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IEOR3608Fall06-lingo-tutorial

Course: IE 3608, Fall 2002
School: Columbia
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8.0 LINGO TUTORIAL Created by: Kris Thornburg Anne Hummel Table of Contents Introduction to LINGO 8.0..2 Creating a LINGO Model3 Solving a LINGO Model.4 Using Sets in LINGO..6 The LINGO Data Section8 Variable Types in LINGO.10 Navigating the LINGO Interface...11 LINGO Operators and Functions...14 Common LINGO Error Messages.16 LINGO Programming Examples...17 Introduction to LINGO 8.0 LINGO is a software tool...

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8.0 LINGO TUTORIAL Created by: Kris Thornburg Anne Hummel Table of Contents Introduction to LINGO 8.0..2 Creating a LINGO Model3 Solving a LINGO Model.4 Using Sets in LINGO..6 The LINGO Data Section8 Variable Types in LINGO.10 Navigating the LINGO Interface...11 LINGO Operators and Functions...14 Common LINGO Error Messages.16 LINGO Programming Examples...17 Introduction to LINGO 8.0 LINGO is a software tool designed to efficiently build and solve linear, nonlinear, and integer optimization models. LINGO 8.0 includes several new features, including: A new global solver to confirm that the solution found is the global optimum, Multistart capability to solve problems more quickly, Quadratic recognition and solver to identify quadratic programming (QP) problems, A faster and more robust Dual Simplex solver, An improved integer solver to enhance performance in solving many types of problems, Linearization capability to transform common nonsmooth functions to a series of linear functions, Infeasible and unbounded analytical tools to help identify model definition problems, A decomposition feature to identify if a model contains independent submodels, A threadsafe DLL for various classes of models, and More fun than ever before! 1 Creating a LINGO Model An optimization model consists of three parts: Objective function This is single formula that describes exactly what the model should optimize. A general manufacturing example of an objective function would be to minimize the cycle time for a given product. Variables These are the quantities that can be changed to produce the optimal value of the objective function. For example, when driving a car, the duration of the trip (t) and the speed at which it is taken (v) determine the distance (d) that can be traveled. Constraints These are formulas that define the limits on the values of the variables. If an ice cream store is determining how many flavors it should offer, only a positive number of flavors is feasible. This constraint could be expressed as Flavors >= 0; A sample model for cookie production by two bakers at a bakery is given by: ! A cookie store can produce drop cookies and decorated cookies, which sell for $1 and $1.50 apiece, respectively. The two bakers each work 8 hours per day and can produce up to 400 drop cookies and 200 decorated cookies. It takes 1 minute to produce each drop cookie and 3 minutes to produce each decorated cookie. What combination of cookies produced will maximize the baker's profit? ; MAX = 1*Drop + 1.5*Deco; Drop <= 400; Deco <= 200; 1/60*Drop + 3/60*Deco <=16; Several other items must be noted about this model: Comments in the model are initiated with an exclamation point (!) and appear in green text. LINGO specified operators and functions appear in blue text. All other text is shown in black. Each LINGO statement must end in a semi-colon (;). Variable names are not case-sensitive and must begin with a letter (A-Z). Other characters in the variable name may be letters, numbers (0-9), or the underscore character (_). Variable names can be up to 32 characters in length. 2 Solving a LINGO Model Once the LINGO model has been entered into the LINGO Model window, the model can be solved by clicking the Solve button on the toolbar, by selecting LINGO | Solve from the menus, or by using the ctrl + s keyboard shortcut. LINGO will notify you of any errors it has encountered. The best way to get information about these errors is to consult the Error Messages section in the softwares proprietary tutorial. If no errors are found, then the LINGO Solver Status window appears. This window provides information on the number of nonlinear, integer, and total variables in the model; the nonlinear and total number of constraints used in the model; and the number of nonlinear and total nonzero variable coefficients used. The Solver Status box in this window details the model classification (LP, QP, ILP, IQP, NLP, etc.), state of the current solution (local or global optimum, feasible or infeasible, etc.), the value of the objective function, the infeasibility of the model (amount constraints are violated by), and the number of iterations required to solve the model. The Extended Solver Status box details similar information for the more advanced branch-and-bound, global, and multistart solvers. By closing this window, you can then view the Solution Report window. 