FORS 4710 / 6710 Forest Planning
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Transcript of FORS 4710 / 6710 Forest Planning
FORS 4710 / 6710 Forest Planning
FORS 8450 • Advanced Forest Planning
Lecture 2
Linear Programming
Mathematical programming
A name given to a group of techniques that take problems explicitlydefined 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 programmingtechnique for forest-level problems.
Others include:
Non-linear programmingMonte Carlo simulationBinary searchHeuristicsDynamic 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 thanwhat 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 otheritems.
Linear Programming Components
D. Policy Constraints
These are constraints that prevent the problem from scheduling activitiesbased 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 linesrepresenting the objective function and constraints.
2. By viewing a linear problem as a set of equations to be solvedwith 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 1000 X1 + 500 X2
Constraints:LandBudget LaborContractNonnegativity
4 X1 + 1.5 X2 24240 X1 + 30 X2 120020 X1 + 20 X2 200X1 2X1 0X2 0
Scott Bauer, USDA ARS, www.forestryimages.org
Image number 1320069
David J. Moorhead, The University of Georgia, www.forestryimages.org
Image number 2651045
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
Land constraint:4 X1 + 1.5 X2 24
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
4 X1 + 1.5 X
2 24
Set one of the two variables to 0,solve for the other.
4 (0) + 1.5 (X2) = 24 1.5 (X2) = 24 X2 = (24 / 1.5) X2 = 16
So X2 = 16 when X1 = 0, creating a coordinate for a end point of the lineof (0, 16).
Solving X1 when X2 = 0 yields thecoordinate (6, 0).
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
4 X1 + 1.5 X
2 24
Feasible region
The set of all points that satisfy theconstraints.
The area of the graph where potentialsolutions do not violate the constraints.
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Budget Constraint:240 X1 + 30 X2 1200
Endpoints of line:
When X1 = 0, XY coords are: (0, 40)
When X2 = 0, XY coords are: (5, 0)
240 X1 + 30 X
2 1200
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Labor Constraint: 20 X1 + 20 X2 200
Endpoints of line:
When X1 = 0, XY coords are: (0, 10)
When X2 = 0, XY coords are: (10, 0)
20 X1 + 20 X
2 200
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Contract Constraint: X1 2
Endpoints of line:
A little different, since X1 is not in theequation. It is a vertical line at X1 = 2.X
1
2
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Nonnegativity Constraint: X1 0 X2 0
The first equation is redundant - we already have an equation X1 2.
X2 0
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Each point within the gray box representsa feasible solution to the problem.
This is the solution space.
The question now is to determine whichof the areas within the feasible regioninclude inefficient solutions, and which represent the optimal solution.
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
To find the most efficient (or optimal)solution, we must now display the objective function.
Objective function: 1000 X1 + 500 X2
This isn't a constraint - there is noright hand side to the equation (noequality or inequality).
Various levels of X1 and X2 can be used,and it is our job to determine how many of each.
$9,000
$6,000$3,000
(9, 0)(6, 0)(3, 0)
(0, 6)
(0, 12)
(0, 18)
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Graphically, we find the place in thefeasible region of the solution spacethat touches the objective functionvalue line of the highest value.
Lines are developed by placing a proposed objective function valueas the right-hand side of theobjective function, for example:
1000 X1 + 500 X2 = 9000
(3, 0)
$9,000
$6,000
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Graphically, we find the place in thefeasible region of the solution spacethat touches the objective functionvalue line of the highest value.
(3, 0)
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Graphically, we find the place in thefeasible region of the solution spacethat touches the objective functionvalue line of the highest value.
(3, 0)
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
(3, 0)3.6
6.4
The optimal solution is foundat the point (3.6, 6.4), which meansthat we should produce:
3.6 steer6.4 lots of trees
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
(3, 0)
What does linear programming do?It defines the solution space, aswe have done here graphically.
Then, it identifies the corners, andsearches from corner to corner forimprovements in the objective functionvalue.
The corner with the highest objective function value is the optimal solution.
FORS 4710 / 6710 Forest Planning
Viewing a linear problem as a graph
Steers and trees problem
5
10
15
20
2 4 6 8 10
X1, number of steers
X2,
nu
mb
er o
f tr
ee lo
ts
Solution space (gray polygon)
Optimal solution
Inefficient, feasible solution
Infeasible solution
FORS 4710 / 6710 Forest Planning
Linear Programming Output
Problem summary
Objective function:maximize 1000 X1 + 500 X2
Constraints:LandBudget LaborContractNonnegativity
4 X1 + 1.5 X2 24240 X1 + 30 X2 120020 X1 + 20 X2 200X1 2X2 0
Linear Programming Output
Problem formulation within LINDO
Max 1000 X1 + 500 X2subject to2) 4 X1 + 1.5 X2 <= 243) 240 X1 + 30 X2 <= 12004) 20 X1 + 20 X2 <= 2005) X1 >= 26) X2 >= 0end
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) 6800.000
VARIABLE VALUE REDUCED COST X1 3.600000 0.000000 X2 6.400000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 200.000000 3) 144.000000 0.000000 4) 0.000000 10.000000 5) 1.600000 0.000000 6) 6.400000 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 amountof 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 themaximum 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 X2subject to2) 4 X1 + 1.5 X2 <= 253) 240 X1 + 30 X2 <= 12004) 20 X1 + 20 X2 <= 2005) X1 >= 26) X2 >= 0end
The objective function valueshould change $200 - - - ifsome other constraint doesnot, in the meantime, becomebinding within the new problem.
Linear Programming Output
OBJECTIVE FUNCTION VALUE
1) 7000.000
VARIABLE VALUE REDUCED COST X1 4.000000 0.000000 X2 6.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 200.000000 3) 60.000000 0.000000 4) 0.000000 10.000000 5) 2.000000 0.000000 6) 6.000000 0.000000
LINDO solution values
We find, in fact, that the objective function value increases $200...
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.
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
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.