements in square meters. Using the fact that readers have a range of 5 meters to 50 meters, the total number of readers will range depending on the quality of the reader. Therefore, the number of readers will depend upon the type of reader used. Starting with Lowe’s:
4 Note: The answers do not attend to significant digits.
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First, convert the square feet to square meters for Lowe’s:
€
138,134 square feet ×0.3048 meters
1 foot×
0.3048 meters
1 foot≈12,833.1 square meters
Assuming the design will be a square with the total area approximately 12,833 square meters, the dimensions of the square can be obtained using several different methods such as guess and check or finding the square root of 12,833 to give the length of a side of the square building.
€
12,833.1 square meters ≈113 meters . Each side of the building will have a length of 113 meters. The reader can be thought of as the center of a circle with a radius of 5 meters to 50 meters, depending on the quality of the reader. The figure below gives an example displaying a part of a building layout with the readers located at the center of each of the circles.
Considering the range of the reader to be 5 meters, the diameter of each circle would be 10 meters. Therefore divide 113 meters by 10 meters to determine that 11.3 or 11 readers would be necessary across the width of the building (113 meters ÷ 10 meters ≈11.3). The same calculation indicates to cover the length of the building requires 11 readers. The total number of readers needed to cover the building is 121 (11 *11=121). However, as seen in the figure above, some of the area is not readable due to the gaps between the circles or to the “extra space” caused by using 11 readers when 11.3 readers were needed. To determine the actual area that the readers are not capable of covering, we can subtract the total area of circles from the area of the rectangle. First, calculate the total area of the rectangle that is Lowe’s footprint: 113m 113m =12,769 square meters. The area each circle covers isπ(5 meters)2 = 25π square meters≈78.54square meters. So, the total area of the circles is the number of readers times the area of coverage for one reader:
€
121 readers ×78.54 square meters
1 reader≈ 9,505 square meters The area of
the building that will not be covered by the RFID readers is 12,769 square meters – 9,505 square meters = 3,264 square meters. Next, consider the range of the reader to be 50 meters; the diameter of each circle would be 100 meters. Dividing 113 meters by 100 meters
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per reader means that approximately 1.3 or 1reader would be necessary across the width of the building. Similarly, 1 reader would be needed to cover the length of the building. So, only one reader is needed to cover the entire building. However, some of the area is not covered due to the gaps between the circles. The amount of area not covered would be the area of Lowe’s footprint less the area covered by one reader. The area of Lowe’s footprint is 12,769 square meters. The area of the circular region covered by the reader is π(50 meters)2 = 2500π square meters ≈ 7853.98 square meters or 7854 square meters. So the area not covered by a reader is approximately 12,769 square meters – 7854 square meters =4915 square meters.For Home Depot:First, the square feet must be converted into square meters.
€
102,000 square feet ×0.3048 meters
1 foot×
0.3048 meters
1 foot≈ 9476 square meters
So, the design will be a square with the total space approximately 9,476 square meters and side length
€
9476 ≈ 97.34 meters.Using tangent circles with diameter 10 meters as we did for Lowe’s case to represent the readers that read up to 5 meters would require 97.34 meters ÷ 10 meters/reader or approximately 10 readers to cover the length or the width of the floor space. It would take 1010=100 readers to cover the floor space.The area not covered would be the area caused by the gaps between the circles. This calculation is harder than the case for Lowe’s because part of the 10th circle along the length (or along the width) actually would go outside of the rectangle. In reality, the readers would probably be spaced across the store so that their circular reading areas overlap. This means the area not covered by the readers would be less than the area of the store minus the area covered by the readers. This area covered is the number of readers times the area of each circular region, 100π(5 meters)2 = 2500π square meters≈7854 square meters. Subtracting this area from the area of the floor, 9476 square meters – 7854 square meters=1622 square meters. So, the readers would not cover less than approximately1622 square meters. If the range of the reader were 50 meters, the diameter of the circles would be 100 meters. Only one reader would be needed to cover the floor. When we place a circle with radius 50 meters to have its center coincide with the intersection of the diagonals of a square with side length 97.34, we have the following image:
