Linear Programming Models: Graphical and Computer Methods Chapter 7 Linear Programming Models: Graphical and Computer Methods

Chapter Outline 7.1 Introduction 7.2 Requirements of a Linear Programming Problem 7.3 Formulating LP Problems 7.4 Graphical Solution to an LP Problem 7.5 Solving LP Problem (QM for Window and Excel) 7.6 Solving Minimization Problems 7.7 Four Special Cases in LP 7.8 Sensitivity Analysis

Linear Programming: An Overview Objectives of business decisions frequently involve maximizing profit or minimizing costs. Linear programming is an analytical technique in which linear algebraic relationships represent a firm’s decisions, given a business objective, and resource constraints. Steps in application: Identify problem as solvable by linear programming. Formulate a mathematical model of the unstructured problem. Solve the model. Implementation

Requirements of LP Problems and LP Basic Assumptions Decision variables - mathematical symbols representing levels of activity of a firm. Objective function - a linear mathematical relationship describing an objective of the firm, in terms of decision variables - this function is to be maximized or minimized. Constraints – requirements or restrictions placed on the firm by the operating environment, stated in linear relationships of the decision variables. Parameters - numerical coefficients and constants used in the objective function and constraints. Assumptions Proportionality - The rate of change (slope) of the objective function and constraint equations is constant. Additivity - Terms in the objective function and constraint equations must be additive. Divisibility -Decision variables can take on any fractional value and are therefore continuous as opposed to integer in nature. Certainty - Values of all the model parameters are assumed to be known with certainty (non-probabilistic). Non-negativity - all answers or variables are greater than or equal to (≥) zero Decision variables: The unknowns to be determined by the solution to the model. Objective function: The goal desired to be achieved by the system. Constraints: Limitation on variables to satisfy the restriction of the modeled system Parameters - numerical coefficients and constants used in the objective function and constraints. Linearity requires the following assumptions: Proportionality - a change in a variable results in a proportionate change in that variable's contribution to the value of the function. Additivity - the function value is the sum of the contributions of each term. Divisibility - the decision variables can be divided into non-integer values, taking on fractional values. Integer programming techniques can be used if the divisibility assumption does not hold. In addition to these linearity assumptions, linear programming assumes certainty; that is, that the coefficients are known and constant.

Minime Restaurant Pho (40K VND/bowl) Hu tieu (50K VND/bowl) Chef (works 40 min./day) Meat (120 grs meat/day) I can prepare 1 bowl of: - Pho in 1 min. - Hu tieu in 2 min. 1 bowl of: - Pho needs 4 gr meat - Hu tieu neads 3 gr. meat

A Maximization Example (1 of 3) Resource Requirements LP Model Formulation A Maximization Example (1 of 3) Product mix problem – Minime Noodle Restaurant How many bowls of Pho and Hu tieu should be prepared to maximize profits given labor and materials constraints? Product resource requirements and unit profit: Products Resource Requirements Profits Labor (min./bowl) Meat (gr./bowl) (K. VND/bowl) Pho 1 4 40 Hu tieu 2 3 50 Resource Availability 40 min. 120 grs.  

A Maximization Example (2 of 3) Resource Requirements LP Model Formulation A Maximization Example (2 of 3) Products Resource Requirements Profits Labor (min./bowl) Meat (gr./bowl) (K. VND/bowl) Pho 1 4 40 Hu tieu 2 3 50 Avail. Resource 120   Trial and error All for Pho: 40 bowls 30 bowls 30 bowls x 40= 1200 (chef’s time surplus) All for Hu tieu: 20 bowls 40 bowls 20 bowls x 50= 1000 (meat surplus) Half for Pho and Half for Hu tieu PHO 20 bowls 15 bowls 15 bowls x 40= 600 (chef’s time surplus) HU TIEU 10 bowls 20 bowls 10 bowls x 50= 500 (meat surplus) Time Meat Profit

