Transcript A.3 Unsolvable Linear Programs

Readings
Readings
Chapter 2
An Introduction to Linear Programming
BA 452 Lesson A.3 Unsolvable Linear Programs
1
Overview
Overview
BA 452 Lesson A.3 Unsolvable Linear Programs
2
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.
BA 452 Lesson A.3 Unsolvable Linear Programs
3
Blending
Blending
BA 452 Lesson A.3 Unsolvable Linear Programs
4
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,
…).
BA 452 Lesson A.3 Unsolvable Linear Programs
5
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 B
Shade tolerance
2
5
Traffic resistance
4
-1
Drought 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.)
BA 452 Lesson A.3 Unsolvable Linear Programs
6
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?
BA 452 Lesson A.3 Unsolvable Linear Programs
7
Blending
Answer: First, formulate the problem as a linear program:
Objective: minimize cost of blending x1 pounds of
Type A seed with x2 pounds of Type B seed
Min
s.t.
5x1 + 2x2
2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x 1, x 2 > 0
Constraint 1: Shade tolerance
Constraint 2: Traffic resistance
Constraint 3: Drought resistance
Non-negativity constraints
BA 452 Lesson A.3 Unsolvable Linear Programs
8
Blending

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
Feasible side
Binding edge
BA 452 Lesson A.3 Unsolvable Linear Programs
9
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
Binding edge
BA 452 Lesson A.3 Unsolvable Linear Programs
10
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
Binding edge
BA 452 Lesson A.3 Unsolvable Linear Programs
11
Blending
Min 5x1 + 2x2
s.t. 2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
Feasible region
4x1 - x2 > 12
x1 + x2 > 4
2x1 + 5x2 > 10
BA 452 Lesson A.3 Unsolvable Linear Programs
12
Blending
Min 5x1 + 2x2
s.t. 2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
Min 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
4x1 - x2 > 12
x1 + x2 > 4
2x1 + 5x2 > 10
BA 452 Lesson A.3 Unsolvable Linear Programs
13
Blending
Min 5x1 + 2x2
s.t. 2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
Min 5x1 + 2x2
Binding at optimum:
4x1 - x2 = 12
Binding at optimum:
x1 + x2 = 4
BA 452 Lesson A.3 Unsolvable Linear Programs
14
Blending


Min 5x1 + 2x2
s.t. 2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
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:
4x1 - x2 = 12
x 1 + x2 = 4
4
1
4
12 -1
x1 = det 4 1 / det 1
-1
1
4
-1
x2 = det
4 12
1 4
/ det
1
-1
1
x1
x2
=
12
4
= (12x1+1x4)/(4x1+1x1) = 16/5
= (4x4-12x1)/(4x1+1x1) = 4/5
1
BA 452 Lesson A.3 Unsolvable Linear Programs
15
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.
BA 452 Lesson A.3 Unsolvable Linear Programs
16
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.
BA 452 Lesson A.3 Unsolvable Linear Programs
17
Blending
Min 5x1 + 2x2
s.t. 2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
Min
5x1 + 2x2 + 0s1 + 0s2 + 0s3
s.t.
2x1 + 5x2 - s1
= 10
4x1 - x2
- s2
= 12
x1 + x 2
- s3 = 4
x 1, x 2, s 1, s 2, s 3 > 0
s1 , s2 , and s3 are
surplus variables
BA 452 Lesson A.3 Unsolvable Linear Programs
18
Blending
Min 5x1 + 2x2
s.t. 2x1 + 5x2 > 10
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
BA 452 Lesson A.3 Unsolvable Linear Programs
19
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
BA 452 Lesson A.3 Unsolvable Linear Programs
20
Non-Unique Optimal Solutions
Non-Unique Optimal Solutions
BA 452 Lesson A.3 Unsolvable Linear Programs
21
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.
BA 452 Lesson A.3 Unsolvable Linear Programs
22
Non-Unique Optimal Solutions



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.
BA 452 Lesson A.3 Unsolvable Linear Programs
23
Non-Unique Optimal Solutions


Alternative optimal solutions
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
s.t.
4x1 + 6x2
x1
< 6
2x1 + 3x2 < 18
x1 + x2 < 7
x1 > 0 and x2 > 0
BA 452 Lesson A.3 Unsolvable Linear Programs
24
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.
x2
x1 + x 2 < 7
7
6
5
A
B
4
x1 < 6
2x1 + 3x2 < 18
3
2
1
1
2
3
4
5
6
7
8
9
x1
10
BA 452 Lesson A.3 Unsolvable Linear Programs
25
Non-Unique Optimal Solutions



Conditions for infeasibility
Infeasibility means no solution to the LP problem
satisfies all the constraints, including the nonnegativity 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).
BA 452 Lesson A.3 Unsolvable Linear Programs
26
Non-Unique Optimal Solutions
Graph of infeasibility
There are no points that satisfy both
constraints of the problem, so the feasible
region is empty (there are no feasible
solutions).
x2
10
2x1 + x2 > 8
8
6
4x1 + 3x2 < 12
4
2
2
4
6
8
Max
2x1 + 6x2
s.t.
2x1 + x2 > 8
4x1 + 3x2 < 12
x1, x2 > 0
x1
10
BA 452 Lesson A.3 Unsolvable Linear Programs
27
Non-Unique Optimal Solutions



Unbounded objective-function values
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:
Max 4x + 5x
1
s.t.
2
3x1 + x2 > 8
x1 + x2 > 5
x1, x2 > 0
BA 452 Lesson A.3 Unsolvable Linear Programs
28
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.

x2
10
3x1 + x2 > 8
8
6
Max
4x1 + 5x2
4
s.t.
3x1 + x2 > 8
x1 + x2 > 5
x1, x2 > 0
x1 + x 2 > 5
2
2
4
6
8
x1
10
BA 452 Lesson A.3 Unsolvable Linear Programs
29
Blending with Mixed Constraints
Blending with Mixed Constraints
BA 452 Lesson A.3 Unsolvable Linear Programs
30
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.
BA 452 Lesson A.3 Unsolvable Linear Programs
31
Blending with Mixed Constraints
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?
BA 452 Lesson A.3 Unsolvable Linear Programs
32
Blending with Mixed Constraints
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).
BA 452 Lesson A.3 Unsolvable Linear Programs
33
Blending with Mixed Constraints
•
•
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).
BA 452 Lesson A.3 Unsolvable Linear Programs
34
Blending with Mixed Constraints

Adding the non-negativity of consumption
completes the formulation.
Max
s.t.
37(42.5243)x1 + 65(85.0485)x2
150x1 +
380x2 < 450
4.5x1 +
17x2 < 25
42.5243x1 + 85.0485x2 > 120
x1 > 0 and x2 > 0
Taste index
Calorie constraint
Fat constraint
Weight constraint
Non-negativity
constraints
BA 452 Lesson A.3 Unsolvable Linear Programs
35
Blending with Mixed Constraints
Max
s.t.


37(42.5243)x1 + 65(85.0485)x2
150x1 +
380x2 < 450
4.5x1 +
17x2 < 25
42.5243x1 + 85.0485x2 > 120
x1 > 0 and x2 > 0
2.1541 servings of Twinkies each day.
0.3339 servings of Ho Ho’s each day.
BA 452 Lesson A.3 Unsolvable Linear Programs
36
BA 452
Quantitative Analysis
End of Lesson I.3
BA 452 Lesson A.3 Unsolvable Linear Programs
37