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A tiny bit of the circular region is outside the square region. More of the square region is outside the circular region. So, the area not covered would be less than the area of the square minus the area of the circle, which is the area of the floor space minus the area of coverage for one reader. The area not covered is less than approximately 9476 square meters – π(50)2 square meters ≈ 1622 square meters.Way : Assume that the stores only care about changes in inventory caused by items leaving the store (e.g., sales, theft). The store might place readers only at the exit of the buildings. One way to think about this would be to have a line of readers across the front of the store where the main doors are. Assuming that the store is a square, as we did in Way , the number of readers needed would be the number of readers that would provide coverage across one side of the square building.For Lowe’s:The length of one of the sides is approximately113 meters. Therefore, if the reader has a range of 5 meters and if the reader is placed at the center of the circles as seen in the figure in Way . To cover 113 meters with tangent circles of diameter 10 meters requires 113 meters ÷ 10 meters/reader or 11.3 readers. To ensure good coverage, use 12 readers and allow their circular coverage area to overlap a bit.Using the reader with a range of 50 meters, 113 meters ÷ 100 meters/reader = 11.3 readers. In this case, another big reader would be overkill. So, we could go with one reader and realize 13 meters would not be covered, or we might try to use one 50-meter reader and two 5-meter readers with slightly overlapping regions.For Home Depot:The length of one of the sides of the square floor is approximately 97.34 meters. If the reader has a range of 5 meters and if the reader is placed at the center of the circles as seen in the figure above in Way , we need 97.34 meters ÷ 10 meters/reader = 9.7 or 10 readers with their regions overlapping slightly.For a reader with a range of 50 meters, 1 reader would cover the 97.34-meter length. So, only one reader would be needed.Explanation of differences and similarities:
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The two ways described above differ in their use of area versus length. This distinction became important, particularly for the smaller store. While the circular region was smaller than the square region in Way , the diameter of the circle was larger than the length of a side in the length model Way . This cause the final answers to be different, with only one reader needed in Way but one reader being insufficient in Way .Enactment: If students choose this strategy, some students may consider that
the circles of readability in the case of the RFID readers with a range of 5 meters are tangent to one another at only one point. This may lead to concern for the reliability of the RFID readers under such extreme circumstances. If students fail to recognize this situation, a discussion might be helpful.
In the case of the Lowe’s store with the RFID readers’ capabilities of 50 meters, the solution above only calls for 1 reader. Students may recognize that only the RFID readers will not cover the four corners. They could address this situation by placing one of the cheaper readers in each of the four corners.
Students may be surprised to see that, in Way 1, the approximate area not covered by either type of reader is the same, 1622 square meters. Draw students’ attention to the fact that the area covered by one circle with a 50-meter radius is the same as the total area of 100 non-overlapping circles with 5-meter radii. Encourage students to reason through the calculations:For the 5-meter circle: Axmall=π(5)2 square meters.For the 50-meter circle: Abig=π(50)2 square meters.Working with the last equation, we have Abig=π(50)2=π(105)2= π(10)2(5)2=π100(5)2=100 π (5)=100 π(5) =100Asmall. Since the area of the big circle is 100 times the area of the small circle, the area covered by the 1 big circle is the same and the area covered by 100 small circles.
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Free-response items
F1. Expecting Cheaper RFID tags
On average, RFID tags cost approximately $0.40 each in 2004. Research companies believe that, approximately seven years later, the price for one RFID tag will be approximately $0.10.
a. Suppose the relationship between the cost of the tags and the number of years is a linear relationship. Write the linear equation for this relationship assuming that 2000 is year 0. Explain how you arrived at your answer.Answer:
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Way : Start with the two points we have. First guess is the year number times 10. 4x10 works but 11x10 does not. Try starting with 11 and 10. Year number minus 1 works for 11-1 but not for 4-1. Seeing no easy solution, it might help to figure out the numbers for another year. The year 2000, with year number 0, looks good. The cost decreased by 40-10=30 over 11-4=7 years. So, the cost decreased 30/7 cents each year. From year 0 to year 4, it would decrease 4 times 30/7 cents. Subtracting (30/7)(4) from 40 gives 57 as the cost in 2000. Now, year 11, we add so much per year times 11 to 57 to get 10. That “so much” is –4.3. Testing this idea for year 4, 4(-4.3)+57=49.8, which is very close to 40. So, y=-4.3x+57 is a fairly good equation of the line. Years since 2000
Cost of RFID tags in cents
First guess
Second guess
Third guess (with (0,57)
4 40 4x10 Not 4-1
Test: 4(-4.3)+57=39.8, which is very close to 40
11 10 Not 11x10
11-1 10=11(_)+57 _ is -4.3
0
€
40 +(40 −10)
(11− 4)(4 − 0) =
40 +30
7(4) ≈ 57
0(_)+57
Way : Using the two point form of a line to get the equation of a line passing through (4,40) and (11,10), the line is given by
€
y − 40 =10 − 40
11− 4(x − 4)
y − 40 =−30
7(x − 4)
7y − 280 = −30x +120
7y = −30x + 400
y =−30
7x +
400
7Way : Plotting the points and generating the line through them gives us the following result:
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Way : Using a TI-73, enter the data in L1 and L2 as shown in the first screen below. Using 2nd STAT, choose CALC and option 5 (see second screen below). The calculator results of this regression appear in the third screen below. The equation of a reasonable line would be y=-0.043x+0.571.
