Linear Programming: Formulations & Graphical Solution

1© 2003 by Prentice Hall, Inc. Upper Saddle River, NJ 07458

Linear Programming: Formulations Linear Programming: Formulations & Graphical Solution& Graphical Solution


Introduction To Linear Programming

• After three decades of experimentation and scrutiny, LP has been applied with impressive success to problems ranging from the familiar cases in industry, military, agriculture, economics, transportation, and health systems to the extreme cases in behavioral and social sciences.



• Today many of the resources needed as inputs to operations are in limited supply.

• Operations managers must understand the impact of this situation on meeting their objectives.

• Linear programming (LP) is one way that operations managers can determine how best to allocate their scarce resources.

• A Linear Programming model seeks to maximize or minimize a linear function, subject to a set of linear constraints.


Linear Programming Definition

• Linear Programming is a mathematical technique for optimum allocation of limited or scarce resources, such as labour, material, machine, money, energy and so on, to several competing activities such as products, services, jobs and so on, on the basis of a given criteria of optimality.



• The maximization or minimization of some quantity is the objective in all linear programming problems.

• All LP problems have constraints that limit the degree to which the objective can be pursued.

• A feasible solution satisfies all the problem's constraints.• An optimal solution is a feasible solution that results in the

largest possible objective function value when maximizing (or smallest when minimizing).

• A graphical solution method can be used to solve a linear program with two variables.



• If both the objective function and the constraints are linear, the problem is referred to as a linear programming problem.

• Linear functions are functions in which each variable appears in a separate term raised to the first power and is multiplied by a constant (which could be 0).

• Linear constraints are linear functions that are restricted to be "less than or equal to", "equal to", or "greater than or equal to" a constant.

• Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.


Construction of the Mathematical Model

• The construction of a mathematical model can be initiated by answering the following three questions:

1. What does the model seek to determine? In other words, what are the variables (unknowns) of the problem?

2. What constraints must be imposed on the variables to satisfy the limitations of the modeled system?

3. What is the objective (goal) that needs to be achieved to determine the optimum (best) solution from among all the feasible values of the variables?

• An effective way to answer these questions is to give a verbal summary of the problem. In terms of the Reddy Mikks example, the situation is described as follows.


• Understand the problem thoroughly.• Define the decision variables.• Describe the objective.• Describe each constraint.• Write the objective in terms of the decision

variables.• Write the constraints in terms of the decision

variables.

Guidelines for Model Formulation


Reddy Mikks Problem (Taha)

• The Reddy Mikks company owns a small paint factory that produces both interior and exterior house paints for wholesale distribution. Two basic raw materials, A and B, are used to manufacture the paints.

• The maximum availability of A is 6 tons a day; that of B is 8 tons a day. The daily requirements of the raw materials per ton of interior and exterior paints are summarized in the following table.

Tons of Raw Material per Ton of Paint

Exterior Interior Maximum Availability (tons)

Raw Material A 1 2 6 Raw Material B 2 1 8


Reddy Mikks Problem (Taha)

• A market survey has established that the daily demand for the interior paint cannot exceed that of exterior paint by more than 1 ton. The survey also showed that the maximum demand for the interior paint is limited to 2 tons daily.

• The wholesale price per ton is $3000 for exterior paint and $2000 per interior paint. How much interior and exterior paint should the company produce daily to maximize gross income?


Reddy Mikks Problem Formulation

• The company seeks to determine the amounts (in tons) of interior and exterior paints to be produced to maximize (increase as much as is feasible) the total gross income (in thousands of dollars) while satisfying the constraints of demand and raw material usage.

• Variables: since we desire to determine the amounts of interior and exterior paints to be produced, the variables of the model can be defined as

• XE = tons produced daily of exterior paint

• XI = tons produced daily of interior paint



• Objective Function: since each ton of exterior paint sells for $3000, the gross income from selling XE tons is 3XE thousand dollars. Similarly, the gross income from XI tons of interior paint is 2XI thousand dollars.

• Under the assumption that the sales of interior and exterior paints are independent, the total gross income becomes the sum of the two revenues.