3 This window shows the values of each variable that will produce the optimal value of the objective function. The reduced cost for any variable that is included in the optimal solution is always zero. For variables not included in the optimal solution, the reduced cost shows how much the value of the objective function would decrease (for a MAX problem) or increase (for a MIN problem) if one unit of that variable were to be included in the solution. For example, if the reduced cost of a certain variable was 5, then the optimal value of the MAX problem would decrease by 5 units if 1 unit of the variable were to be added. The Slack or Surplus column in the Solution Report shows how tight the constraint is. If a constraint is completely satisfied as an equality, then slack/surplus is zero. If slack/surplus is positive, then this tells you how many more units of the variable could be added to the optimal solution before the constraint becomes an equality. If slack/surplus is negative, then the constraint has been violated. The Dual Price column describes the amount to which the value of the objective function would improve if the constraining value is increased by one unit. 4 Using Sets in LINGO LINGO allows you to group many instances of the same variable into sets. For example, if a model involved 27 delivery trucks, then these 27 trucks could be described more simply as a single set. Sets may also include attributes for each member, such as the hauling capacity for each delivery truck. Sets may be either primitive or derived. A primitive set is one that contains distinct members. A derived set, however, contains other sets as its members. To use sets in a model, the special section called the SETS section must be defined before any of the set members are used in the models constraints. This section begins with the tag SETS: and ends with the tag ENDSETS. A primitive set could be defined as follows: SETS: Trucks/TR1..TR27/:Capacity; ENDSETS This set is given the setname Trucks and contains 27 members, identified by TR1 TR27. The attributes for each member are called Capacity. The derived set is defined similarly, but must also include the parent set list. An example of a derived set could be: SETS: Product/X Y/; Machine/L M/; Make(Product Machine)/X L, X M, Y M/; ENDSETS This set declaration defines two primitive sets, Product and Machine, and one derived set called Make. The Make set is derived from the parent sets Product and Machine. Members are specified as shown. Notice that a fourth Product-Machine combination, Y L, could be theoretically possible. This example does not allow for such a combination. If all combinations of the parent sets are possible, then no member set need be defined. An attribute list for the derived set can also be included in the same way as for a primitive set. 5 Several set looping functions are also available for use in LINGO. These functions are as follows: @FOR generates constraints over members of a set. @SUM sums an expression over all members of the set. @MIN computes the minimum of an expression over all members of the set. @MAX computes the maximum of an expression over all members of the set. Each of the above looping functions has a similar form of syntax and the looping functions can even be nested. Examples of expressions using each type of looping function are as follows: This @FOR statement sets the hauling capacity for all 27 delivery trucks in the Trucks set to at most 3000 pounds: @FOR(Trucks(T): Capacity(T)<=3000); This @SUM statement calculates the total hauling capacity from the individual trucks: TOTAL_HAUL=@SUM(Trucks(J): Capacity(J)); These @MIN and @MAX statements find the extreme hauling capacity levels from the individual delivery trucks: MIN_HAUL = @MIN(Trucks(J): Capacity(J)); MAX_HAUL = @MAX(Trucks(J): Capacity(J)); 6 The LINGO Data Section LINGO provides a separate section called the DATA section in which values can be defined for different variables. Set members can be initialized in this section, attributes of the sets can be defined, or scalar variable parameters can be assigned values as well. The DATA section is defined after the SETS section is defined in the model. The section begins with the tag DATA: