Linear programming is a procedure for deciding the best way to allocate resources.
"Best" may mean fastest, most profitable, least expensive, or best by whatever criterion is appropriate. For linear programming to be applicable, the problem must have some special features. These will be illustrated by an example.
In a primitive economy, a man decides to earn a living by making hinges and gate latches. He is able to obtain a supply of 25 kilograms a week of suitable metal at a price of 2 cowrie shells per kilogram. His design requires 500 grams to make a hinge and 250 grams to make a gate latch. With his primitive tools, he finds that he can make a hinge in 1 hour, and it takes 3/4 hour to make a gate latch. He is willing to work 60 hours a week. The going price is 3 cowrie shells for a hinge and 2 cowrie shells for a gate latch. How many hinges and how many gate latches should he produce each week in order to maximize his net income?
To analyze the problem, let x be the number of hinges produced per week and let y be the number of gate latches. Then the amount of metal used is (O.Sx + 0.25y) kilograms. Clearly, this must be less than or equal to 25 kilograms:
O.Sx + 0.25y s 25 time taken making gate latches cannot exceed 60 hours. Therefore,
lx + 0.75y s 60
The mathematical problem can now be stated as follows: Find the point (x, y) that max
imizes the objective function R(x, y) = 3x + 2y- 50, subject to the linear constraints feasible set either meets the set in a single line segment or only touches the set on its boundary. In particular, because of the way the feasible set is defined in terms of linear inequalities, it turns out that it is impossible for one line to meet the feasible set in two separate pieces.
For example, the shaded region if Figure 2.4.10 cannot possibly be the feasible set for a linear programming problem, because some lines meet the region in two line segments. (This property of feasible sets is not difficult to prove, but since this is only a brief illustration, the proof is omitted.)
y
50 x
lines
R(x, y)
= constant"-- 2x
+y
=100
Figure
2.4.9 The feasible region for the linear programming example. The grey lines are level sets of the objective functionR.
Now consider sets of the form
R(x,
y) = k, where k is a constant; these are called the level sets ofR.
These sets obviously form a family of parallel lines, and some of them are shown in Figure2.4.9.
Choose some point in the feasible set: check that(20, 20)
is such a point. Then the lineR(x,y)
=R(20,20)
=50
y
x
Figure
2.4.10 The shaded region cannot be the feasible region for a linear programming problem because it meets a line in two segments.
Section 2.4 Applications of Systems of Linear Equations 109
meets the feasible set in a line segment. (30, 30) is also a feasible point (check), and
R(x, y) = R(30, 30) = 100
also meets the feasible set in a line segment. You can tell that (30, 30) is not a boundary point of the feasible set because it satisfies all the constraints with strict inequality;
boundary points must satisfy one of the constraints with equality.
As we move further from the origin into the first quadrant, R(x, y) increases. The biggest possible value for R(x, y) will occur at a point where the set R(x, y) = k (for some constant k to be determined) just touches the feasible set. For larger values of R(x, y), the set R(x, y) = k does not meet the feasible set at all, so there are no feasible points that give such bigger values of R. The touching must occur at a vertex-that is, at an intersection point of two of the boundary lines. (In general, the line R(x,y) = k for the largest possible constant could touch the feasible set along a line segment that makes up part of the boundary. But such a line segment has two vertices as endpoints, so it is correct to say that the touching occurs at a vertex.)
For this particular problem, the vertices of the feasible set are easily found to be (0, 0), (50, 0), and (0, 80), and the solution of the system of equations is
2x + y = 100 4x + 3y = 240
For this particular problem, the vertices of the feasible set are (0, 0), (50, 0), (0, 80), and (30, 80). Now compare the values of R(x, y) at all of these vertices: R(O, 0), R(50, 0), R(O, 80) = 110, and R(30, 40) = 120. The vertex (30, 40) gives the best net revenue, so the producer should make 30 hinges and 40 gate latches each week.
General Remarks
Problems involving allocation of resources are common in business and government. Problems such as scheduling ship transits through a canal can be analyzed this way. Oil companies must make choices about the grades of crude oil to use in their refineries, and about the amounts of various refined products to produce. Such problems often involve tens or even hundreds of variables-and similar numbers of constraints. The boundaries of the feasible set are hyperplanes in some IR", where n is large. Although the basic principles of the solution method remain the same as in this example (look for the best vertex), the problem is much more com
plicated because there are so many vertices. In fact, it is a challenge to find vertices;
simply solving all possible combinations of systems of boundary equations is not good enough. Note in the simple two-dimensional example that the point (60, 0) is the inter
section point of two of the lines (y = 0 and 4x + 3y = 240) that make up the boundary, but it is not a vertex of the feasible region because it fails to satisfy the constraint 2x + y $ 100. For higher-dimension problems, drawing pictures is not good enough, and an organized approach is called for.