• If we let Z represents the total gross revenue (in thousands of dollars), the objective function may be written mathematically as

• Z = 3XE +2XI

The goal is to determine the (feasible) values of XE and XI that will maximize this criterion.



• Constraints: The Reddy Mikks problem imposes restrictions on the usage of raw materials and on demand. The usage restriction may be expressed verbally as

(usage of raw material by both paints) ≤(maximum raw material availability)

• This leads to the following restrictions (see the data for the problem):

XE + 2XI ≤ 6 (raw material A)

2XE + XI ≤ 8 (raw material B)



• The demand restrictions are expressed verbally as

(excess amount of interior over exterior paint) ≤ 1 ton per day (demand for interior paint) ≤ 2 tons per day

• Mathematically, these are expressed, respectively, as

XI - XE ≤ 1

XI ≤ 2



• An implicit (or "understood-to-be) constraint is that the amount produced of each paint cannot be negative (less than zero). To avoid obtaining such a solution, we impose the nonnegativity restrictions, which are normally written

XI ≥ 0

XE ≥ 0

• The values of the variables XE and XI are said to constitute a feasible solution if they satisfy all the constraints of the model.



• The complete mathematical model for the Reddy Mikks problem may now be summarized as follows:


This is a typical optimization problem. Any values of x1, x2 that satisfy all the constraints of the model is called a feasible solution. We are interested in finding the optimum feasible solution that gives the maximum profit while satisfying all the constraints.


More generally, an optimization problem looks as follows:

Determine the decision variables x1, x2, …, xn so as to optimize an objective function f (x1, x2, …, xn) satisfying the constraints

gi (x1, x2, …, xn) ≤ bi (i=1, 2, …, m).


An optimization problem is called a Linear Programming Problem (LPP) when the objective function and all the constraints are linear functions of the decision variables, x1, x2, …, xn. We also include the “non-negativity restrictions”, namely xj ≥ 0 for all j=1, 2, …, n. Thus a typical LPP is of the form:

Linear Programming Problems(LPP)


Optimize (i.e. Maximize or Minimize) z = c1 x1 + c2 x2+ …+ cn xn subject to the constraints: a11 x1 + a12 x2 + … + a1n xn ≤ b1 a21 x1 + a22 x2 + … + a2n xn ≤ b2 . . . am1 x1 + am2 x2 + … + amn xn ≤ bm x1, x2, …, xn 0


LP Assumptions• When we use LP as an approximate representation of a real-life

situation, the following assumptions are inherent:• Proportionality. - The contribution of each decision variable to

the objective or constraint is directly proportional to the value of the decision variable.

• Additivity. - The contribution to the objective function or constraint for any variable is independent of the values of the other decision variables, and the terms can be added together sensibly.

• Divisibility. - The decision variables are continuous and thus can take on fractional values.

• Deterministic (Certainty). - All the parameters (objective function coefficients, right-hand side coefficients, left-hand side, coefficients) are known with certainty.


Example

• Cycle Trends is introducing two new lightweight bicycle frames, the Deluxe and the Professional, to be made from aluminum and steel alloys. The anticipated unit profits are $10 for the Deluxe and $15 for the Professional.

• The number of pounds of each alloy needed per frame is summarized on the next slide. A supplier delivers 100 pounds of the aluminum alloy and 80 pounds of the steel alloy weekly. How many Deluxe and Professional frames should Cycle Trends produce each week?


Pounds of each alloy needed per frame

Aluminum Alloy Steel Alloy Deluxe 2 3 Professional 4 2


Example: LP Formulation

• Define the objective• Maximize total weekly profit

• Define the decision variables• x1 = number of Deluxe frames produced weekly

• x2 = number of Professional frames produced weekly

• Write the mathematical objective function• Max Z = 10x1 + 15x2


Example: LP Formulation

• LP in Final Form• Max Z = 10x1 + 15x2• Subject To

• 2x1 + 4x2 < 100 ( aluminum constraint)• 3x1 + 2x2 < 80 ( steel constraint)• x1 , x2 > 0 (non-negativity constraints)