and ends with the tag ENDDATA. Statements within the DATA section follow the syntax: object_list = value_list; The object list contains the names of the attributes or of the set whose values are being initialized. The value list assigns the values to the specified members of the object list. The following examples show two ways to use the DATA section in LINGO. In each example, the X and Y attributes of SET1 are being initialized to [1, 2, 3] and [4, 5, 6], respectively. The first example defines values for each attribute separately: SETS: SET1 /A, B, C/: X, Y; ENDSETS DATA: X = 1, 2, 3; Y = 4, 5, 6; ENDDATA The next example shows how one statement can be used to assign values to the two attributes simultaneously. Each row assigns different values to the X, Y pair: SETS: SET1 /A, B, C/: X, Y; ENDSETS DATA: X, Y = 1, 4, 2, 5, 3, 6; ENDDATA 7 When parameters or attributes are defined in the DATA section of a model, a feature called Whatif Analysis can be used to examine the effects of varying the value of the parameter or attribute. For example, if the inflation rate is most likely going to fall between 2% and 6%, the parameter can be defined as follows in the DATA section: DATA: INFLATION_RATE = ?; ENDDATA When LINGO encounters the ? in the DATA section, it will prompt the user to enter a value for the parameter. The user can then enter values between 2% and 6%, and LINGO will solve the model using that what-if value. All the elements of an attribute can be initialized to a single value using the DATA section as well. following The example shows how to assign the value of 20 to all seven members of the NEEDS attribute and 100 to all seven members of the COST attribute: SETS: DAYS / MO, TU, WE, TH, FR, SA, SU/: NEEDS, COST; ENDSETS DATA: NEEDS, COST = 20, 100; ENDDATA Data values can also be omitted from the DATA section of a LINGO model to indicate that LINGO is free to determine the values of those attributes itself. The following example shows that the first two values of the attribute CAPACITY have been initialized to 34, but the last three variables have not been defined: SETS: YEARS /1..5/: CAPACITY; ENDSETS DATA: CAPACITY = 34, 34, , , ; ENDDATA 8 Variable Types in LINGO All variables in a LINGO model are considered to be non-negative and continuous unless otherwise specified. LINGOs four variable domain functions can be used to override the default domain for given variables. These variable domain functions are: @GIN any positive integer value @BIN a binary value (ie, 0 or 1) @FREE any positive or negative real value @BND any value within the specified bounds Similar syntax is used for the @GIN, @BIN, and @FREE variable domain functions. The general form for the declaration of a variable x using any of these functions is @FUNCTION(X); The @BND function has a slightly modified syntax, which includes the upper and lower bounds for the acceptable variable values. The general form for the declaration of a variable x between a lower bound and an upper bound is given by @BND(lowerBound, X, upperBound); 9 Navigating the LINGO Interface Operations in LINGO can be carried out using commands from the menus, toolbars, or shortcut keys. There are five menus in the main LINGO window. They are the File, Edit, LINGO, Window, and Help menus. The following list details the commands in the File menu. Shortcut keys are included in parentheses when available: New (F2) Open (F3) Save (F4) Save As (F5) Close (F6) Print (F7) Print Setup (F8) Print Preview (Shift+F8) Log Output (F9) Take Commands (F11) Import LINDO File (F12) Export File License Database User Info Exit (F10) Open a new model window Open a saved file Save a current model Save a current model to a new filename Close the current model Prints the current windows content Configures printer preferences Displays the window content as it would look if printed Opens a log file to log output to the command window Runs a command script contained in a file Converts a LINDO file into a LINGO model Exports a model in MPS or MPI file format Prompts user for new license password to upgrade system Prompts for a user id and password for database access via the @ODBC() function Closes LINGO The Edit menu contains the following commands: Undo (ctrl+z) Redo (ctrl+y) Cut (ctrl+x) Copy (ctrl+c) Paste (ctrl+v) Paste Special Undoes the last action Redoes the last undo command Copies and deletes highlighted text Copies highlighted text to the clipboard Pastes the clipboards contents into the document Pastes the clipboards content into the document, in a userspecified manner Select All (ctrl+a) Selects all of the contents of the current window