Chapter 6 Lectures

Report
6
Linear Programming: A Geometric Approach
 Graphing Systems of Linear
Inequalities in Two Variables
 Linear Programming Problems
 Graphical Solutions of Linear
Programming Problems
 The Simplex Method:
Standard Maximization Problems
 The Simplex Method:
Standard Minimization Problems
6.1
Graphing Systems of Linear Inequalities
in Two Variables
y
4x + 3y = 12
4
4 x  3 y  12
x y 0
3
x–y=0
2
12 )
P( 127 , 12
7
1
–1
1
2
3
x
Graphing Linear Inequalities
 We’ve seen that a linear equation in two variables x and y
ax  by  c  0
has a solution set that may be exhibited graphically as points
on a straight line in the xy-plane.
 There is also a simple graphical representation for linear
inequalities of two variables:
ax  by  c  0
ax  by  c  0
ax  by  c  0
ax  by  c  0
Procedure for Graphing Linear Inequalities
1. Draw the graph of the equation obtained for the given
inequality by replacing the inequality sign with an
equal sign.
✦ Use a dashed or dotted line if the problem involves a
strict inequality, < or >.
✦ Otherwise, use a solid line to indicate that the line
itself constitutes part of the solution.
2. Pick a test point lying in one of the half-planes
determined by the line sketched in step 1 and substitute
the values of x and y into the given inequality.
✦ Use the origin whenever possible.
3. If the inequality is satisfied, the graph of the inequality
includes the half-plane containing the test point.
✦ Otherwise, the solution includes the half-plane not
containing the test point.
Examples
 Determine the solution set for the inequality 2x + 3y  6.
Solution
 Replacing the inequality  with an equality =, we obtain
the equation 2x + 3y = 6, whose graph is:
y
7
5
2x + 3y = 6
3
1
–5
–3
–1
–1
1
3
5
x
Examples
 Determine the solution set for the inequality 2x + 3y  6.
Solution
 Picking the origin as a test point, we find 2(0) + 3(0)  6,
or 0  6, which is false.
 Thus, the solution set is:
y
7
5
2x + 3y  6
2x + 3y = 6
3
1
–5
–3
–1
–1
(0, 0)
1
3
5
x
Graphing Systems of Linear Inequalities
 The solution set of a system of linear inequalities in two
variables x and y is the set of all points (x, y) that satisfy
each inequality of the system.
 The graphical solution of such a system may be obtained
by graphing the solution set for each inequality
independently and then determining the region in common
with each solution set.
Examples
 Graph x – 3y > 0.
Solution
 Replacing the inequality > with an equality =, we obtain
the equation x – 3y = 0, whose graph is:
y
3
x – 3y = 0
1
–5
–3
1
–1
–3
3
5
x
Examples
 Graph x – 3y > 0.
Solution
 We use a dashed line to indicate the line itself will not be
part of the solution, since we are dealing with a strict
inequality >.
y
3
x – 3y = 0
1
–5
–3
1
–1
–3
3
5
x
Examples
 Graph x – 3y > 0.
Solution
 Since the origin lies on the line, we cannot use the origin
as a testing point:
y
3
x – 3y = 0
1
(0, 0)
–5
–3
1
–1
–3
3
5
x
Examples
 Graph x – 3y > 0.
Solution
 Picking instead (3, 0) as a test point, we find (3) – 2(0) > 0,
or 3 > 0, which is true.
 Thus, the solution set is:
y
3
x – 3y = 0
1
(3, 0)
–5
–3
1
3
–1
x – 3y > 0
–3
5
x
Graphing Systems of Linear Inequalities
 The solution set of a system of linear inequalities in two
variables x and y is the set of all points (x, y) that satisfy
each inequality of the system.
 The graphical solution of such a system may be obtained
by graphing the solution set for each inequality
independently and then determining the region in common
with each solution set.