A Maximization Example (3 of 3) LP Model Formulation A Maximization Example (3 of 3) Decision x1 = # of bowls of Pho to prepare per day Variables: x2 = # of bowls of Hu tieu to prepare per day Objective Maximize Z = 40x1 + 50x2 Function: Where Z = profit per day subject to (s.t.): Resource 1x1 + 2x2  40 minutes of labor Constraints: 4x1 + 3x2  120 grs. of meat Non-Negativity x1  0; x2  0 Constraints:

Feasible and Infeasible Solutions A feasible solution does not violate any of the constraints: Example x1 = 5 bowls of Pho x2 = 10 bowls of Hu tieu Z = 40x1 + 50x2 = 700 (K.VND) Labor constraint check: 1(5) + 2(10) = 25 < 40 minutes, within constraint Meat constraint check: 4(5) + 3(10) = 70 < 120 grams, within constraint An infeasible solution violates at least one of the constraints: Example x1 = 10 bowls of Pho x2 = 20 bowls of Hu tieu Z = 1400 (K.VND) 1(10) + 2(20) = 50 > 40 minutes, violates constraint

Graphical Solution of LP Models Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty). Graphical methods provide visualization of how a solution for a linear programming problem is obtained.

Graphical Solution of Maximization Model (1 of 11) Coordinate Axes Graphical Solution of Maximization Model (1 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.2 Coordinates for Graphical Analysis

Graphical Solution of Maximization Model (2 of 11) Labor Constraint Graphical Solution of Maximization Model (2 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.3 Graph of Labor Constraint

Graphical Solution of Maximization Model (3 of 11) Labor Constraint Area Graphical Solution of Maximization Model (3 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.4 Labor Constraint Area

Graphical Solution of Maximization Model (4 of 11) Meat Constraint Area Graphical Solution of Maximization Model (4 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.5 Clay Constraint Area

Graphical Solution of Maximization Model (5 of 11) Both Constraints Graphical Solution of Maximization Model (5 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.6 Graph of Both Model Constraints

Feasible Solution Area Graphical Solution of Maximization Model (6 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.7 Feasible Solution Area

Graphical Solution Methods ISOPROFIT METHOD 1. Graph all constraints and find the feasible region. 2. Select a specific profit (or cost) line and graph it to find the slope. 3. Move the objective function line in the direction of increasing profit (or decreasing cost) while maintaining the slope. The last point it touches in the feasible region is the optimal solution. 4. Find the values of the decision variables at this last point and compute the profit (or cost). CORNER POINT METHOD 2. Find the corner points of the feasible reason. 3. Compute the profit (or cost) at each of the feasible corner points. 4. select the corner point with the best value of the objective function found in Step 3. This is the optimal solution.

Graphical Solution of Maximization Model (7 of 11) Isoprofit Line Method Graphical Solution of Maximization Model (7 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Construct the isoprofit line: Z = 40x1 + 50x2 x2 = Z/50 - 40/50x1 = Z/50 -4/5x1 Figure 2.8 Objection Function Line for Z = $800

Graphical Solution of Maximization Model (8 of 11) Isoprofit Line Method Graphical Solution of Maximization Model (8 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.9 Alternative Objective Function Lines

Graphical Solution of Maximization Model (9 of 11) Isoprofit Line Method Graphical Solution of Maximization Model (9 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.10 Identification of Optimal Solution

Optimal Solution Coordinates Graphical Solution of Maximization Model (10 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.11 Optimal Solution Coordinates

Extreme (Corner) Point Method Graphical Solution of Maximization Model (11 of 11) Maximize Z = 40x1 + 50x2 subject to: 1x1 + 2x2  40 4x1 + 3x2  120 x1, x2  0 Figure 2.12 Solutions at All Corner Points

Using Excel’s Solver to Solve LP Problems The Solver tool in Excel can be used to find solutions to LP problems Integer programming problems Noninteger programming problems Solver may be sensitive to the initial values it uses Solver is limited to 200 variables and 100 constraints It can be used for small real world problems Add-ins like What’s Best! Can be used to expand Solver’s capabilities

