FORS 8450 • Advanced Forest Planning
Lecture 2
Linear Programming
FORS 4710 / 6710 Forest Planning
Mathematical programming
A name given to a group of techniques that take problems explicitly
defined with an objective function and a set of constraints, and
efficiently search through solutions to try to find the one that is
optimum, with respect to the objective function.
Linear programming is the most widely used mathematical programming
technique for forest-level problems.
Others include:
Non-linear programming
Monte Carlo simulation
Binary search
Heuristics
Dynamic programming
FORS 4710 / 6710 Forest Planning
Linear Programming
Although developed independently a few years earlier by Russian and
Dutch scientists, "Linear programming" was formalized in a general form
in 1947 when colleagues of George Dantzig suggested he use the term
as a descriptor for the solution technique he had developed for the
U.S. Air Force.
The reasoning behind the choice of words was: (a) Dantzig's problem-solving
methods were based on linear equations, and (b) the term "programming"
was (in 1947) synonymous with "planning" in the military.
Many, if not all of the U.S. National Forest plans developed from about
1970 to today have used linear programming.
In addition, a number of industrial companies, state forest management
organizations, and Canadian Provinces also use linear programming for
the development of forest plans.
Linear Programming Components
A. Objective Function
An objective function for a linear programming model is a linear
function which helps you evaluate the quality of a solution to a problem.
It is used by the model to evaluate all potential combinations of
management actions.
Associated with objective functions is that notion that something is
either being maximized or minimized.
As you begin the development of the objective function, you must first
remember what you are attempting to do.....
Linear Programming Components
B. Accounting Rows
Accounting rows are used to aggregate values that may be needed for
reporting purposes, although the values accumulated through an
accounting row could also be used to constrain the solution to a problem.
As an example, a natural resource management organization may be
developing a plan that maximizes the net present value of activities over
a 20-year time horizon and subject to a variety of constraints.
The management organization may also desire to understand how much
habitat, or how much timber volume is produced each time period.
To do so, they would need to somehow add up the habitat or volume
produced in each time period.
To develop an accounting row, you must understand what values
you are trying to accumulate.
Linear Programming Components
C. Resource Constraints
These are constraints that prevent the problem from using more than
what might be available to the organization.
For example, stand-level resource constraints might indicate that the
sum of the area assigned for harvest over the time horizon should not
exceed the total size of each stand.
Resource constraints could also reflect limiting budgets and other
items.
Linear Programming Components
D. Policy Constraints
These are constraints that prevent the problem from scheduling activities
based on landowner policies.
• Constraints on harvested area
• Constraints on harvested volume
• Constraints on habitat development or maintenance
Linear Programming
We will examine LP two ways:
1. By viewing a linear problem as a graph with a set of lines
representing the objective function and constraints.
2. By viewing a linear problem as a set of equations to be solved
with a piece of LP software (either LINDO or SOLVER).
FORS 4710 / 6710 Forest Planning
Viewing a linear
problem as a
graph
Steers and Trees from Davis et al. (2001)
FORS 4710 / 6710 Forest Planning
Steers and Trees Problem
Problem summary
Objective function:
maximize
Constraints:
Land
Budget
Labor
Contract
Nonnegativity
Scott Bauer, USDA ARS, www.forestryimages.org
Image number 1320069
FORS 4710 / 6710 Forest Planning
1000 X1 + 500 X2
4 X1 + 1.5 X2  24
240 X1 + 30 X2  1200
20 X1 + 20 X2  200
X1  2
X1  0
X2  0
David J. Moorhead, The University of Georgia, www.forestryimages.org
Image number 2651045
Viewing a linear problem as a graph
Steers and trees problem
Land constraint:
4 X1 + 1.5 X2  24
Set one of the two variables to 0,
solve for the other.
X2, number of tree lots
20
4 (0) + 1.5 (X2) = 24
1.5 (X2) = 24
X2 = (24 / 1.5)
X2 = 16
15
10
So X2 = 16 when X1 = 0, creating a
coordinate for a end point of the line
of (0, 16).
5
Solving X1 when X2 = 0 yields the
coordinate (6, 0).
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Feasible region
The set of all points that satisfy the
constraints.