Example

The Burroughs garment company manufactures men's shirts and women’s blouses for Walmark Discount stores. Walmark will accept all the production supplied by Burroughs. The production process includes cutting, sewing and packaging. Burroughs employs 25 workers in the cutting department, 35 in the sewing department and 5 in the packaging department. The factory works one 8-hour shift, 5 days a week. The following table gives the time requirements and the profits per unit for the two garments:


Garment Cutting Sewing Packaging Unit profit($)

Shirts 20 70 12 8.00

Blouses 60 60 4 12.00

Minutes per unit

Determine the optimal weekly production schedule for Burroughs.


Solution

Assume that Burroughs produces x1 shirts and x2 blouses per week.

Profit got = 8 x1 + 12 x2

Time spent on cutting = 20 x1 + 60 x2 mts

Time spent on sewing = 70 x1 + 60 x2 mts

Time spent on packaging = 12 x1 + 4 x2 mts


The objective is to find x1, x2 so as to maximize the profit z = 8 x1 + 12 x2

satisfying the constraints:

20 x1 + 60 x2 ≤ 25 40 60

70 x1 + 60 x2 ≤ 35 40 60

12 x1 + 4 x2 ≤ 5 40 60

x1, x2 ≥ 0, integers


The Nutrition Problem


The Nutrition Problem• Each fruit contains different nutrients• Each fruit has different cost• An apple a day keeps the doctor away – but apples are costly!• A customer’s goal is to fulfill daily nutrition requirements at lowest

cost.• Lets take a simpler case of just apples and bananas.• Must take at least 100 units of Calories & 90 units of Vitamins for good

nutrition.• A customer’s goal is to buy fruits in such a quantity that it minimizes

cost but fulfills nutrition.

Calories Vitamins Cost($)2 3 54 3 7


The Nutrition Problem Formulation

Objective Function

Min. Z = 5x1 + 7x2

Constraint Functions

2x1 + 4x2 100

3x1 + 3x2 90

x1, x2 0


An Electric Company Problem• An electric company manufacturers two radio models,

each on a separate rate production line. The daily capacity of the first line is 60 radios and that of the second is 75 radios. Each unit of the first model uses 10 pieces of a certain electronic component, whereas each unit of the second model requires 8 pieces of the same component. The maximum daily availability of the special component is 800 pieces. The profit per unit of models 1 and 2 is $30 and $20, respectively. Determine the optimum daily production of each model.


An Electric Company Problem Formulation

X1 = number of radios of model 1X2 = number of radios of model 2 Objective Function max Z =30X1 + 20X2

� � Subject To X1 60 X2 75

10 X1+8X2 800

X1 0, X2 0


Furniture Factory Problem• A small furniture factory manufacturers tables and chairs.

It takes 2 hours to assemble a table and 30 minutes to assemble a chair. Assembly is carried out by four workers on the basis of a single 8-hour shift per day. Customers usually buy at least four chairs with each table, meaning that the factory must produce at least four times as many chairs as tables. The sale price is $150 per table and $50 per chair. Determine the daily production mix of chairs and tables that would maximize the total daily revenue to the factory.


Furniture Factory Problem Formulation

X1 = number of tablesX2 = number of chairs Objective Function max Z =150X1 + 50X2

�� Subject To X2 - 4X1 0

120 X1+30X2 4x8x60

X1 0, X2 0


Graphical Solution of an LP Problem

• Used to solve LP problems with two (and sometimes three) decision variables

• Consists of two phases• Finding the values of the decision variables for which

all the constraints are met (feasible region of the solution space)

• Determining the optimal solution from all the points in the feasible region (from our knowledge of the nature of the optimal solution)


Finding the Feasible Region (2D)

• Steps• Use the axis in a 2-dimensional graph to represent the

values that the decision variables can take• For each constraint, replace the inequalities with

equations and graph the resulting straight line on the 2-dimensional graph

• For the inequality constraints, find the side (half-space) of the graph meeting the original conditions (evaluate whether the inequality is satisfied at the origin)

• Find the intersection of all feasible regions defined by all the constraints. The resulting region is the (overall) feasible region.


