Linear Programming (LP)
A mathematical model consists of linear relationships representing the objective and
resource constraints.
Terminology
•
Decision Variables: a set of input variables whose values you can change.
•
e.g. The number of units of a product we should produce.
•
e.g. The amount of material we should buy.
•
Objective Function: an output function that you want to maximize or
minimize.
•
e.g. Maximize profit, yield, contribution.
•
e.g. Minimize cost, space used, time taken, or risk.
•
Constraints: a set of restrictions on the permissible values (or
combinations of values) of the input variables.
•
e.g. Cannot produce less than 0 units of a product.
•
e.g. Cannot use more than 1,000 yards of material.
Terminology for Solutions
Feasible solution: A solution for which all the constraints are satisfied. It is
possible for a problem to have no feasible solutions.
Optimal solution: A feasible solution that has the most favorable value of the
objective function.
one optimal solution
multiple optimal solutions
no optimal solutions
Solving LP Problems
Graphical Solution – used mainly as a teaching tool
Simplex Method – most common analytic tool
Steps in Formulating a Linear Programming (LP) Model
1. Understand the problem.
2. Identify the decision variables.
3. State the objective function as a linear combination of the decision variables.
4. State the constraints:
upper or lower bounds on the decision variables, including non-negativity constraints if
applicable.
linear combinations of the decision variables.
LP Model Format
Maximize or Minimize
z = c1x1 + c2x2 + ... + cnxn
subject to:
a11x1 + a12x2 + ... + a1nxn (≤, =, ≥) b1
a21x1 + a22x2 + ... + a2nxn (≤, =, ≥) b2
:
am1x1 + am2x2 + ... + amnxn (≤, =, ≥) bm
z = objective
xj = decision variables
cj = objective function coefficients
aij = constraint coefficients
bi = constraint levels
Integer Program: Requires decision variables to be integers.
LP Model: Example
Determine the number of each product to produce each day in order to maximize
revenue subject to the available resources
There are 40 hours of labor and 120 pounds of clay available each day
Decision variables
x1 = number of bowls to produce
x2 = number of mugs to produce
LP Model
Maximize Z = $40 x1 + 50 x2
Subject to
x1
+
2x2
£ 40 hr
(labor constraint)
4x1
+
3x2
£ 120 lb
(clay constraint)
x1 , x2 ³ 0
Solution is
x1 = 24 bowls, x2 = 8 mugs
Total revenue each day = 40 x 24 + 50 x 8 = $1,360
Graphical Solution Method
Plot model constraint on a set of coordinates in a plane
Identify the feasible solution space on the graph where all constraints are satisfied
simultaneously
Plot objective function to find the point on boundary of this space that maximizes (or
minimizes) value of objective function
Graphical Solution
Find the intersection of any two linear constraint lines.
Optimal Solution Always Located on an intersection point.
Solution may change when the coefficients in the objective function change.
Minimization Problem
A farmer wants to blend two brands of fertilizers (Gro-Plus, $6/bag and Crop-fast,
$3/bag) to minimize cost while getting 16 lbs. of nitrogen and 24 lbs. of phosphate.
Graphical Solution
0