Example
 Determine the solution set for the system
4 x  3 y  12
x y 0
Solution
 The intersection of the solution regions of the two
inequalities represents the solution to the system:
y
4x + 3y = 12
4
4x + 3y  12
3
2
1
–1
1
2
3
x
Example
 Determine the solution set for the system
4 x  3 y  12
x y 0
Solution
 The intersection of the solution regions of the two
inequalities represents the solution to the system:
y
4
3
x–y0
x–y=0
2
1
–1
1
2
3
x
Example
 Determine the solution set for the system
4 x  3 y  12
x y 0
Solution
 The intersection of the solution regions of the two
inequalities represents the solution to the system:
y
4x + 3y = 12
4
4 x  3 y  12
x y 0
x–y=0
3
2
P( 127 , 127 )
1
–1
1
2
3
x
Bounded and Unbounded Sets
 The solution set of a system of linear inequalities
is bounded if it can be enclosed by a circle.
 Otherwise, it is unbounded.
Example
 The solution to the problem we just discussed is
unbounded, since the solution set cannot be
enclosed in a circle:
y
4x + 3y = 12
4
4 x  3 y  12
x y 0
3
x–y=0
2
P( 127 , 127 )
1
–1
1
2
3
x
Example
 Determine the solution set for the system
x y 6 0
2x  y  8  0
x0
y0
Solution
 The intersection of the solution regions of the four
inequalities represents the solution to the system:
y
7
2x  y  8  0
5
P(2, 4)
3
x y 6 0
1
–1
x
1
3
5
9
Example
 Determine the solution set for the system
x y 6 0
2x  y  8  0
x0
y0
Solution
 Note that the solution to this problem is bounded, since it
can be enclosed by a circle:
y
7
2x  y  8  0
5
P(2, 4)
3
x y 6 0
1
–1
x
1
3
5
9
6.2
Linear Programming Problems
Maximize
P  x  1.2 y
Subject to
2 x  y  180
x  3 y  300
x0
y0
Linear Programming Problem
 A linear programming problem consists of a
linear objective function to be maximized or
minimized subject to certain constraints in the
form of linear equations or inequalities.
Applied Example 1: A Production Problem
 Ace Novelty wishes to produce two types of souvenirs:




type-A will result in a profit of $1.00, and type-B in a
profit of $1.20.
To manufacture a type-A souvenir requires 2 minutes on
machine I and 1 minute on machine II.
A type-B souvenir requires 1 minute on machine I and 3
minutes on machine II.
There are 3 hours available on machine I and 5 hours
available on machine II.
How many souvenirs of each type should Ace make in
order to maximize its profit?
Applied Example 1: A Production Problem
Solution
 Let’s first tabulate the given information:
Type-A
Type-B
Time Available
Profit/Unit
$1.00
$1.20
Machine I
2 min
1 min
180 min
Machine II
1 min
3 min
300 min
 Let x be the number of type-A souvenirs and y the number
of type-B souvenirs to be made.
Applied Example 1: A Production Problem
Solution
 Let’s first tabulate the given information:
Type-A
Type-B
Time Available
Profit/Unit
$1.00
$1.20
Machine I
2 min
1 min
180 min
Machine II
1 min
3 min
300 min
 Then, the total profit (in dollars) is given by
P  x  1.2 y
which is the objective function to be maximized.
Applied Example 1: A Production Problem
Solution
 Let’s first tabulate the given information:
Type-A
Type-B
Time Available
Profit/Unit
$1.00
$1.20
Machine I
2 min
1 min
180 min
Machine II
1 min
3 min
300 min
 The total amount of time that machine I is used is
2x  y
and must not exceed 180 minutes.
 Thus, we have the inequality
2 x  y  180
Applied Example 1: A Production Problem
Solution
 Let’s first tabulate the given information:
Type-A
Type-B
Time Available
Profit/Unit
$1.00
$1.20
Machine I
2 min
1 min
180 min
Machine II
1 min
3 min
300 min
 The total amount of time that machine II is used is
x  3y
and must not exceed 300 minutes.
 Thus, we have the inequality
x  3 y  300
Applied Example 1: A Production Problem
Solution
 Let’s first tabulate the given information:
Type-A
Type-B
Time Available
Profit/Unit
$1.00
$1.20
Machine I
2 min
1 min
180 min
Machine II
1 min
3 min
300 min
 Finally, neither x nor y can be negative, so
x0
y0
Applied Example 1: A Production Problem
Solution
 In short, we want to maximize the objective function
P  x  1.2 y
subject to the system of inequalities
2 x  y  180
x  3 y  300
x0
y0
 We will discuss the solution to this problem in section 6.4.