Flair Furniture Company Data - Table 7.1

Hours Required to Produce One UnitDepartment T

TablesC

Chairs

AvailableHours This

Week• Carpentry• Painting &Varnishing

42

31

240100

Profit Amount $7 $5

Constraints: 4T + 3C 240 (Carpentry) 2T + 1C 100 (Paint & Varnishing)

T ≥ 0 (1st nonnegative cons) C ≥ 0 (2nd nonnegative cons)

Max. Objective, z: 7T + 5C

Mathematical formulation:Mathematical formulation:


Flair Furniture Company Constraints

The easiest way to solve a small LP problem, such as that of the Flair Furniture Company, is with the graphical solution approach.graphical solution approach.

The graphical method works only when there are twotwo decision variables, but it provides valuable insight into how larger problems are structured.

When there are more than two variables, it is not possible to plot the solution on a two-dimensional graph; a more complex approach is needed.

But the graphical method is invaluable in providing us with insights into how other approaches work.


Flair Furniture Company Constraints

Number of Tables

120

100

80

60

40

20

0

Num

ber o

f Cha

irs

20 40 60 80 100

Painting/Varnishing

Carpentry4T + 3C ≤ 240

2T + 1C ≤ 100


Flair Furniture Company Feasible Region

120

100

80

60

40

20

0

Num

ber o

f Cha

irs

20 40 60 80 100Number of Tables

Painting/Varnishing

CarpentryFeasibleRegion


Isoprofit Lines Steps1.1. Graph all constraints and find the feasible region.2.2. Select a specific profit (or cost) line and graph it

to find the slope.3.3. 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.

4.4. Find the values of the decision variables at this last point and compute the profit (or cost).


Isoprofit Line Solution MethodIsoprofit Line Solution Method Start by letting profits equal some arbitrary but small

dollar amount. Choose a profit of, say, $210.

- This is a profit level that can be obtained easily without violating either of the two constraints.

The objective function can be written as $210 = 7T + 5C.

Flair Furniture Company Isoprofit Lines


Isoprofit Line Solution MethodIsoprofit Line Solution Method• The objective function is just the equation of a line called an

isoprofit line. - It represents all combinations of (T, C) that would yield a total profit of $210.

To plot the profit line, proceed exactly as done to plot a constraint line: - First, let T = 0 and solve for the point at which the line crosses the C axis.- Then, let C = 0 and solve for T.$210 = $7(0) + $5(C) C = 42 chairs

Then, let C = 0 and solve for T.$210 = $7(T) + $5(0) T = 30 tables



Isoprofit Line Solution MethodIsoprofit Line Solution Method Next connect these two points with a straight line. This

profit line is illustrated in the next slide. All points on the line represent feasible solutions that

produce an approximate profit of $210 Obviously, the isoprofit line for $210 does not produce

the highest possible profit to the firm. Try graphing more lines, each yielding a higher profit. Another equation, $420 = $7T + $5C, is plotted in the

same fashion as the lower line.



Isoprofit Line Solution MethodIsoprofit Line Solution Method When T = 0,

$420 = $7(0) + 5(C) C = 84 chairs

When C = 0, $420 = $7(T) + 5(0) T = 60 tables

This line is too high to be considered as it no longer touches the feasible region.

The highest possible isoprofit line is illustrated in the second following slide. It touches the tip of the feasible region at the corner point (T = 30, C = 40) and yields a profit of $410.




Number of Tables

Num

ber o

f Cha

irs120

100

80

60

40

20

0

20 40 60 80 100

Painting/Varnishing

Carpentry

7T + 5C = 2107T + 5C = 420


Flair Furniture Company Optimal Solution

Num

ber o

f Cha

irs

120

100

80

60

40

20

0


Painting/Varnishing

CarpentryCarpentry

Solution(T = 30, C = 40)

Isoprofit LinesIsoprofit Lines


Corner Point Solution MethodCorner Point Solution Method A second approach to solving LP problems It involves looking at the profit at every corner point

of the feasible region The mathematical theory behind LP is that the

optimal solution must lie at one of the corner points in the feasible region

Flair Furniture Company Corner Point


Corner Point Solution Method, SummaryCorner Point Solution Method, Summary1.1. Graph all constraints and find the feasible

region.2.2. Find the corner points of the feasible region.3.3. Compute the profit (or cost) at each of the

feasible corner points.4.4. Select the corner point with the best value of

the objective function found in step 3. This is the optimal solution.