Using Excel’s Solver to Solve LP Problems Maximize Z = 40 x1 + 50 x2 Subject to x1 + 2x2 40 mins. (labor constraint) 4x1 + 3x2 120 grs. (meat constraint) x1 , x2 0 Click on “Tools” to invoke “Solver.” Objective function =E6-F6 =E7-F7 =C6*B10+D6*B11 =C7*B10+D7*B11 Decision variables – Pho (x1)=B10; Hu tieu (x2)=B11

Using Excel’s Solver to Solve LP Problems After all parameters and constraints have been input, click on “Solve.” Objective function Decision variables C6*B10+D6*B11≤40 C7*B10+D7*B11≤120 Click on “Add” to insert constraints

Using Excel’s Solver to Solve LP Problems

Using QM for Windows and Excel Solving Flair Furniture’s LP Problem Maximize profit = $70T + $50C subject to 4T + 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting and varnishing constraint) T, C ≥ 0 (nonnegativity constraint) Most organizations have access to software to solve big LP problems While there are differences between software implementations, the approach each takes towards handling LP is basically the same Once you are experienced in dealing with computerized LP algorithms, you can easily adjust to minor changes

Using QM for Windows Computer screen for input of data First select the Linear Programming module Specify the number of constraints (non-negativity is assumed) Specify the number of decision variables Specify whether the objective is to be maximized or minimized For the Flair Furniture problem there are two constraints, two decision variables, and the objective is to maximize profit Computer screen for input of data

Using QM for Windows Computer screen for input of data Computer screen for output of solution

Using QM for Windows Graphical output of solution

A Minimization Example (1 of 6) LP Model Formulation A Minimization Example (1 of 6) Two brands of fertilizer available - Super-Gro, Crop-Quick. Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate. Super-Gro costs $6 per bag, Crop-Quick $3 per bag. Problem: How much of each brand to purchase to minimize total cost of fertilizer given following data ?

Fertilizing Farmer’s Field

A Minimization Example (2 of 6) LP Model Formulation A Minimization Example (2 of 6) Decision Variables: x1 = bags of Super-Gro x2 = bags of Crop-Quick The Objective Function: Minimize Z = 6x1 + 3x2 Model Constraints: 2x1 + 4x2  16 lb (nitrogen constraint) 4x1 + 3x2  24 lb (phosphate constraint) x1, x2  0 (non-negativity constraint)

LP Model Formulation and Constraint Graph A Minimization Example (3 of 6) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x1 + 3x2  24 x1, x2  0 Figure 2.16 Graph of Both Model Constraints

Feasible Solution Area A Minimization Example (4 of 6) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x1 + 3x2  24 x1, x2  0 Figure 2.17 Feasible Solution Area

Optimal Solution Point A Minimization Example (5 of 6) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x1 + 3x2  24 x1, x2  0 Construct the objective line: Z = 6x1 + 3x2 x2 = Z/3 – 6/3x1 x2 = Z/3 -2x1 Figure 2.18 Optimum Solution Point

A Minimization Example (6 of 6) Graphical Solutions A Minimization Example (6 of 6) Minimize Z = $6x1 + $3x2 subject to: 2x1 + 4x2  16 4x1 + 3x2  24 x1, x2  0 Figure 2.19 Graph of Fertilizer Example

Four Special Cases in LP Four special cases and difficulties arise at times when using the graphical approach to solving LP problems Infeasibility Unboundedness Redundancy Alternate Optimal Solutions

Four Special Cases in LP Infeasibility-A problem with no feasible solution Exists when there is no solution to the problem that satisfies all the constraint equations No feasible solution region exists 8 – – 6 – 4 – 2 – 0 – X2 | | | | | | | | | | 2 4 6 8 X1 Region Satisfying Third Constraint Region Satisfying First Two Constraints

Four Special Cases in LP Unboundedness- A solution region unbounded to the right In a graphical solution, the feasible region will be open ended This usually means the problem has been formulated improperly 15 – 10 – 5 – 0 – X2 | | | | | 5 10 15 X1 X1 ≥ 5 X2 ≤ 10 Feasible Region X1 + 2X2 ≥ 15

