BA 452 Lesson A.3 Unsolvable Linear Programs 1 1ReadingsReadings Chapter 2 An Introduction to Linear...

BA 452 Lesson A.3 Unsolvable Linear Programs 11

Readings

Readings

Chapter 2An Introduction to Linear Programming


Overview

Overview


Overview

Blending Problems are Linear Programming Profit Maximization problems when additional inputs may be bought. Blending Problems thus help blend resources to maximize profit or to minimize cost.

Non-Unique Optimal Solutions of a linear program mean there are either alternative optimal solutions, no solutions because it is infeasible, or no solutions because the objective can be infinite.

Blending with Mixed Constraints help blend resources (grass seeds, desserts, horse food, …) when constraints restrict permissible blends by a mixture of minimal and maximal characteristics.


Blending

Blending


Overview

Blending Problems are Linear Programming Profit Maximization problems when additional inputs may be bought. Blending Problems thus help production managers blend resources (grass seeds, desserts, horse food, …) to produce goods either to maximize profit or to minimize cost. Constraints restrict permissible blends by product characteristics (shade tolerance, calories, vitamin content, …).

Blending


Question: The Sanders Garden Shop blends two types of grass seed. Each type has been rated (per pound) according to its shade tolerance, ability to stand up to traffic, and drought resistance, as shown in the table: Type A Type BShade tolerance 2 5Traffic resistance 4 -1Drought resistance 1 1

For example, if 1 pound of Type A is the blend, then that blend scores 2 points of Shade tolerence; if 2 pounds of Type A, then 4 points. (The points could measure how many blades of green grass survive being in the shade.)For another example, if 1 pound of Type B is the blend, then that blend scores -1 points of Traffic resistance. (The points could measure the number of blades of green grass minus blades of dead grass when there is traffic on the grass. A negative score means there is more dead grass.)

Blending


Type A seed costs $5 and Type B seed costs $2. Suppose the blend needs to score at least 10 points for shade tolerance, at least 12 points for traffic resistance, and at least 4 points for drought resistance.

How many pounds of each seed should be in the blend? Which targets will be exceeded? How much will the blend cost?

Blending


Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Objective: minimize cost of blending x1 pounds of Type A seed with x2 pounds of Type B seed

Constraint 1: Shade tolerance

Constraint 2: Traffic resistance

Constraint 3: Drought resistance

Non-negativity constraints

Blending

Answer: First, formulate the problem as a linear program:


Constraint 1: 2x1+5x2 > 10. Binding edge: 2x1+5x2 = 10. On the edge, when x1 = 0, then x2 = 2; when x2 = 0, then x1 = 5. Connect (5,0) and (0,2) for the binding edge of the constraint plus non-negativity. The feasible ">" side is above the binding edge.

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Binding edgeBinding edge

Feasible side

Blending


Constraint 2: 4x1-x2 > 12. Binding edge: 4x1-x2 = 12. On the edge, when x2 = 0, then x1 = 3. But setting x1 to 0 yields x2 = -12, which violates non-negativity. Thus, to get a second point on the binding edge, set x1 to any number larger than 3 and solve for x2: when x1 = 4, then x2 = 4. Pass a line through (3,0) and (4,4) for the binding edge of the constraint. The feasible ">" side is to the right.

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0


Feasible side

Blending


Constraint 3: x1 + x2 > 4. Binding edge: x1 + x2 = 4. On the edge, When x1 = 0, then x2 = 4; when x2 = 0, then x1 = 4. Connect (4,0) and (0,4) for the binding edge of the constraint. The feasible ">" side is above that line.

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0


Feasible side

Blending


4x1 - x2 > 124x1 - x2 > 12

2x1 + 5x2 > 102x1 + 5x2 > 10

Feasible region

x1 + x2 > 4x1 + x2 > 4

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Blending


4x1 - x2 > 124x1 - x2 > 12

2x1 + 5x2 > 102x1 + 5x2 > 10

x1 + x2 > 4x1 + x2 > 4

Min 5x1 + 2x2Min 5x1 + 2x2 Set the objective function equal to an arbitrary

constant (say, 20) and graph it. For 5x1 + 2x2 = 20, when x1 = 0, then x2 = 10; when x2= 0, then x1 = 4. Connect (4,0) and (0,10).