Explanation of differences and similarities: The calculator values for the coefficients in Way are the decimal equivalents of the coefficients found using the two-point form of the equation of the line in Way .The value for the slope found by using the graph in Way is close to the slope values in Ways and . However, the intercept in Way is a bit different from the intercept in Ways and because the value in Way was estimated from the graph while the values in Ways and were computed exactly.Multiplying the values from Way or Way by 100 yields approximately the values found by using the chart in Way . The factor of 100 makes sense because the values using the chart in Way are in cents while the values in Ways and are in dollars.Putting these observations together, we see that the methods produce the same line, generally speaking. The differences in the slope and
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intercept arise from one of three differences in the methods: using cents versus dollars, estimating a value (from a graph) versus computing the value, or expressing non-integer rational numbers as fractions versus decimals,Enactment:Students should be encouraged to note that their linear results are the same if they assume there are exactly two valid data points.
b. Does a linear relationship between year and cost of the tags make sense to you? Explain.Answer: Acceptable answers would have roots in reality and in mathematics. What follows is a sample answer. If we think about the reality of the situation, a linear relationship does not make sense. The data we have suggests the line would not be a horizontal line, so the line we determine would have a non-zero slope and its graph would have to intersect the x-axis at some x value, say x0. The horizontal intercept means that in the year x0 years after 2000, the cost of the tags would be $0. After that year the cost would be negative! We can also reason from the slope. A line has a constant slope, but the cost of an RFID tag probably does not decrease by exactly the same amount each year.Based on the line we create, we might get a false sense of security. For example, with the calculator use in Way , press 2nd VARS to get the first screen below and then choosing EQ r as shown, shows the correlation coefficient is r=-1. This is a perfect result, or so it seems at first. However, since we entered only two distinct data points, and exactly one line passes through two distinct points, the answer had to be exact! [The same is true for Way and probably is true for each of the other options. We see this if set up a scatter plot (see second screen below) and we enter the regression equation expression for y1 (see third screen below). The graph passes directly through the two points (see fourth screen below).
Enactment:It would be important it have students think about what the horizontal intercept for this linear relationship would mean. Students should talk about the likelihood of the price coming down over time in a somewhat consistent manner resulting in a linear relationship for that time span.
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F2. Cheaper Option
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A typical bar code costs less than 1 cent. A cheap RFID tag costs at least 35 cents. An expensive RFID tag may cost $250.
a. How many cheap RFID tags can a company buy for $100?Answer:Dividing $100 by the cost of the cheap RFID would give us the answer, if that quotient were an integer. However, 100÷0.35≈2857.1 is not an integer. This means the company could buy 2857 tags (and have a few cents left).
b. Which of the following statements is the best description of the relationship between the cost of the expensive RFID and the typical bar code? Explain your thinking.
An expensive RFID tag costs 250 times as much as a typical bar code.
An expensive RFID tag costs at most 250 times as much as a typical bar code.
An expensive RFID tag costs 2500 times as much as a typical bar code.
An expensive RFID tag costs 25000 times as much as a typical bar code.
An expensive RFID tag costs at least 25000 times as much as a typical bar code.
Answer:$250 is 25,000 times 1 cent. One of the expensive RFID tags would cost more than $250 or more than 25,000 cents, and one of the bar codes would cost less than 1 cent. The comparison would be greater than it first appears. So, an expensive RFID would cost at least 25,000 times as much as a typical bar code.Enactment:Students may try numerical cases to make sense of this situation. For example, an expensive RFID tag could cost $300 dollars. The bar code could cost 0.3 cents. Comparing $300 to 0.3 cents is like comparing 30,000 cents to 0.3 cents. 30,000÷0.3=100,000. [Or, reasoning proportionally, students may reason through the following $300:$0.003::30000:0.3::300000:3::100000:1.] So, 30,000 is more than 25,000 times 0.3, which means the expensive RFID tag costs at least 25,000 times as much as a typical bar code.