Applied Example 2: A Nutrition Problem
 A nutritionist advises an individual who is suffering from




iron and vitamin B deficiency to take at least 2400
milligrams (mg) of iron, 2100 mg of vitamin B1, and 1500
mg of vitamin B2 over a period of time.
Two vitamin pills are suitable, brand-A and brand-B.
Each brand-A pill costs 6 cents and contains 40 mg of iron,
10 mg of vitamin B1, and 5 mg of vitamin B2.
Each brand-B pill costs 8 cents and contains 10 mg of iron
and 15 mg each of vitamins B1 and B2.
What combination of pills should the individual purchase
in order to meet the minimum iron and vitamin
requirements at the lowest cost?
Applied Example 2: A Nutrition Problem
Solution
 Let’s first tabulate the given information:
Brand-A
Brand-B
6¢
8¢
Iron
40 mg
10 mg
2400 mg
Vitamin B1
10 mg
15 mg
2100 mg
Vitamin B2
5mg
15 mg
1500 mg
Cost/Pill
Minimum Requirement
 Let x be the number of brand-A pills and y the number of
brand-B pills to be purchased.
Applied Example 2: A Nutrition Problem
Solution
 Let’s first tabulate the given information:
Brand-A
Brand-B
6¢
8¢
Iron
40 mg
10 mg
2400 mg
Vitamin B1
10 mg
15 mg
2100 mg
Vitamin B2
5mg
15 mg
1500 mg
Cost/Pill
Minimum Requirement
 The cost C (in cents) is given by
C  6x  8 y
and is the objective function to be minimized.
Applied Example 2: A Nutrition Problem
Solution
 Let’s first tabulate the given information:
Brand-A
Brand-B
6¢
8¢
Iron
40 mg
10 mg
2400 mg
Vitamin B1
10 mg
15 mg
2100 mg
Vitamin B2
5mg
15 mg
1500 mg
Cost/Pill
Minimum Requirement
 The amount of iron contained in x brand-A pills and y
brand-B pills is given by 40x + 10y mg, and this must be
greater than or equal to 2400 mg.
 This translates into the inequality
40 x  10 y  2400
Applied Example 2: A Nutrition Problem
Solution
 Let’s first tabulate the given information:
Brand-A
Brand-B
6¢
8¢
Iron
40 mg
10 mg
2400 mg
Vitamin B1
10 mg
15 mg
2100 mg
Vitamin B2
5mg
15 mg
1500 mg
Cost/Pill
Minimum Requirement
 The amount of vitamin B1 contained in x brand-A pills and
y brand-B pills is given by 10x + 15y mg, and this must be
greater or equal to 2100 mg.
 This translates into the inequality
10 x  15 y  2100
Applied Example 2: A Nutrition Problem
Solution
 Let’s first tabulate the given information:
Brand-A
Brand-B
6¢
8¢
Iron
40 mg
10 mg
2400 mg
Vitamin B1
10 mg
15 mg
2100 mg
Vitamin B2
5mg
15 mg
1500 mg
Cost/Pill
Minimum Requirement
 The amount of vitamin B2 contained in x brand-A pills and
y brand-B pills is given by 5x + 15y mg, and this must be
greater or equal to 1500 mg.
 This translates into the inequality
5x  15 y  1500
Applied Example 2: A Nutrition Problem
Solution
 In short, we want to minimize the objective function
C  6x  8 y
subject to the system of inequalities
40 x  10 y  2400
10 x  15 y  2100
5x  15 y  1500
x0
y0
 We will discuss the solution to this problem in section 6.4.
6.3
Graphical Solutions
of Linear Programming Problems
40 x  10 y  2400
10 x  15 y  2100
5x  15 y  1500
y
A(0, 240)
200
S
B(30, 120)
100
C(120, 60)
D(300, 0)
x
100
200
300
Feasible Solution Set and Optimal Solution
 The constraints in a linear programming problem form a
system of linear inequalities, which have a solution set S.
 Each point in S is a candidate for the solution of the linear
programming problem and is referred to as a feasible
solution.