The standard method for solving linear programming problems has been the sim
plex method, which finds an initial vertex and then prescribes a method (very similar to row reduction) for moving to another vertex, improving the value of the objective function with each step.
Again, it has been possible to hint at major application areas for linear algebra, but to pursue one of these would require the development of specialized mathematical tools and information from specialized disciplines.
PROBLEMS 2.4
Practice Problems
Al Determine the system of equations for the reaction forces and axial forces in members for the truss shown in the diagram.
E
CHAPTER REVIEW
Suggestions for Student Review
1 Explain why elimination works as a method for solv
ing systems of linear equations. (Section 2.1)
2 When you row reduce an augmented matrix
[A
Ib]
tosolve a system of linear equations, why can you stop when the matrix is in row echelon form? How do you use this form to decide if the system is consistent and if it has a unique solution? (Section 2.1)
3 How is reduced row echelon form different from row echelon form? (Section 2.2)
4 (a) Write the augmented matrix of a consistent non
homogeneous system of three linear equations in four variables, such that the coefficient matrix is
A2 Determine the augmented matrix of the system of linear equations, and determine the loop currents indicated in the diagram.
i E1 R1 l1 R1 R3
R6 Rs
R1 E2 l
A3 Find the maximum value of the objective function x + y subject to the constraints 0 $ x $ 100, 0 $ y $ 80, and 4x + Sy $ 600. Sketch the feasible region.
in row echelon form (but not reduced row echelon form) and of rank 3.
(b) Determine the general solution of your system.
(c) Perform the following sequence of elementary row operations on your augmented matrix:
(i) Interchange the first and second rows.
(ii) Add the (new) second row to the first row.
(iii) Add twice the second row to the third row.
(iv) Add the third row to the second.
(d) Regard the result of (c) as the augmented matrix of a system and solve that system directly. (Don't just use the reverse operation in (c).) Check that your general solution agrees with (b ).
5 For homogeneous systems, how can you use the row echelon form to determine whether there are non
trivial solutions and, if there are, how many parame
ters there are in the general solution? Is there any case where we know (by inspection) that a homogeneous system has non-trivial solutions? (Section 2.2) 6 Write a short explanation of how you use informa
tion about consistency of systems and uniqueness of solutions in testing for linear independence and in de
termining whether a vector x belongs to a given sub
space of JR11• (Section 2.3)
Chapter Quiz
El Determine whether the following system is consis
tent by row reducing its augmented matrix:
X2 -2x3 + X4 = 2 such that the system is inconsistent.
Chapter Review 111
have if the set is a linearly independent spanning set?
(Section 2.3) why there must be non-zero vectors orthogonal to all of a, v, and w. non-homogeneous system of linear equations, then the system must be inconsistent.
(c) Some homogeneous systems of linear equa
tions have unique solutions.
(d) If there are more variables than equations in a system of linear equations, then the system can
not have a unique solution.
Further Problems
These problems are intended to be challenging. They may not be of interest to all students.
Fl The purpose of this exercise is to explore the relationship between the general solution of the system
[A
Ib]
and the general solution of the corresponding homogeneous system[A
I0 ].
This relation will be studied with different tools in Section 3.4. We begin by considering some examples where the coefficient matrix is in reduced row echelon form.
solution of the homogeneous system
where each of v1 and v2 can be expressed in
The pattern should now be apparent; if it is not, try again with another special case of R. In the next part of this exercise, create an effective la
belling system so that you can clearly indicate what you want to say.
consistent and show that the general solution is
where
jJ
is expressed in terms of the compo important in many industrial and commercial applications. It also tends to improve accuracy: every arithmetic operation is an opportunity to lose accu
racy through truncation or round-off, subtraction of two nearly equal numbers, and so on.
We want to count the number of multiplications and/or divisions in solving a system by elimination.
We focus on these operations because they are more time-consuming than addition or subtraction, and the number of additions is approximately the same as the number of multiplications. We make certain assumptions: the system
[A
Ib]
has n equations and n variables, and it is consistent with a unique solution. (Equivalently,A
has n rows, n columns, and rank n.) We assume for simplicity that no row interchanges are required. (If row interchanges are required, they can be handled by renaming"addresses" in the computer.)
(a) How many and divisions are re
(b) Determine how many multiplications and divi
sions are required to solve the system with the
substitution. Conclude that the Gauss-Jordan procedure is as efficient as Gaussian elim
ination with back-substitution. For large n, the number of multiplications and divisions is of multiplications and divisions required in this procedure is roughly so that this procedure
requires 50% more operations
than the more efficient procedures.