Corner Point


Corner Point Solution MethodCorner Point Solution Method The feasible region for the Flair Furniture Company problem

is a four-sided polygon with four corner, or extreme, points. These points are labeled 1 ,2 ,3 , and 4 on the next graph. To find the (T, C) values producing the maximum profit,

find the coordinates of each corner point and test their profit levels.

Point 1:(T = 0,C = 0) profit = $7(0) + $5(0) = $0Point 2:(T = 0,C = 80) profit = $7(0) + $5(80) = $400Point 3:(T = 30,C = 40) profit = $7(30) + $5(40) = $410Point 4 : (T = 50, C = 0) profit = $7(50) + $5(0) = $350

Flair Furniture Company Corner Point


Flair Furniture Company Optimal Solution

Num

ber o

f Cha

irs

120

100

80

60

40

20

0


Painting/Varnishing

Carpentry

Solution(T = 30, C = 40)

Corner PointsCorner Points

1

2

3

4


Solving Minimization ProblemsMany LP problems minimize an objective,such as cost, instead of maximizing aprofit function. For example,

• A restaurant may wish to develop a work schedule to meet staffing needs while minimizing the total number of employees.

Or, A manufacturer may seek to distribute its products from several

factories to its many regional warehouses in such a way as to minimize total shipping costs.

Or, A hospital may want to provide a daily meal plan for its patients

that meets certain nutritional standards while minimizing food purchase costs.


Solving Minimization ProblemsMinimization problems can be solvedgraphically by first setting up the feasible solution region and then

using either the corner point method or an isocost line approach (which is analogous to the

isoprofit approach in maximization problems) to find the values of the decision variables(e.g., X1 and X2) that yield the minimum cost.


Solving Minimization ProblemsHoliday Meal Turkey Ranch exampleHoliday Meal Turkey Ranch exampleMinimize: 2X1 + 3X2

Subject to:5X1 + 10X2 90 oz. (A)

4X1 + 3X2 48 oz. (B)

½ X1 1 ½ oz. (C)

X1, X2 0 (D)where,X 1 = # of pounds of brand 1 feed purchased

X 2 = # of pounds of brand 2 feed purchased

(A) = ingredient A constraint(B) = ingredient B constraint(C) = ingredient C constraint(D) = non-negativity constraints


Holiday Meal Turkey RanchUsing the Corner Point MethodUsing the Corner Point MethodTo solve this problem: 1. Construct the feasible solution region.

This is done by plotting each of the three constraint equations.2. Find the corner points.

This problem has 3 corner points, labeled a, b, and c.

- Minimization problems are often unboundoutward (i.e., to the right and on top), but this causes no difficulty in solving them.

- As long as they are bounded inward (on the left side and the bottom), corner points may be established.

- The optimal solution will lie at one of the corners as it would in a maximizationproblem.


Holiday Meal Turkey Problem

Corner PointsCorner Points


Solving Minimization ProblemsUsing the Isocost Line ApproachUsing the Isocost Line Approach As with isoprofit lines, there is no need to compute the

cost at each corner point, but instead draw a series of parallel cost lines.

The lowest cost line (i.e., the one closest in toward the origin) to touch the feasible region provides the optimal solution corner.

1. Start, for example, by drawing a 54-cent cost line, namely 54 = 2X1 + 3X2. - Obviously, there are many points in the feasible region that would

yield a lower total cost.


Solving Minimization Problems (continued)

2. Proceed to move the isocost line toward the lower left, in a plane parallel to the 54-cent solution line.

3. The last point touched while still in contact with the feasible region is the same as corner point b of the Corner Point diagram in the previous slide.