 The set S itself is referred to as a feasible set.
 Among all the points in the set S, the point(s) that
optimizes the objective function of the linear programming
problem is called an optimal solution.
Theorem 1
Linear Programming
 If a linear programming problem has a solution,
then it must occur at a vertex, or corner point, of
the feasible set S associated with the problem.
 If the objective function P is optimized at two
adjacent vertices of S, then it is optimized at every
point on the line segment joining these vertices, in
which case there are infinitely many solutions to
the problem.
Theorem 2
Existence of a Solution
 Suppose we are given a linear programming
problem with a feasible set S and an objective
function P = ax + by.
a. If S is bounded, then P has both a maximum and
a minimum value on S.
b. If S is unbounded and both a and b are
nonnegative, then P has a minimum value on S
provided that the constraints defining S include
the inequalities x  0 and y  0.
c. If S is the empty set, then the linear
programming problem has no solution: that is, P
has neither a maximum nor a minimum value.
The Method of Corners
1. Graph the feasible set.
2. Find the coordinates of all corner points
(vertices) of the feasible set.
3. Evaluate the objective function at each corner
point.
4. Find the vertex that renders the objective
function a maximum or a minimum.
✦ If there is only one such vertex, it constitutes a
unique solution to the problem.
✦ If there are two such adjacent vertices, there
are infinitely many optimal solutions given by
the points on the line segment determined by
these vertices.
Applied Example 1: A Production Problem
 Recall Applied Example 1 from the last section (3.2), which
required us to find the optimal quantities to produce of
type-A and type-B souvenirs in order to maximize profits.
 We restated the problem as a linear programming problem
in which we wanted to maximize the objective function
P  x  1.2 y
subject to the system of inequalities
2 x  y  180
x  3 y  300
x0
y0
 We can now solve the problem graphically.
Applied Example 1: A Production Problem
 We first graph the feasible set S for the problem.
✦ Graph the solution for the inequality
2 x  y  180
considering only positive values for x and y:
y
200
100
(0, 180)
2 x  y  180
(90, 0)
100
x
200
2 x  y  180
300
Applied Example 1: A Production Problem
 We first graph the feasible set S for the problem.
✦ Graph the solution for the inequality
x  3 y  300
considering only positive values for x and y:
y
200
100
(0, 100)
x  3 y  300
x  3 y  300
(300, 0)
100
200
300
x
Applied Example 1: A Production Problem
 We first graph the feasible set S for the problem.
✦ Graph the intersection of the solutions to the inequalities,
yielding the feasible set S.
(Note that the feasible set S is bounded)
y
200
100
x  3 y  300
S
x
100
200
2 x  y  180
300
Applied Example 1: A Production Problem
 Next, find the vertices of the feasible set S.
✦ The vertices are A(0, 0), B(90, 0), C(48, 84), and D(0, 100).
y
200
D(0, 100)
100
C(48, 84)
S
A(0, 0)
x  3 y  300
B(90, 0)
100
x
200
2 x  y  180
300
Applied Example 1: A Production Problem
 Now, find the values of P at the vertices and tabulate them:
Vertex
P = x + 1.2 y
A(0, 0)
0
B(90, 0)
90
200
C(48, 84)
148.8
D(0, 100)
100
D(0, 100)
120
y
C(48, 84)
S
A(0, 0)
x  3 y  300
B(90, 0)
100
x
200
2 x  y  180
300
Applied Example 1: A Production Problem
 Finally, identify the vertex with the highest value for P:
✦ We can see that P is maximized at the vertex C(48, 84)
and has a value of 148.8.
Vertex
P = x + 1.2 y
A(0, 0)
0
B(90, 0)
90
200
C(48, 84)
148.8
D(0, 100)
100
D(0, 100)
120
y
C(48, 84)
S
A(0, 0)
x  3 y  300
B(90, 0)
100
x
200
2 x  y  180
300
Applied Example 1: A Production Problem
 Finally, identify the vertex with the highest value for P:
✦ We can see that P is maximized at the vertex C(48, 84)
and has a value of 148.8.
✦ Recalling what the symbols x, y, and P represent, we
conclude that ACE Novelty would maximize its profit at
$148.80 by producing 48 type-A souvenirs and 84 type-B
souvenirs.