X2, number of tree lots
20
The area of the graph where potential
solutions do not violate the constraints.
15
10
5
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Budget Constraint:
240 X1 + 30 X2  1200
Endpoints of line:
X2, number of tree lots
20
When X1 = 0, XY coords are: (0, 40)
15
When X2 = 0, XY coords are: (5, 0)
10
5
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Labor Constraint:
20 X1 + 20 X2  200
Endpoints of line:
X2, number of tree lots
20
When X1 = 0, XY coords are: (0, 10)
15
When X2 = 0, XY coords are: (10, 0)
10
5
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Contract Constraint:
X1  2
Endpoints of line:
15
X1  2
X2, number of tree lots
20
A little different, since X1 is not in the
equation. It is a vertical line at X1 = 2.
10
5
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Nonnegativity Constraint:
X1  0
X2  0
X2, number of tree lots
20
The first equation is redundant we already have an equation X1  2.
15
10
5
X2  0
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
X2, number of tree lots
Each point within the gray box represents
a feasible solution to the problem.
20
This is the solution space.
15
The question now is to determine which
of the areas within the feasible region
include inefficient solutions, and which
represent the optimal solution.
10
5
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
To find the most efficient (or optimal)
solution, we must now display the
objective function.
X2, number of tree lots
20
(0, 18)
Objective function: 1000 X1 + 500 X2
This isn't a constraint - there is no
right hand side to the equation (no
equality or inequality).
15
(0, 12)
Various levels of X1 and X2 can be used,
and it is our job to determine how many
of each.
10
(0, 6)
5
(3, 0)
2
4
(9, 0)
(6, 0)
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Graphically, we find the place in the
feasible region of the solution space
that touches the objective function
value line of the highest value.
X2, number of tree lots
20
15
Lines are developed by placing a
proposed objective function value
as the right-hand side of the
objective function, for example:
10
1000 X1 + 500 X2 = 9000
5
(3, 0)
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Graphically, we find the place in the
feasible region of the solution space
that touches the objective function
value line of the highest value.
X2, number of tree lots
20
15
10
5
(3, 0)
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Graphically, we find the place in the
feasible region of the solution space
that touches the objective function
value line of the highest value.
X2, number of tree lots
20
15
10
5
(3, 0)
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
The optimal solution is found
at the point (3.6, 6.4), which means
that we should produce:
X2, number of tree lots
20
3.6 steer
6.4 lots of trees
15
10
6.4
5
3.6
(3, 0)
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
What does linear programming do?
It defines the solution space, as
we have done here graphically.
X2, number of tree lots
20
Then, it identifies the corners, and
searches from corner to corner for
improvements in the objective function
value.
15
The corner with the highest objective
function value is the optimal solution.
10
5
(3, 0)
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Viewing a linear problem as a graph
Steers and trees problem
Solution space (gray polygon)
X2, number of tree lots
20
15
Optimal solution
Inefficient, feasible solution
10
5
Infeasible solution
2
4
6
8
X1, number of steers
FORS 4710 / 6710 Forest Planning
10
Linear Programming Output
Problem summary
Objective function:
maximize
1000 X1 + 500 X2
Constraints:
Land
Budget
Labor
Contract
Nonnegativity
4 X1 + 1.5 X2  24
240 X1 + 30 X2  1200
20 X1 + 20 X2  200
X1  2
X2  0
Linear Programming Output
Problem formulation within LINDO
Max 1000 X1 + 500 X2
subject to
2) 4 X1 + 1.5 X2 <= 24
3) 240 X1 + 30 X2 <= 1200
4) 20 X1 + 20 X2 <= 200
5) X1 >= 2
6) X2 >= 0
end
Objective function
Land constraint
Budget constraint
Labor constraint
Contract constraint
Nonneg. constraint
Linear Programming Output
LINDO solution values
LP OPTIMUM FOUND AT STEP
3
OBJECTIVE FUNCTION VALUE
1)
VARIABLE
X1
X2
ROW
2)
3)
4)
5)
6)
6800.000
VALUE
3.600000
6.400000
SLACK OR SURPLUS
0.000000
144.000000
0.000000
1.600000
6.400000
REDUCED COST
0.000000
0.000000
DUAL PRICES
200.000000
0.000000
10.000000
0.000000
0.000000
Linear Programming Output
LINDO solution values
Reduced costs: These relate to the decision variables. If a decision variable
had a value of 0 in the optimal solution, this is the amount that the
value of a product (in the objective function) would have to increase before
the product would appear in the optimal solution.