- It has the coordinates (X1 = 8.4, X2 = 4.8) and an associated cost of 31.2 cents.


Holiday Meal Turkey Problem

Isoprofit LinesIsoprofit Lines


Example: Find the maximum value of z, given:

Max. z = 2x + 3y and

2 5 253 2 21

00

x yx y

xy

objective function

constraints


x

y

2552 yx

2123 yx

y = 0x = 0

feasible region


x

y

Example continued: Maximize the value of z = 2x +3y over the feasible region.

feasible region

(0, 0)

(0, 5)

(7, 0)

(5, 3)

z = 2(0) + 3(0) = 0

z = 2(0) + 3(5) = 15

z = 2(7) + 3(0) = 14

z = 2(5) + 3(3) = 19

Test the value of z at each of the vertices. The maximum value of z is 19. This occurs at the point (5, 3) or when x =5 and y = 3.

Maximum value of z

If a linear programming problem has a solution, then the solution is at a vertex of the feasible region.


Solving Linear Programming Problems Graphically

1. Graph the feasible region. 2. Find the vertices of the region. 3. Evaluate the objective function at each vertex.

4. Select the vertices that optimize the objective function.

a) If the feasible region is bounded the objective function will have both a maximum and a minimum.

b) If the feasible region is unbounded and the objective function has an optimal value, the optimal value will occur at a vertex of the feasible region.

Note: If the optimum value occurs at two vertices, its value is the same at both vertices and along the line segment joining them.


Example(1)

• Find the maximum and minimum value of

z = x + 3y

subject to the constraints –x + 3y 6 x + 3y 6 x + y 6 x , y 0


Example(1)\solution

x

y63 yx

63 yx6 yx

(0, 2)(3, 3)

(6, 0) z = 6 + 3(0) = 6

z = 3 + 3(3) = 12Maximum value

of z

z = 0 + 3(2) = 6

•The maximum value of z is 12 and occurs at (3, 3).•The minimum value of z is 6 and occurs at both (0, 2) and (6, 0) and at every point along the line joining them.


Example(2)

• Objective Function Minimize z = 3x – y • Subject to

x – y 1 x + y 5 x 0, and y 0.


Example(2)\solution

y

x

1 yx

5 yx(1, 0)

(3, 2)

(0, 5)vertex value of z at vertex

(0, 0) z = 3(0) – (0) = 0(1, 0) z = 3(1) – (0) = 3(3, 2) z = 3(3) – (2) = 7(0, 5) z = 3(0) – (5) = –5

The minimum value of z is –5 and this occurs at (0, 5).


Flair Furniture - QM for Windows

To use QM for Windows, 1. Select the Linear Programming module.2. Then specify

- the number of constraints (other than the non-negativity constraints, as it is assumed that the variables must be nonnegative),

- the number of variables, and - whether the objective is to be maximized or minimized.

For the Flair Furniture Company problem, there aretwo constraints and two variables.

3. Once these numbers are specified, the input window opens as shown on the next slide.


Flair Furniture - QM for Windows (continued)

4. Next, the coefficients for the objective function and the constraints can be entered.- Placing the cursor over the X1 or X2 and typing a new name

such as Tables and Chairs will change the variable names. - The constraint names can be similarly modified.- When you select the SolveSolve button, you get the output shown

in the next slide.



5. Modify the problem by selecting the EditEdit button and returning to the input screen to make any desired changes.


7. Once the problem has been solved, a graph may be displayed by selecting Window—Graph from the menu bar in QM for Windows.

8. The next slide shows the output for the graphical solution. Notice that in addition to the graph, the corner points and the original problem are also shown.


Flair Furniture - QM for Windows

To use QM for Windows, 1. Select the Linear Programming module.2. Then specify

- the number of constraints (other than the non-negativity constraints, as it is assumed that the variables must be nonnegative),

- the number of variables, and - whether the objective is to be maximized or minimized.

For the Flair Furniture Company problem, there aretwo constraints and two variables.


Linear Programming: Formulations & Graphical Solution

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