Applied Example 2: A Nutrition Problem
 Recall Applied Example 2 from the last section (3.2), which
asked us to determine the optimal combination of pills to
be purchased in order to meet the minimum iron and
vitamin requirements at the lowest cost.
 We restated the problem as a linear programming problem
in which we wanted to minimize the objective function
C  6x  8 y
subject to the system of inequalities
40 x  10 y  2400
10 x  15 y  2100
5x  15 y  1500
x, y  0
 We can now solve the problem graphically.
Applied Example 2: A Nutrition Problem
 We first graph the feasible set S for the problem.
✦ Graph the solution for the inequality
40 x  10 y  2400
considering only positive values for x and y:
40 x  10 y  2400
y
(0, 240)
200
100
(60, 0)
100
x
200
300
Applied Example 2: A Nutrition Problem
 We first graph the feasible set S for the problem.
✦ Graph the solution for the inequality
10 x  15 y  2100
considering only positive values for x and y:
y
10 x  15 y  2100
200
(0, 140)
100
(210, 0)
100
200
x
300
Applied Example 2: A Nutrition Problem
 We first graph the feasible set S for the problem.
✦ Graph the solution for the inequality
5x  15 y  1500
considering only positive values for x and y:
y
200
5x  15 y  1500
100
(0, 100)
(300, 0)
100
200
300
x
Applied Example 2: A Nutrition Problem
 We first graph the feasible set S for the problem.
✦ Graph the intersection of the solutions to the inequalities,
yielding the feasible set S.
(Note that the feasible set S is unbounded)
40 x  10 y  2400
10 x  15 y  2100
5x  15 y  1500
y
200
S
100
x
100
200
300
Applied Example 2: A Nutrition Problem
 Next, find the vertices of the feasible set S.
✦ The vertices are A(0, 240), B(30, 120), C(120, 60), and
D(300, 0).
40 x  10 y  2400
10 x  15 y  2100
5x  15 y  1500
y
A(0, 240)
200
S
B(30, 120)
100
C(120, 60)
D(300, 0)
x
100
200
300
Applied Example 2: A Nutrition Problem
 Now, find the values of C at the vertices and tabulate them:
Vertex
40 x  10 y  2400
10 x  15 y  2100
5x  15 y  1500
y
A(0, 240)
C = 6x + 8y
A(0, 240)
1920
B(30, 120)
1140
C(120, 60)
1200
D(300, 0)
1800
200
S
B(30, 120)
100
C(120, 60)
D(300, 0)
x
100
200
300
Applied Example 2: A Nutrition Problem
 Finally, identify the vertex with the lowest value for C:
✦ We can see that C is minimized at the vertex B(30, 120)
and has a value of 1140.
Vertex
40 x  10 y  2400
10 x  15 y  2100
5x  15 y  1500
y
A(0, 240)
C = 6x + 8y
A(0, 240)
1920
B(30, 120)
1140
C(120, 60)
1200
D(300, 0)
1800
200
S
B(30, 120)
100
C(120, 60)
D(300, 0)
x
100
200
300
Applied Example 2: A Nutrition Problem
 Finally, identify the vertex with the lowest value for C:
✦ We can see that C is minimized at the vertex B(30, 120)
and has a value of 1140.
✦ Recalling what the symbols x, y, and C represent, we
conclude that the individual should purchase 30 brand-A
pills and 120 brand-B pills at a minimum cost of $11.40.
6.4
The Simplex Method:
Standard Maximization Problems
x
y
u
v
P
Constant
1
0
3/5 –1/5
0
48
0
1 –1/5
2/5
0
84
0
0 9/25 7/25
1
148 4/5
The Simplex Method
 The simplex method is an iterative procedure.
 Beginning at a vertex of the feasible region S, each
iteration brings us to another vertex of S with an improved
value of the objective function.
 The iteration ends when the optimal solution is reached.
A Standard Linear Programming Problem
 A standard maximization problem is one in which
1. The objective function is to be maximized.
2. All the variables involved in the problem are
nonnegative.
3. All other linear constraints may be written so that
the expression involving the variables is less than
or equal to a nonnegative constant.
