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Operations Research I Chapter 02 Modeling with Linear Programming. Dr. Ayham Jaaron First semester 2013/2014 August 2013 . Two-Variable LP Model. This section deals with formulation of a two-variable LP.
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Operations Research IChapter 02Modeling with Linear Programming Dr. AyhamJaaron First semester 2013/2014 August 2013
Two-Variable LP Model • This section deals with formulation of a two-variable LP. • We will learn and practice the formulation process before we can try to solve real-life problems.
LP Model Formulation: Two variable LP Model A Maximization Example (1 of 4) • Product mix problem - Beaver Creek Pottery Company • How many bowls and mugs should be produced to maximize profits given labor and materials constraints? • Product maximum daily resource requirements availability and unit profit is given by:
LP Model Formulation A Maximization Example (2 of 4)
LP Model Formulation A Maximization Example (3 of 4) Resource 40 hrs of labor per day Availability: 120 lbs of clay Decision x1 = number of bowls to produce per day Variables: x2 = number of mugs to produce per day Objective Maximize Z = $40x1 + $50x2 Function: Where Z = profit per day Resource 1x1 + 2x2 40 hours of labor Constraints: 4x1 + 3x2 120 pounds of clay Non-Negativity x1 0; x2 0 Constraints:
LP Model Formulation A Maximization Example (4 of 4) Complete Linear Programming Model: Maximize Z = $40x1 + $50x2 subject to: 1x1 + 2x2 40 4x 1+ 3x2 120 x1, x2 0 Can we solve it? What do you think? Shall we try?
Feasible Solutions A feasible solution does not violate any of the constraints: Example: x1 = 5 bowls x2 = 10 mugs Z = $40x1 + $50x2 = $700 Labor constraint check: 1(5) + 2(10) = 25 < 40 hours Clay constraint check: 4(5) + 3(10) = 50<120 pounds
Infeasible Solutions An infeasible solution violates at least one of the constraints: Example: x1 = 10 bowls x2 = 20 mugs Z = $40x1 + $50x2 = $1400 Labor constraint check: 1(10) + 2(20) = 50 > 40 hours
Example 2: The Reddy Mikks Company • Construction of the LP model (from book page 12) Reddy Mikks produces both interior and exterior paints from two raw materials, M1&M2. The following table provides the basic data of the problem.
Example 2 .. Cont’d (class task: 10 minutes) • A market survey indicates that the daily demand for interior paint cannot exceed that of exterior paint by more than 1 ton. Also, the maximum daily demand of interior paint is 2 ton. • Reddy Mikks wants to determine the optimum (best) product mix of interior and exterior paints that maximizes the total daily profit. CLASS TASK: TRY
Problem Formulation • Decision variables X1= Tons produced daily of exterior paint. X2= Tons produced daily of interior paint. • Objective Function Maximize Z= 5 X1 + 4 X2 • Constraints 6 X1+4 X2 24 X1+2 X2 6 - X1+ X2 1 X2 2 X1, X2 0
TRY AT CALSSExample 3: Calculators Company (1 of 4) • A calculator company produces scientific calculators and graphing calculators. There is a demand forecast of at least 100 scientific calculators and 80 graphing calculators per day. • Manufacturing capacity says that a maximum of 200 scientific calculators and 170 graphing calculators can be produced each day. There is a contract that at least 200 calculators would be produced each day. • If each scientific calculator results in the loss of $2 and each graphing calculator results in a profit of $5, how many of each should be produced each day for the maximum profit?
Problem Formulation (2 of 4) • Let x be the number of scientific calculators. • Let Y be the number of graphing calculators. • Each scientific calculator yields a loss of $2 = -2 • Each graphing calculator yields a profit of $5 = 5 • we have to maximize the profit : Maximize : z= -2x + 5y
Problem Formulation..Con’t (3 of 4) • now, let us check the constraints : • minimum of 100 scientific calculators and 80 graphing calculators • maximum of 200 scientific calculators and 170 graphing calculators 100 ≤ x ≤ 200 80 ≤ y ≤ 170 • contract of at least 200 calculators per day x + y ≥ 200
Complete LP Model..Con’t (4 of 4) • Decision variables X= no. of Scientific Calculators Y= no. of Graphical calculators • Objective Function Maximize Z= -2X + 5 Y • Constraints 100 ≤ x ≤ 200 80 ≤ y ≤ 170 x + y ≥ 200 X, Y ≥ 0
More than just mathematics • In some cases, a "common sense" solution may be reached through simple observations. Examples below (see the book for details) • Slow elevator • Check-in facilities at a large airport (British culture) • In a steel mill (Employees delays)