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Variety items These items are short-answer tasks that may be used alone or in
combination. For each item, the student should be encouraged to articulate the answer in the RFID context, and not to settle for a numerical or other mathematical result devoid of the context. In some cases, this means the students
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may have to interpret an answer (e.g., round to the nearest unit) to give the mathematical result meaning in the context.
V1. More Suppliers
It took more than 20 years for the use of bar codes to catch on. In 1962, 15,000 suppliers were using bar codes. By 1987, 75,000 suppliers were using bar codes. What is the percent increase in bar code usage by suppliers from 1962 to 1987?5
Answer:The number of suppliers increased by 75000-15000=60,000. The percent
increase would be
€
60,000
15,000×100% = 400% . There were four times as
many suppliers in 1987 as there were in 1962.Enactment:Students may be asked to think about whether 400% seems reasonable. Since 75,000 is clearly greater than several times 15,000, the percent increase should be greater than 100%.
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V2. Falling Prices6
Right now, typical RFID tags may cost $0.50 each. Analysts say that these tags will not be cost effective until the price reaches $0.05 each. If a company uses approximately 1,286,000 RFID tags per year, how much money would the company save if the cost of the tags drops from $0.50 to $0.05? Answer:The current cost of the tags would be 1,286,000 * $0.50 = $643,000. The cost of the tags at the lower price would be 1,286,000 * $0.05 = $64,300. The company would save $643,000 - $64,300 = $578,700.
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V3. Seaport Inspection
RFID tags can be used to keep track of containers entering a seaport. However, the seaport staff still needs to inspect the containers. A seaport operator states that 17,000 containers of items come into its port daily. They are only able to inspect 2% of these containers each day. How many containers does the seaport need to inspect each day?7
Answer:We need to calculate 2% of 17,000: 0.02 17,000 = 340. The seaport staff would need to inspect 340 containers per day. Enactment:Does this answer seem reasonable? If there is only one inspector and that inspector works an 8-hour day, the inspector would have to inspect
5 Data from (http://www.theregister.co.uk/2003/06/27/rfid_chips_are_here/).6 Data from (http://www.theregister.co.uk/2003/06/27/rfid_chips_are_here/).7 Information from (http://www.theregister.co.uk/2003/06/27/rfid_chips_are_here/).
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340÷8=42.5 containers per hour. This seems to be a considerable number of inspections for one person to perform in an hour! However, this answer makes sense if the inspector has help, either from people or machines. For example, the inspector may have helpers or the inspections may be done through the use of technology such as x-ray devices to scan the contents of the containers.
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V4. Tired of Percent
An RFID-using Michelin tire plant manufactures 800,000 tires each day. How many tires does it manufacture each hour?8
Answer:Assuming the plant runs 24 hours per day, we have
€
800,000 tires
1 day ×
1 day
24 hours = 33,333.3 tires per hour .
Enactment:It may be tempting to force students to round this result to a whole number. That would be the case if the inspection line stopped at the end of each hour. But, the unit of measure for 33,333.3… is not “tires” but “tires per hour.” This is not a count of completed tires but a measure of how quickly the process moves. Since the process may run in a rather continuous manner, the “.333…” as a “fraction of a tire” may not be a problem. Students could think of this way: At the end of the hour, 33,333 inspections and one-third of another inspection are done and the worker(s) doing the inspections will not quit simply because an hour passed.
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V5. Reading Distance
An RFID chip can be read anywhere from a couple of inches to as far as 30 feet from a particular reader. Could a package containing an RFID chip be read if the package were 372 inches from the reader? Why or why not?9
Answer:We need to convert 372 inches to feet or 30 feet to inches.
Converting inches to feet, we have
€
372 inches × 1 foot
12 inches= 31 feet . The
reader would not be able to read the tag because the package would be more than 30 feet from the reader.If we convert feet to inches, the reader can read tags that are less than
€
30 feet × 12 inches
1 foot= 360 inches from the reader. The tag that is 371
inches from the reader would be too far away. The reader could not read it.
8 Details appeared at (http://www.theregister.co.uk/2003/06/27/rfid_chips_are_here/).9 Information from (http://www.spychips.com/what-is-rfid.html).