Four Special Cases in LP Redundancy- A problem with a redundant constraint A redundant constraint is one that does not affect the feasible solution region 30 – 25 – 20 – 15 – 10 – 5 – 0 – X2 | | | | | | 5 10 15 20 25 30 X1 2X1 + X2 ≤ 30 Redundant Constraint X1 ≤ 25 X1 + X2 ≤ 20 Feasible Region

Four Special Cases in LP Alternate optimal solutions Occasionally two or more optimal solutions may exist Graphically this occurs when the objective function’s isoprofit or isocost line runs perfectly parallel to one of the constraints 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 – 0 – X2 | | | | | | | | 1 2 3 4 5 6 7 8 X1 A Optimal Solution Consists of All Combinations of X1 and X2 Along the AB Segment Isoprofit Line for $8 B Isoprofit Line for $12 Overlays Line Segment AB Feasible Region

Sensitivity Analysis Sensitivity analysis is used to determine effects on the optimal solution within specified ranges for the objective function coefficients, constraint coefficients, and right hand side (RHS) values. Sensitivity analysis provides answers to certain what-if questions.

Range of Optimality A range of optimality of an objective function coefficient is found by determining an interval for the objective function coefficient in which the original optimal solution remains optimal while keeping all other data of the problem constant. The value of the objective function may change in this range. Graphically, the limits of a range of optimality are found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines. (This would also apply to simultaneous changes in the objective coefficients.) The slope of an objective function line, Max c1x1 + c2x2, is -c1/c2 The slope of a constraint, a1x1 + a2x2 = b, is -a1/a2

Shadow Price A shadow price for a RHS value (or resource limit) is the amount the objective function will change per unit increase in the right hand side value of a constraint. Graphically, a shadow price is determined by adding +1 to the right hand side value in question and then resolving for the optimal solution in terms of the same two binding constraints. The shadow price is equal to the difference in the values of the objective functions between the new and original problems. The shadow price for a nonbinding constraint is 0.

Example: Sensitivity Analysis Solve graphically for the optimal solution: Max z = 5x1 + 7x2 s.t. x1 < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1, x2 > 0

Example: Sensitivity Analysis Graphical Solution x2 8 7 6 5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 x1 + x2 < 8 Max 5x1 + 7x2 x1 < 6 Optimal x1 = 5, x2 = 3 z = 46 2x1 + 3x2 < 19 x1

Example: Sensitivity Analysis Range of Optimality for c1 The slope of the objective function line is -c1/c2. The slope of the 1st binding constraint, x1 + x2 = 8, is -1, the slope of the 2nd binding constraint, 2x1 + 3x2 = 19, is -2/3. Find the range of values for c1 (with c2 staying 7) such that the objective function line slope lies between that of the two binding constraints: -1 < -c1/7 < -2/3 Multiplying through by -7 (and reversing the inequalities): 14/3 < c1 < 7

Example: Sensitivity Analysis Range of Optimality for c2 Find the range of values for c2 ( with c1 staying 5) such that the objective function line slope lies between that of the two binding constraints: -1 < -5/c2 < -2/3 Multiplying by -1: 1 > 5/c2 > 2/3 Inverting, 1 < c2/5 < 3/2 Multiplying by 5: 5 < c2 < 15/2

Example: Sensitivity Analysis Shadow Prices Constraint 1: Since x1 < 6 is not a binding constraint, its shadow price is 0. Constraint 2: Change the RHS value of the 2nd constraint to 20 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 20 and x1 + x2 = 8. The solution is x1 = 4, x2 = 4, z = 48. Hence, the shadow price = znew - zold = 48 - 46 = 2. Constraint 3: Change the RHS value of the 3rd constraint to 9 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 19 and x1 + x2 = 9. The solution is: x1 = 8, x2 = 1, z = 47. Hence, the shadow price = znew - zold = 47 - 46 = 1.

Homework 06 7.15, 7.16, 7.20, 7.22, 7.24, 7.26, 7.28