Setting Up the Initial Simplex Tableau
1. Transform the system of linear inequalities
into a system of linear equations by
introducing slack variables.
2. Rewrite the objective function
P  c1x1  c2 x2    cn xn
in the form
c1x1  c2 x2    cn xn  P  0
where all the variables are on the left and the
coefficient of P is +1. Write this equation
below the equations in step 1.
3. Write the augmented matrix associated with
this system of linear equations.
Applied Example 1: A Production Problem
 Recall the production problem discussed in section 6.3,
which required us to maximize the objective function
6
P  x  1.2 y or equivalently, P  x  y
5
subject to the system of inequalities
2 x  y  180
x  3 y  300
x, y  0
 This is a standard maximization problem and may be
solved by the simplex method.
 Set up the initial simplex tableau for this linear
programming problem.
Applied Example 1: A Production Problem
Solution
 First, introduce the slack variables u and v into the
inequalities
2 x  y  180
x  3 y  300
and turn these into equations, getting
2x  y  u
x  3y
 180
 v  300
 Next, rewrite the objective function in the form
6
x  y  P  0
5
Applied Example 1: A Production Problem
Solution
 Placing the restated objective function below the system of
equations of the constraints we get
2x  y  u
 180
x  3y
v
 300
6
x  y
P 0
5
 Thus, the initial tableau associated with this system is
x
y
u
v
P
Constant
2
1
1
0
0
180
1
3
0
1
0
300
–1 – 6/5
0
0
1
0
The Simplex Method
1. Set up the initial simplex tableau.
2. Determine whether the optimal solution has
been reached by examining all entries in the last
row to the left of the vertical line.
a. If all the entries are nonnegative, the optimal
solution has been reached. Proceed to step 4.
b. If there are one or more negative entries, the
optimal solution has not been reached.
Proceed to step 3.
3. Perform the pivot operation. Return to step 2.
4. Determine the optimal solution(s).
Applied Example 1: A Production Problem
 Recall again the production problem discussed previously.
 We have already performed step 1 obtaining the initial
simplex tableau:
x
y
u
v
P
Constant
2
1
1
0
0
180
1
3
0
1
0
300
–1 – 6/5
0
0
1
0
 Now, complete the solution to the problem.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
2
1
1
0
0
180
1
3
0
1
0
300
–1
– 6/5
0
0
1
0
Step 2. Determine whether the optimal solution has been
reached.
✦ Since there are negative entries in the last row of the
tableau, the initial solution is not optimal.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
2
1
1
0
0
180
1
3
0
1
0
300
–1
– 6/5
0
0
1
0
Step 3. Perform the pivot operation.
✦ Since the entry – 6/5 is the most negative entry to the left
of the vertical line in the last row of the tableau, the
second column in the tableau is the pivot column.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
2
1
1
0
0
180
180
1
 180
1
3
0
1
0
300
300
3
 100
–1
– 6/5
0
0
1
0
Step 3. Perform the pivot operation.
✦ Divide each positive number of the pivot column into the
corresponding entry in the column of constants and
compare the ratios thus obtained.
✦ We see that the ratio 300/3 = 100 is less than the ratio
180/1 = 180, so row 2 is the pivot row.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
2
1
1
0
0
180
1
3
0
1
0
300
–1
– 6/5
0
0
1
0
Step 3. Perform the pivot operation.
✦ The entry 3 lying in the pivot column and the pivot row
is the pivot element.
Applied Example 1: A Production Problem
Solution
1
3
R2
x
y
u
v
P
Constant
2
1
1
0
0
180
1
3
0
1
0
300
–1
– 6/5
0
0
1
0
Step 3. Perform the pivot operation.
✦ Convert the pivot element into a 1.
Applied Example 1: A Production Problem
Solution
1
3
R2
x
y
u
v
P
Constant
2
1
1
0
0
180
1/3
1
0
1/3
0
100
–1
– 6/5
0
0
1
0
Step 3. Perform the pivot operation.
✦ Convert the pivot element into a 1.
Applied Example 1: A Production Problem
Solution
R1  R2
R3  65 R2
x
y
u
v
P
Constant
2
1
1
0
0
180
1/3
1
0
1/3
0
100
–1
– 6/5
0
0
1
0
Step 3. Perform the pivot operation.