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There are 12 digits in a UPC. Each digit is separated by 7 same-size slices. How many slices are there in one UPC code?10
Answer:Between 12 digits there are 11 spaces; so there are 11 sets of 7 slices each in one UPC code. Each space contains 7 slices. The total number of slices in a UPC code would be 711=77 slices.Enactment:Students may want to use 12 rather than 11 and answer 7x12=84 slices. Teacher may want to help students to reason through why there are only 11 spaces. This use of 11 rather than 12 is also seen when we use a rectangle to model a fraction. For example, to model 7/12 requires using 11 segments (like spaces) to create 12 parts (like digits).
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V7. Check (the) Digit
10 Details found at (www.beakman.com/beakman/news/072102/072102.html).
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The last digit of the 12 digits in a UPC code is the check digit. You can always determine the check digit using a few calculations. Find the sum of the digits in the odd positions, the 1st, 3rd, 5th, 7th, 9th, and 11th digits. Multiply the sum by three. Call this value x. Next, sum the even-position digits (but not the 12th one!). Add this value to x. Call this new value n. The check digit is the value you need to add to your answer, n, to get to the next multiple of 10. Calculate the check digit for the following UPC code found on a box of Chocolate Honey Grahams (shown below with a question mark in place of the check digit): 6 88267 01800 [The check digit is covered.]11
Answer:The odd-position digits are 6, 8, 6, 0, 8 and 0. The sum of these digits is 6+8+6+0+8+0=28. Three times this value is x=328=84. The even-position digits are 8, 2, 7, 1, and 0. Their sum is 8+2+7+1+0=18. n=84+18=102. The next multiple of 10 that is greater than 102 is 110. The check digit is 110-102=8.
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V8. UPC Song
A laser can read UPC values much faster than we can. If a laser can read 41,000 numbers per second from a compact disk, how many numbers will it read during a song that lasts 3 minutes and 34 seconds?12
Answer:We can work with seconds or with minutes. Working with seconds, we
have
€
41,000 numbers
1 second× 214 seconds = 8,774,000 numbers. If we use
minutes, the calculation is longer but the result is the same:
€
41,000 numbers
1 second×
60 seconds
1 minute× 3+
34
60
⎛
⎝ ⎜
⎞
⎠ ⎟ minutes = 8,774,000 numbers
Enactment:This may seem like a lot of numbers to read, but we need to keep in mind how fast a laser can read information, particularly information on a CD.
11 This computational information comes from (http://electronics.howstuffworks.com/upc.htm).12 The facts come from (www.beakman.com/beakman/news/072102/072102.html).
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There are 12 digits in a UPC code. If a laser can read 41,000 UPC digits per second, how many UPC codes can the laser read in one second?13
Answer:We start with the quotient
€
41,000 ÷12 = 3416.3. This means that the laser reads slightly more than 3416 UPC codes in one second.
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V10. RFID Chip Width
On average, an RFID chip can be as narrow as 1/3 millimeter. Express this measurement in inches.14 Answer:To convert cm into an approximate number of inches, we divide by 2.54. 1/3 mm=(1/3÷10) cm=1/30 cm. The width of the RFID chip is approximately 1/30 ÷2.54 ≈0.013 inches. Enactment:Since inch is a larger unit than centimeter and centimeter is a larger unit than millimeter, the number of inches should be smaller than the initial number of millimeters.
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V11. Average Military Shipment
The U.S. Military has already put RFID chips on 270,000 cargo containers. These containers are sent to 40 different countries.15 On average, how many containers will be sent to each of these countries?Answer:The average number of containers sent to each of the countries is 270,000÷40=6750 containers. Enactment;This is a great opportunity to help students enrich their conceptual understanding of average. Ask students, “Do you think that every country would receive exactly 40 containers? Why or why not?” Students may answer these questions based on their knowledge of the world, connecting perhaps with what they learn in social studies. In this shipping case, they may note there is nothing in the problem statement that suggests each country receives the same number of containers. Students may also see differences between countries that may warrant different size shipments to different countries such as: the size of the country’s ports, the countries’ populations or numbers of consumers to buy the products, and the wealth or economies of the countries. Students likely will agree that each country
13 The information comes from (www.beakman.com/beakman/news/072102/072102.html).14 Fact appeared at (http://www.theregister.co.uk/2003/06/27/rfid_chips_are_here/).15 Information from (http://www.theregister.co.uk/2003/06/27/rfid_chips_are_here/).