Find (ctrl+f) Searches the document for a specific text string Find Next (ctrl+n) Searches the document for the next occurrence of a specific text string Replace (ctrl+h) Replaces a specific text string with a new string Go To Line (ctrl+t) Moves the cursor to a certain line number Match Parenthesis (ctrl+p) Finds the selected parenthesiss mate 10 Paste Function Select Font (ctrl+j) Insert New Object Links Object Properties Pastes a syntax template for a specific LINGO @function Configures the font for a selected portion of text Puts an OLE object in the document Creates links to external objects Defines the properties of an embedded object The LINGO menu contains the following commands: Solve (ctrl+s) Solution (ctrl+o) Range (ctrl+r) Options (ctrl+i) Generate Picture (ctrl+k) Debug (ctrl+d) Model Statistics (ctrl+e) Look (ctrl+l) Solves the model Makes a solution report window for the current model Creates a range analysis report for the current window Sets system options Generates the algebraic form of the model Displays a graphic of the models matrix Identifies errors in infeasible and unbounded models Reports the technical details the model Creates a formulation report for the current window The Window menu contains the following commands: Command Window (ctrl+1) Status Window (ctrl+2) Send to Back (ctrl+b) Close All (ctrl+3) Tile (ctrl+4) Cascade (ctrl+5) Arrange Icons (ctrl+6) Opens a window for command-line operation of LINGO Opens the solver's status window Places the current window behind all other open windows Closes all open windows Places open windows in a tiled arrangement Places all open windows in a cascaded arrangement Aligns icon windows at the bottom of the main LINGO window The Help window contains the following commands: Help Topics Register AutoUpdate About LINGO Opens LINGOs manual Registers your version of LINGO online Provides prompts to download software updates Displays software information 11 A single toolbar located at the top of the main LINGO window contains many of the same commands as listed above. These commands can be accessed simply by using the mouse to click on the icon representing them. The following pictures detail which icons correspond to which commands. New Save Open Find Cut Print Paste Copy Redo Undo Match Close Parenthesis Solution Options All Go To Line Solve Matrix Picture Help Topics Send to Tile Back Help 12 LINGO Operators and Functions LINGO provides a vast array of operators and functions, making it a useful problem-solving tool. A selection of the primary operators and functions is given below. There are three types of operators that LINGO uses: arithmetic, logical, and relational operators. The arithmetic operators are as follows: Exponentiation: ^ Multiplication: * Division: / Addition: + Subtraction: - The logical operators are used in set looping functions to define true/false conditions: #LT#: TRUE if the left argument is strictly less than the right argument, else FALSE #LE#: TRUE if the left argument is less-than-or-equal-to the right argument, else FALSE #GT#: TRUE if the left argument is strictly greater than the right argument, else FALSE #GE#: TRUE if the left argument is greater-than-or-equal-to the right argument, else FALSE #EQ#: TRUE if both arguments are equal, else FALSE #NE#: TRUE if both arguments are not equal, else FALSE #AND#: TRUE only if both arguments are TRUE, else FALSE #OR#: FALSE only if both arguments are FALSE, else TRUE #NOT#: TRUE if the argument immediately to the right is FALSE, else FALSE The relational operators are used when defining the constraints for a model. They are as follows: The expression is equal: = The left side of the expression is less than or equal to the right side: <= The left side of the expression is greater than or equal to the right side: >= 13 The following list contains a sampling of mathematical functions that can be used in LINGO: @ABS(X) returns the absolute value of X @SIGN(X) returns -1 if X is negative and +1 if X is positive @EXP(X) calculates eX @LOG(X) calculates the natural log of X @SIN(X) returns the sine of X, where X is the angle in radians @COS(X) returns the cosine of X @TAN(X) returns the tangent of X LINGO also contains a plethora of financial, probability, and import/export functions. These are commonly used in more advanced models, which are beyond the intended scope of this tutorial. 