Move the objective function line in the direction that lowers its value (down), since we are minimizing, until it touches the last point of the feasible region, determined in Example 2 by the first two constraints

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Blending


Binding at optimum: 4x1 - x2 = 12Binding at optimum: 4x1 - x2 = 12

Binding at optimum: x1 + x2 = 4Binding at optimum: x1 + x2 = 4

Min 5x1 + 2x2Min 5x1 + 2x2

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Blending


x2 = det / det = (4x4-12x1)/(4x1+1x1) = 4/5

4 121 4

12 -1 4 1

The optimal solution (x1 = 5, x2 = 3) is where the second and third constraints bind:

4x1 - x2 = 12 and x1 + x2 = 4 Solve those equalities using linear algebra,

matricies, determinates (det), and Cramer’s rule:

4 -11 1

x1

x2

124 =

x1 = det / det = (12x1+1x4)/(4x1+1x1) = 16/5

4 -11 1

4 -11 1

4x1 - x2 = 12 x1 + x2 = 4

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Blending


Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

How many pounds of each seed should be in the blend?

16/5 pounds of Type A seed; 4/5 pounds of Type B seed.

Which targets will be exceeded? The traffic resistance and draught resistance constraints are binding at the optimum, but the shade tolerance target is exceeded (the constraint is slack).

How much will the blend cost?

5x1 + 2x2 = 5(16/5) + 2(4/5) = 88/5 = 17.6 dollars.

Blending


Graph the feasible solutions for each of the constraints.

Determine the feasible region that satisfies all the constraints simultaneously.

Graph an objective function line. Move parallel objective function lines toward smaller

objective function values without entirely leaving the feasible region.

Any feasible solution on the objective function line with the smallest value is an optimal solution.

Blending


Min 5x1 + 2x2 + 0s1 + 0s2 + 0s3

s.t. 2x1 + 5x2 - s1 = 10

4x1 - x2 - s2 = 12

x1 + x2 - s3 = 4

x1, x2, s1, s2, s3 > 0s1 , s2 , and s3 are surplus variables

Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Blending


Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Blending


Min 5x1 + 2x2

s.t. 2x1 + 5x2 > 10

4x1 - x2 > 12

x1 + x2 > 4

x1, x2 > 0

Optimal Solution: Minimized cost = 17.6 Optimal x1 = 3.2, x2 = 0.8 0.4 surplus in the first constraint

Blending


Non-Unique Optimal Solutions



Overview

Non-Unique Optimal Solutions of a linear program mean there are either alternative optimal solutions, no solutions because it is infeasible, or no solutions because the objective function can be feasibly improved without bound.



The feasible region for a two-variable LP problem can be empty (nonexistent), a single point, a line, a polygon, or an unbounded area.

Any linear program falls in one of four categories:• has a unique optimal solution• has alternative optimal solutions• has no solution because it is infeasible • has no solution because the objective function can

be feasibly improved without bound A feasible region may be unbounded and yet there may

be optimal solutions. This is common in practical minimization problems and sometimes occurs in practical maximization problems.



In the graphical method, if the objective function line is parallel to a boundary constraint in the direction of optimization, there are alternate optimal solutions, with all points on this line segment being optimal.

For example, consider the following LP problem:

Max 4x1 + 6x2

s.t. x1 < 6

2x1 + 3x2 < 18

x1 + x2 < 7

x1 > 0 and x2 > 0

Non-Unique Optimal Solutions Alternative optimal solutions


x1

x2

7

6

5

4

3

2

1

1 2 3 4 5 6 7 8 9 10

2x1 + 3x2 < 18

x1 + x2 < 7

x1 < 6

Max 4x1 + 6x2A

B

Non-Unique Optimal Solutions Graph of alternative optimal solutions

The binding edge of constraint 2x1 + 3x2 < 18 is parallel to all objective-function lines of 4x1 + 6x2

All points on segment A – B are optimal solutions.