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should not receive 40 containers. This observation and discussion should help students to understand that the average is an overall view of a set of numbers and the value of the average need not be a data point.
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V12. Three Seaports Rule
Three different seaports in the U.S. handle 70% of the RFID-marked containers shipped through the country. These U.S. seaports use RFID tags to track 17,000 containers each day. How many containers do the other seaports in the U.S. track with RFID tags?16
Answer:17,000 is 70% of the total. The total is 17,000÷0.70≈24,286 containers. The other seaports are responsible for approximately 24,286-17,000=7286 containers.Alternatively, we could work with the percentages. Three seaports account for 70% of the containers. The other seaports account for (100-
70)%=30%. We now have the proportion
€
70
17,000=
30
n. The other
seaports are responsible for approximately
€
n =30 ×17,000
70≈ 7286
containers. Enactment:To extend the context to make sense of the situation or to connect with social studies, students might discuss possible reasons as to why some ports might handle a larger volume of containers than other ports. Some possible conclusions may be the size of the port, the location of the port, and if that port is responsible for shipping cargo to another location other than receiving containers only.
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V13. In 6 to 8 Years
On average, RFID tags cost approximately $0.40 each in 2004. Research companies believe that, in 6 to 8 years, the price for one RFID tag will be approximately $0.10. What is the percent decrease in cost over the 6 to 8 years? Answer:
The percent decrease is
€
0.40 − 0.10
0.40×100% = 75%.
Enactment:With the development of better versions of new technology, cheaper ways to make technology, and more people or companies using the technology, one would expect the cost of manufacturing items to decrease.
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16 Facts appeared at (http://www.theregister.co.uk/2003/06/27/rfid_chips_are_here/).
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On average, an RFID chip can be as narrow as 1/3 millimeter. How many of these RFID chips would it take to make a 1-meter string of RFID chips lying side by side?Answer:1/3 mm is 1/30 cm. 1/30 cm is 1/300 m. We can use a proportion to determine how many RFID chips are needed to stretch for 1 meter:
€
1
300meters
1 chip=
1 meter
? chips. If each chip were 1/300 meter wide, it would take
€
1 chip × 1 meter1
300 meter
= 300 chips to stretch 1 meter.
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V15. Bits of Information
One RFID can contain up to 4k bits of information. A bar code contains at most 2K characters. Which type of technology, RFID or bar code, is capable of containing more information? {Note: Abbreviations used include 1k for 1000 and 1K for 1024. One character requires 8 bits.}Answer:The RFID contains as many as 4k bits, or 4 x 1000 = 4000 bits. The bar code contains 2K characters. Each character requires 8 bits, and so the bar code would require 2 x 1024 ÷ 8 = 256 bits. The RFID clearly is capable of containing more information.
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V16. Early RFID Users
Some of the first companies using RFID tags reduced their costs of supplies by 3% to 5%. Suppose the costs of supplies for one of these companies after it started using RFID tags was $678,926. What were the company’s costs of supplies before it started using RFID?Answer:The original cost is the percent decrease divided by 100% times the original cost. If the percent decrease was 3%, $678,926 would be (100-3)%=97% of the pre-RFID costs. So, if the percent decrease is 3%, the pre-RFID costs would be 678,926÷(100-3)%≈$699,923.71. If the percent decrease was 5%, the pre-RFID costs would be 678,926÷(100-5)%≈$714,658.95. The company’s pre-RFID costs would be between approximately $699,923.71 and $714,658.95. Enactment:It seems that the companies were able to save more than $20,000 by switching to RFID tags. Because the percent is given, this item may be a novel or challenging problem for students. However, thinking like this is
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needed to interpret some reports or news stories, such as the source from which these numbers came.
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V17. Food Tray Tags
Mark & Spencer puts RFID tags on its 3.5 million food trays. Suppose the RFID tag for each tray costs $0.05. What would be the cost of placing an RFID tag on each of the trays?Answer:3.5 million x $0.05 yields the cost of $175,000. Enactment:Students may think $175,000 seems like a large sum of money to spend on RFID tags. If so, it may help to talk about how the company might benefit from this choice. They should see an increase in inventory accuracy which will benefit the companies that purchase Mark & Spencer food trays.