14 Common LINGO Error Messages LINGO provides a variety of error messages useful for debugging a developing model. The most common errors include the following: 7: Unable to open file: filename o Retype filename correctly 11: Invalid input: A syntax error has occurred o Check the line LINGO suggests for missing semi-colons, etc. 12: Unmatched parenthesis o Close the parenthesis set 15: No relational operator found o Make sure all constraints contain =, <=, >= 44: Unterminated condition o Put a colon at the end of each conditional statement in a set operator 50: Improper use of the @FOR() function o @FOR() functions cannot be nested inside other set operators 68: Multiple objective functions in model o Only one is allowed, please 71: Improper use of a variable domain function (eg, @GIN, @BIN, @FREE, @BND) o Check the syntax 81: No feasible solution found o Check models consistency and constraints 82: Unbounded solution o Add constraints 102: Unrecognized variable name: variable name o Check spelling 108: The models dimensions exceed the capacity of this version o Upgrade to full version or use Excel 164: Invalid LINGO name o Create a name to conform to LINGOs naming conventions 15 LINGO Programming Examples A common programming model is the Knapsack problem, which deals with maximizing the utility of loading capacity. This example shows how to properly set up a knapsack problem. SETS: ITEMS / ANT_REPEL, BEER, BLANKET, BRATWURST, BROWNIES, FRISBEE, SALAD, WATERMELON/: INCLUDE, WEIGHT, RATING; ENDSETS DATA: WEIGHT RATING = 1 2 3 9 4 3 3 8 3 10 1 6 5 4 10 10; KNAPSACK_CAPACITY = 15; ENDDATA MAX = @SUM( ITEMS: RATING * INCLUDE); @SUM( ITEMS: WEIGHT * INCLUDE) <= KNAPSACK_CAPACITY; @FOR( ITEMS: @BIN( INCLUDE)); 16 Another common programming model is the transportation problem. Transportation problems deal with transporting goods from one location to another at minimal cost. This example shows how to model a simple transportation problem. MODEL: ! A 6 Warehouse 8 Vendor Transportation Problem; SETS: WAREHOUSES / WH1 WH2 WH3 WH4 WH5 WH6/: CAPACITY; VENDORS / V1 V2 V3 V4 V5 V6 V7 V8/ : DEMAND; LINKS( WAREHOUSES, VENDORS): COST, VOLUME; ENDSETS ! The objective; MIN = @SUM( LINKS( I, J): COST( I, J) * VOLUME( I, J)); ! The demand constraints; @FOR( VENDORS( J): @SUM( WAREHOUSES( I): VOLUME( I, J)) = DEMAND( J)); ! The capacity constraints; @FOR( WAREHOUSES( I): @SUM( VENDORS( J): VOLUME( I, J)) <= CAPACITY( I)); ! Here is the data; DATA: CAPACITY = 60 55 51 43 41 52; DEMAND = 35 37 22 32 41 32 43 38; COST = 6 2 6 7 4 2 5 9 49538582 52197433 76739271 23957265 5 5 2 2 8 1 4 3; ENDDATA END 17
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Q1- Mega-Mart, a discount store chain, is to build a new store in Rock Springs. The parcel of land thecompany has purchased is large enough to accommodate a store with 140,000 square feet of floor space.Based on marketing and demographic surveys of the
UNSW - ACCT - 2542
TUTORIAL 1 SOLUTIONSChapter 1, Discussion Question 1414. As part of its national economic development programme, the Australian government isworking vigorously to promote the growth of small high-tech companies. Becausethese companies tend to have lim
UNSW - ACCT - 2542
TUTORIAL 2 SOLUTIONSExercise 19.2 Current asset and liability classificationsCurrent AssetsCash and cash equivalentsTrade and other receivablesInventoriesOther current assetsCurrent LiabilitiesTrade and other payablesFinancial liabilitiesCurrent
UNSW - ACCT - 2542
TUTORIAL 3 SOLUTIONSProblem 3.6 Share issue, options, statement of changes in equityDAWSON LTDPart (a) General Journal EntriesDATEDETAILS20/09/09 Dividend declaredDividend payableDrCr7 20027/09/09 Dividend payableCash(120 000 x 6c = $7 200)D
UNSW - ACCT - 2542
TUTORIAL 4 SOLUTIONS27.2Potential ordinary shares are dilutive if the EPS is recalculated on the basis that thesecurities were converted to ordinary shares, and the recalculated diluted EPS figureis less than the basic EPS (which is based on the ordin
UNSW - ACCT - 2542
1SOLUTIONS TUTORIAL 7Chapter 22 Controlled entities: the consolidation methodDISCUSSION QUESTIONS21.RequiredDiscuss the potential for Gilder Ltd being classified as a subsidiary of Croc Ltd.Part ACroc LtdGlider Ltd40%- NCI = 60%- no other part
UNSW - ACCT - 2542
1TUTORIAL 8 SOLUTIONSChapter 24 Consolidation: intragroup transactionsDISCUSSION QUESTIONS1. Why is it necessary to make adjustments for intragroup transactions?The consolidated financial statements are the statements of the group, an economic entity
UNSW - ACCT - 2542
1TUTORIAL 9 SOLUTIONSChapter 25: Consolidation: Non-controlling interestDISCUSSION QUESTIONS1.If a step approach is used in the calculation of the NCI share of equity, what are the stepsinvolved?Step 1: share of subsidiary equity at acquisition dat
UNSW - ACCT - 2542