Infeasibility means no solution to the LP problem satisfies all the constraints, including the non-negativity conditions.

Graphically, this means the feasible region is empty (does not exist).

Causes include:• A formulation error has been made.• Management’s expectations are too high.• Too many restrictions have been placed on the

problem (that is, the problem is over-constrained).

Non-Unique Optimal Solutions Conditions for infeasibility


x2

x1

4x1 + 3x2 < 12

2x1 + x2 > 8

2 4 6 8 10

4

8

2

6

10

Max 2x1 + 6x2

s.t. 2x1 + x2 > 8

4x1 + 3x2 < 12

x1, x2 > 0

There are no points that satisfy both constraints of the problem, so the feasible region is empty (there are no feasible solutions).

Non-Unique Optimal Solutions Graph of infeasibility


The value of the solution to a maximization LP problem is unbounded if the value of the solution may be made indefinitely large without violating any of the constraints.

For practical problems, this is the result of improper formulation. Most often, a constraint has been inadvertently omitted.

Here is an example:

Non-Unique Optimal Solutions Unbounded objective-function values

Max 4x1 + 5x2

s.t. 3x1 + x2 > 8

x1 + x2 > 5

x1, x2 > 0


Max 4x1 + 5x2

s.t. 3x1 + x2 > 8

x1 + x2 > 5

x1, x2 > 0

x2

x1

3x1 + x2 > 8

x1 + x2 > 5

Max 4x1 + 5x

2

6

8

10

2 4 6 8 10

4

2

Non-Unique Optimal Solutions Graph of unbounded function values

The feasible region is unbounded and the objective function line can be moved outward from the origin without bound, infinitely increasing the objective function.


Blending with Mixed Constraints



Overview

Blending with Mixed Constraints help production managers blend resources (grass seeds, desserts, horse food, …) when constraints restrict permissible blends by a mixture of minimal and maximal product characteristics (shade tolerance, calories, vitamin content, …). For example, a blend of desserts should have at least a minimal weight to be satisfying and at most a maximal number of calories and fat to be healthful.




Question: Maggie Stewart loves twinkies and ho ho’s, but due to weight and cholesterol concerns she has decided she must plan her desserts carefully. Maggie allows herself no more then 450 calories and 25

grams of fat in her daily desserts. She requires at least 120 grams of desserts a day. Each dessert also has a “taste index”: 37 for each twinkie gram, 65 for each ho ho gram.

Plan her daily desserts to stay within her constraints and maximize the total taste index? What further information do you need for your answer?


Answer: First, formulate the linear program: Step 1. Identify decision

variables: x1 = daily servings of Twinkies, x2 = daily servings of Ho Ho’s.

Step 2. Identify remaining information needed:• Each Twinkie serving has

150 calories, has 4.5 grams of fat, weighs 1.5 ounces.

• 1.5 ounces = 42.5243 grams (since 1 ounce = 28.3495 grams).



• Each Ho Ho serving has 380 calories, has 17 grams of fat, weighs 3 ounces.

• 3 ounces = 85.0485 grams (since 1 ounce = 28.3495 grams).



Adding the non-negativity of consumption completes the formulation.

Max 37(42.5243)x1 + 65(85.0485)x2

s.t. 150x1 + 380x2 < 450

4.5x1 + 17x2 < 25

42.5243x1 + 85.0485x2 > 120

x1 > 0 and x2 > 0

Taste index

Non-negativity constraints

Calorie constraint

Fat constraint

Weight constraint



2.1541 servings of Twinkies each day. 0.3339 servings of Ho Ho’s each day.

Max 37(42.5243)x1 + 65(85.0485)x2

s.t. 150x1 + 380x2 < 450

4.5x1 + 17x2 < 25

42.5243x1 + 85.0485x2 > 120

x1 > 0 and x2 > 0



End of Lesson I.3

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BA 452 Lesson A.3 Unsolvable Linear Programs 1 1ReadingsReadings Chapter 2 An Introduction to Linear...

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