Chapter 2 Linear Programming Models: Graphical and Computer Methods

Outline LP Meaning LP Models Model Formulation Solving LP

LP Meaning LP is a powerful quantitative tool used by the bp to obtain optimal solutions to problem that involve restrictions or limitations. LP techniques consists of sequence of steps that will lead to an optimal solutions to linear-constrained problems. LP involves the use of straight line relationships to maximize or minimise some functions usually called objective functions.

Properties of LP Models Seek to minimize or maximize Include “constraints” or limitations There must be alternatives available All equations are linear

LP models LP models are mathematical representations of constrained optimization problems. Four components provide the structure of a LP models Objective functions Decision variables Constraints Parameters

Objective function: mathematical statement of profit(or cost) for a given solution Decision variables: Amount of either inputs or outputs Constraints: Limitations that restrict the available alternatives.3 types of constraints ie <=,>= or=

Solving the LP:Outline of Graphical procedure Set up the objective function and the constraints in mathematical format Plot the constraints Identify the feasible solution space Plot the objective function Determine the optimum solution

Steps in Developing a Linear Programming (LP) Model Formulation Solution Interpretation and Sensitivity Analysis

Example LP Model Formulation: The Product Mix Problem Decision: How much to make of > 2 products? Objective: Maximize profit Constraints: Limited resources

Example: Flair Furniture Co. Two products: Chairs and Tables Decision: How many of each to make this month? Objective: Maximize profit

Flair Furniture Co. Data Tables (per table) Chairs (per chair) Hours Available Profit Contribution $7 $5 Carpentry 3 hrs 4 hrs 2400 Painting 2 hrs 1 hr 1000 Other Limitations: Make no more than 450 chairs Make at least 100 tables

Decision Variables: T = Num. of tables to make C = Num. of chairs to make Objective Function: Maximize Profit Maximize $7 T + $5 C

Have 2400 hours of carpentry time available Constraints: Have 2400 hours of carpentry time available 3 T + 4 C < 2400 (hours) Have 1000 hours of painting time available 2 T + 1 C < 1000 (hours) Nonnegativity: Cannot make a negative number of chairs or tables T > 0 C > 0

Model Summary Max 7T + 5C (profit) 3T + 4C < 2400 (carpentry hrs) Subject to the constraints: 3T + 4C < 2400 (carpentry hrs) 2T + 1C < 1000 (painting hrs) T, C > 0 (nonnegativity)

Graphical Solution Graphing an LP model helps provide insight into LP models and their solutions. While this can only be done in two dimensions, the same properties apply to all LP models and solutions.

Carpentry Constraint Line 3T + 4C = 2400 Intercepts (T = 0, C = 600) Carpentry Constraint Line 3T + 4C = 2400 Intercepts (T = 0, C = 600) (T = 800, C = 0) Infeasible > 2400 hrs 3T + 4C = 2400 Feasible < 2400 hrs 0 800 T

Painting Constraint Line 2T + 1C = 1000 Intercepts (T = 0, C = 1000) Painting Constraint Line 2T + 1C = 1000 Intercepts (T = 0, C = 1000) (T = 500, C = 0) 2T + 1C = 1000 0 500 800 T

Corner point solution Value of Z at corner point a Value of Z at d Z= 7T+5C Za =7.0+5*600 Za=3000 Value of Z at d Z=7T+5C Zd=7*500+5*0=3500

Use simultaneous equation Value of Z at b Z=7T+5C Use simultaneous equation (1) 3T + 4C =2400 (carpentry hrs) (2) 2T + 1C = 1000 (painting hrs) Multiply equation 2 by 4 (3) 8T+4C =4000 1-3= -5T= -1600 T= 320 now substitute the value of T in any equation 2*320+ 1C=1000 1C= 1000-640 C= 360 Z= 7*320+5*360=4040

LP Characteristics Feasible Region: The set of points that satisfies all constraints Corner Point Property: An optimal solution must lie at one or more corner points Optimal Solution: The corner point with the best objective function value is optimal

Special Situation in LP Redundant Constraints - do not affect the feasible region Example: x < 10 x < 12 The second constraint is redundant because it is less restrictive.

Problem-1 A frim can produce two types of a certain product, namely basic B and deluxeD.The estimated profit per unit is $10 Basic and $15 Deluxe.The firm has 60 hours of labour available per day and 100 units of capital.Each unit of basic product B requires two hrs of labor and four units of capital.Each unit de-luxe product D requires four hrs of labor and five units of capital. How much of each type of product should the firm produce in order to maximise total profit?

Model Summary Maximise Z= 10B+15D …objective function Subject to: 2B+4D<= 60… labour constraint 4B+5D<=100...Capital constraint B>=0,D>= non-negative constraints

Plotting the constraints The labour constraint line:if we use all our 60 hrs of labor on the basic B we can produce 30 units of B but zero D.At the other extreme if use all our labor to produce D we can produce 15 D but Zero B The capital Constraint line same way we can get B= 25 and D=20

Identify the feasible solution space Feasible region of production area OVWX Value of Z at corner point V Zv=10B+15D=10*25+15*0=250 Value of Z at corner point X Zx=10*0+15*5=225 Value of Z at corner point W (1)2B+4D=60 (2)4B+5D=100 Here multiply equation 1 by 2 and subtract the equation 4B+8D=120 4B+5D=100 3D= 20 D= 20/3 then substitute D & get the Value of B=16 2/3 and we can solve for Z at W= 266 2/3

Problem2 A firm uses three types of factor input, namely labour, capital and raw material in producing two products X and Y. Each unit of X contributes $20 to profit, each unit of Y contributes $30 to profit. To produce one unit of X requires one unit of labour, one unit of capital and two units of raw material. To produce one unit of Y requires one unit of labour, two units of capital and one unit of raw material. The firm has 50 units of labour,80 units of capital and 80 units of raw material available. How much of each type of product should the firm produce in order to maximise total profit?

Minimisation Problem-3 A travel company operates two types of vehicle A and B. Vehicle A can carry 40 passengers and 30 tons of baggage; Vehicle B can carry 60 passengers but only 15 tons of baggage. The travel company is contracted to carry at least 960 passengers and 360 tons of baggage per journey. If vehicle A cost $1000 to operate per journey and vehicle B costs $1200 to operate per journey. What choice of vehicles will minimise the total cost

Minimise Z= 1000A+1200B Subject to 40A+60B>= 960…Passengers constraint 30A+15B>=360…baggage constraint A>=0,B>=0… non negative constraints

Using Excel’s Solver for LP Recall the Flair Furniture Example: Max 7T + 5C (profit) Subject to the constraints: 3T + 4C < 2400 (carpentry hrs) 2T + 1C < 1000 (painting hrs) C < 450 (max # chairs) T > 100 (min # tables) T, C > 0 (nonnegativity) Go to file 2-1.xls