1TUTORIAL 10 SOLUTIONSChapter 26: Consolidation: indirect ownership interestsDISCUSSION QUESTIONS1.What is the difference between direct and indirect NCI?Direct NCI (DNCI) own shares directly in an entity. Indirect NCI are DNCIshareholders in an en
UNSW - ACCT - 2542
TUTORIAL 11 SOLUTIONSExercise 28.2 Accounting for an associate by an investorJASMINE LTD HAYLEY LTD40%Jasmine LtdHayley LtdAt 1 July 2009:Net fair value of identifiable assetsand contingent liabilities of Hayley LtdNet fair value acquiredCost of
UNSW - ACCT - 2542
TUTORIAL 12 SOLUTIONSDQ5. AASB 131 Interest in Joint Ventures uses joint control as a characteristic of a jointventure. Discuss what is meant by the term joint control?Paragraph 3 of AASB 131 contains the following definition of joint control:Joint co
Georgia Tech - ME - 3322
COE3001 ZamirPage 1 of 4Exam #1 Solutions1. (25 pts) You use the mechanical testing apparatus shown above to nd the Youngsmodulus for a section of bone. You have already measured the bone cross-sectional2area A = 0.8 cm and length L = 1.5 cm before
Georgia Tech - ME - 3322
COE3001ZamirExam #21. (25 pts) Draw the shear and moment diagrams for the beam. Make sure to calculateand label the numerical values for V and M at each end of the beam and at minima/maxima.Page 1 of 4Wednesday, October 21, 2009COE3001Zamir
Georgia Tech - ME - 3322
COE3001ZamirExam 31. (45 pts) The left ventricle of the heart can be modeled as a cylindrical pressure vessel. At the outerwall, the muscle bers are wrapped helically around the longitudinal axis, making an angle = 60with respect to the equator. Assu
Georgia Tech - ME - 3322
COE3001ZamirHW#6 Solutions1. (10 pts) If the wide-ange beam is subjected to a shear of V = 25 kip, nd themaximum shear stress and determine the shear force resisted by the web of thebeam.Last modied on Monday, October 19, 2009COE3001Zamir2. (10 p
Georgia Tech - ME - 3322
COE3001ZamirHW#5 Solutions1. A copper wire having a diameter d = 3 mm is bent into a circle and held with theends just touching. If the maximum permissible strain in the copper ismax = 0.0016, what is the shortest length L of wire that can be used?L
Georgia Tech - ME - 3322
COE3001ZamirHW#4 SolutionsDue 9/30/091. (10 pts) Draw and label the reaction forces and moments that are potentiallygenerated at the ends of each beam:a.BABFx , Fy , Fxb.BABFx , Fy , Fx , M A , M Bc.BABFy , Fy , Fx , M Bd.BBAAFx
Georgia Tech - ME - 3322
COE3001 ZamirHW#7Due Wed Nov 41. (10 pts) Determine and draw (all) the stresses acting on an element located aty = 1.3 cm above the neutral axis, given b = 3 cm, h = 4 cm , M = 15 N-m ,V = 600 N, and = 24 CCW.Wednesday, November 4, 2009COE3001 Zami
DeVry NY - NETW - 206
Week 3, Assignment #21Standardized Configurations for NetworkGrading RubricCategoriesPoints/GradingContentPoints/GradingContentPoints/GradingContentPoints/GradingContent20The network reportcontains TCO withextensive detail for thenetwork
Keller Graduate School of Management - MM - 522
Assignment # 06Part - 11. Define local area network?A local area network (LAN) is a group of microcomputers located in thesame general area. A LAN covers a clearly defined small area, such as one flooror work area, a single building, or a group of bu
Keller Graduate School of Management - MM - 522
Chapter TwoAPPLICATION LAYERThe application layer (also called layer 5) is the software that enables the user to perform usefulwork. The software at the application layer is the reason for having the network because it is thissoftware that provides th
Keller Graduate School of Management - MM - 522
Review Questions for Lecture-18 : Analog Transmission and Modulation1. How does analog data differ from digital data?Computers produce digital data that are binary, either on or off. In contrast, telephones produceanalog data whose electrical signals a
Keller Graduate School of Management - MM - 522
CSN200 Introduction to Telecommunications, Winter 2000Review Questions for Test-1 (with Answers)Review Questions for Test-1 (with Answers):Chapter 1: Introduction to Data CommunicationsOutline1.1 Network Basics1.2 Network Layer Model (most important
Keller Graduate School of Management - MM - 522
CSN200 Introduction to Telecommunications, Winter 2000Test-2, Part-AReview Questions for Test-2 (Partial List) Part-AEnd-of-Chapter-4 Questions (match questions, overlook number mismatch)1. How does analog data differ from digital data?Computers prod
Keller Graduate School of Management - MM - 522
3.5 PositioningProduct positioning is closely related to market segment focus. Product positioning involvescreating a unique, consistent, and recognized customer perception about a firm's offering. Aproduct or service may be positioned on the basis of