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V18. Pallet Tags
In its initial phase of using RFID, Wal-Mart put RFID tags on pallets (collections of items) rather than on each item. To use an RFID system, suppose Wal-Mart needs a new warehouse. The warehouse building costs approximately 2 million dollars. The cost of a tag is five cents. The equipment costs $2500. Which of the following expressions represents the total cost of the RFID project if there are n pallets in the warehouse? Explain.
(a) 2+2500+5n
(b) 0.05n+25+5n
(c) 0.05n+2500+2000
(d) 0.05n+2500000+2000000
(e) 0.05n+2500+2000000Answer:The answer would be (e). The variable n represents the number of pallets. The coefficient for n must be the cost per pallet, five cents. The cost of the equipment and the cost of the warehouse are one-time costs; these costs are not incurred once for each pallet. So, the cost of the equipment ($2500) and the cost of the warehouse ($2,000,000) would be constants or fixed costs. The cost of the project would be the sum of the fixed costs and the costs based on the number of pallets. This sum expressed in dollars is 2500+2,000,000+0.05n, which is equivalent to 0.05n+2500+2000000.
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V19. More Pallet Tags
A different business is beginning to use an RFID system for pallets. The business buys equipment for $3,200. The business pays $0.20 for each tag. Which of the following expressions represents the cost of this project if the business has n pallets? Explain.
(a) 20n +32
(b) 0.20n + 32
(c) 20n + 3200
(d) 0.20n + 3200
(e) 0.20n + 3200000Answer: The business pays $3,200 only once. It is the constant term. The cost per tag, $0.20, would be the coefficient of n. The expression would be something like 0.20n+3200. This is expression (d).
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V20. Tags for Another Warehouse
Suppose Wal-Mart has an old warehouse in which the company wants to put an RFID system. The cost of a tag is 40 cents. The equipment costs $2500. Which of the following expressions represents the total cost of the RFID project if there are n pallets in the warehouse? Explain.
(a) 40n + 2500
(b) 40n + 25
(c) 0.40n + 25
(d) 0.40n + 250000
(e) 0.40n +2500Answer: $2500 is the fixed costs, paid only once. The 40-cent price per tag expressed in dollars to match the fixed costs is $0.40. So, the expression 0.40n+2500 (choice e) would be correct.
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Particular materials
Mathematics tasksNone for this module
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Mathematics handoutsNone for this module
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Technology guidesQuick TricksPlotting Data Points and Fitting Curves
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Visual componentsRFID tag removed from shirt purchased at the GAP
Four parts of an EPC number
Four possible descriptions
An expensive RFID tag costs 250 times as much as a typical bar code.
An expensive RFID tag costs at most 250 times as much as a typical bar code.
An expensive RFID tag costs 2500 times as much as a typical bar code.
An expensive RFID tag costs 25000 times as much as a typical bar code.
An expensive RFID tag costs at least 25000 times as much as a typical bar code.
UPC sample
Pallet Tags Expressions
(a) 2+2500+5n
(f) 0.05n+25+5n
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Web in-sites http://www.theregister.co.uk/2003/06/27/rfid_chips_are_here/ (Note:
This website is very informative and has additional informative links on RFIDs and bar codes.)
http://www.packagingdigest.com/articles/200407/36.php (Note: this website has particular information on RFIDs and Wal-Mart.
http://www.idtechex.com/products/en/article.asp?articleid=40 (Note: This website would be great for a student that is interested stereo systems as it includes detailed information on the frequencies involved with RFIDs.
http://electronics.howstuffworks.com/upc.htm (Note: This website addresses bar codes.)
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Materials and equipment neededStudents would need the following materials and equipment to work on
the items in this module:Item Materials and equipmentAll Paper and pencil for notes and calculations; Calculators for computations
Graphing calculator with ability to plot data points (Plot Data Points Lab Guide)
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MotivationThe idea to look at RFID was motivated by the potential of RFID tags to
replace UPC codes. UPC codes seem to be common. We hope students will think about how UPC codes work while comparing them to RFID and understanding why change to RFID may be slow and costly.
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Credits/disclaimerThis module was developed under the GE Math Excellence: Math in a
“New Technology” Context project, funded by the GE Foundation. Ramandeep Chowdhary provided technology context details about RFIDs. Rose May Zbiek, Shari Reed and Tracy Boone generated the mathematical tasks, solutions, and enactment ideas. The materials in this module are the authors, and they do not necessarily reflect the interests or ideas of the GE Foundation.
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