In other words it is....
The amount (in objective function value terms, which in your lab's case is
net present value) that the coefficient for each decision variable in the
objective function must increase before that decision becomes competitive
enough, given the other choices available, to enter the solution.
Decision variables with values other than 0 have no reduced cost.
Linear Programming Output
LINDO solution values
Slack or surplus: These relate to the constraints. They represent the amount
of the right hand side value of a constraint that is not being used in the
optimal solution.
Example: The slack for constraint equation 3 (Budget constraint) is $144.
If we check the budget constraint using the values of the optimal solution,
we find that we have not used all of the budget, in fact, we have $144 left over not used (the budget was $1,200, we used $1,056).
$240 per steer (3.6 steer) + $30 per lot of trees (6.4 lots) = $1,056
Linear Programming Output
LINDO solution values
Dual prices (shadow prices): These relate to the constraints. This is the
maximum amount that the value of the objective function would change if
one more unit of a constraint (the right hand side) were added or subtracted.
For example, if the problem were changed to this:
Max 1000 X1 + 500 X2
subject to
2) 4 X1 + 1.5 X2 <= 25
3) 240 X1 + 30 X2 <= 1200
4) 20 X1 + 20 X2 <= 200
5) X1 >= 2
6) X2 >= 0
end
The objective function value
should change $200 - - - if
some other constraint does
not, in the meantime, become
binding within the new problem.
Linear Programming Output
LINDO solution values
We find, in fact, that the objective function value increases $200...
OBJECTIVE FUNCTION VALUE
1)
VARIABLE
X1
X2
ROW
2)
3)
4)
5)
6)
7000.000
VALUE
4.000000
6.000000
SLACK OR SURPLUS
0.000000
60.000000
0.000000
2.000000
6.000000
REDUCED COST
0.000000
0.000000
DUAL PRICES
200.000000
0.000000
10.000000
0.000000
0.000000
Linear Programming Assumptions
1. Proportionality
Each variable in a linear programming problem is associated with a
coefficient (e.g., revenue per unit produced, cost per mile of road built).
The contribution of each product produced (or road built, etc.) to the
objective function is proportional to the number of units of each product
produced (or number of miles of roads of each type developed).
If one unit of X1 accounts for a change in $100 in the objective function,
then two units of X1 account for a change of $200 in the objective function.
If one unit of X1 accounts for $100 in the objective function,
and one unit of X2 accounts for $200 in the objective function,
then together they account for $300 in the objective function.
Linear Programming Assumptions
2. Additivity
Each variable in an objective function contributes to the objective function
value in a way that is independent of the other variables.
For example, the value of product A ($50 of revenue per unit produced)
does not increase or decrease with the number of units of products B and
C that are produced.
Linear Programming Assumptions
3. Certainty
With this linear programming assumption, the coefficients associated
with each variable are assumed to be known with certainty.
This implies that there are no stochastic (random) variations in the coefficients
within a linear programming model.
For example, the revenue for each unit of product A is $50,
not $50  some random interaction (e.g., $49.35 in some instances,
$52.94 in others, and so on).
In addition, when we discuss constraints to linear models, all of the
right-hand side values of the equations (the goals) are known with certainty.
Linear Programming Assumptions
4. Divisibility
The values (solution) for each variable can take on any positive number,
integer or non-integer.
The value assigned to each variable in a linear programming model is
assumed to be a continuous real number, thus can be assigned fractional
values.
For example, a solution to a maximization problem could be A = 3.24,
B = 112.94, and C = 12.
Each value must be zero or greater, if this is not apparent at first glance.
In some real-life problems, this assumption may not be valid, and require
another type of solution technique to be employed.
For example, if one were attempting to maximize the number of fish
structures (let's assume either logs or boulders) inserted into a stream
system, then a solution of 12.56 logs and 9.23 boulders would seem
unreasonable.