✦ Use elementary row operations to convert the pivot
column into a unit column.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
R1  R2
5/3
0
1 –1/3
0
80
R3  65 R2
1/3
1
0
1/3
0
100
–3/5
0
0
2/5
1
120
Step 3. Perform the pivot operation.
✦ Use elementary row operations to convert the pivot
column into a unit column.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
5/3
0
1 –1/3
0
80
1/3
1
0
1/3
0
100
–3/5
0
0
2/5
1
120
Step 3. Perform the pivot operation.
✦ This completes an iteration.
✦ The last row of the tableau contains a negative number,
so an optimal solution has not been reached.
✦ Therefore, we repeat the iteration step.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
5/3
0
1 –1/3
0
80
1/3
1
0
1/3
0
100
–3/5
0
0
2/5
1
120
Step 3. Perform the pivot operation again.
✦ Since the entry – 3/5 is the most negative entry to the left
of the vertical line in the last row of the tableau, the first
column in the tableau is now the pivot column.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
5/3
0
1 –1/3
0
80
80
5/3
 48
1/3
1
0
1/3
0
100
100
1/3
 300
–3/5
0
0
2/5
1
120
Ratio
Step 3. Perform the pivot operation.
✦ Divide each positive number of the pivot column into the
corresponding entry in the column of constants and
compare the ratios thus obtained.
✦ We see that the ratio 80/(5/3) = 48 is less than the ratio
100/(1/3) = 300, so row 1 is the pivot row now.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
5/3
0
1 –1/3
0
80
1/3
1
0
1/3
0
100
–3/5
0
0
2/5
1
120
Step 3. Perform the pivot operation.
✦ The entry 5/3 lying in the pivot column and the pivot
row is the pivot element.
Applied Example 1: A Production Problem
Solution
3
5
R1
x
y
u
v
P
Constant
5/3
0
1 –1/3
0
80
1/3
1
0
1/3
0
100
–3/5
0
0
2/5
1
120
Step 3. Perform the pivot operation.
✦ Convert the pivot element into a 1.
Applied Example 1: A Production Problem
Solution
3
5
R1
x
y
u
v
P
Constant
1
0
3/5 –1/5
0
48
1/3
1
0
1/3
0
100
–3/5
0
0
2/5
1
120
Step 3. Perform the pivot operation.
✦ Convert the pivot element into a 1.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
R2  13 R1
1
0
3/5 –1/5
0
48
R3  53 R1
1/3
1
0
1/3
0
100
–3/5
0
0
2/5
1
120
Step 3. Perform the pivot operation.
✦ Use elementary row operations to convert the pivot
column into a unit column.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
R2  13 R1
1
0
3/5 –1/5
0
48
R3  53 R1
0
1 –1/5
2/5
0
84
0
0 9/25 7/25
1
148 4/5
Step 3. Perform the pivot operation.
✦ Use elementary row operations to convert the pivot
column into a unit column.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
1
0
3/5 –1/5
0
48
0
1 –1/5
2/5
0
84
0
0 9/25 7/25
1
148 4/5
Step 3. Perform the pivot operation.
✦ The last row of the tableau contains no negative
numbers, so an optimal solution has been reached.
Applied Example 1: A Production Problem
Solution
x
y
u
v
P
Constant
1
0
3/5 –1/5
0
48
0
1 –1/5
2/5
0
84
0
0 9/25 7/25
1
148 4/5
Step 4. Determine the optimal solution.
✦ Locate the basic variables in the final tableau.
In this case, the basic variables are x, y, and P.
 The optimal value for x is 48.
 The optimal value for y is 84.
 The optimal value for P is 148.8.
✦ Thus, the firm will maximize profits at $148.80 by
producing 48 type-A souvenirs and 84 type-B souvenirs.
This agrees with the results obtained in section 6.3.
6.5
The Simplex Method:
Standard Minimization Problems
u
v
1
w
x
y
P
Constant
0
–3/20 3/100 –1/50
0
1/50
0
1
11/10
–1/50
2/25
0
13/25
0
0
450
30
120
1
1140
Solution for the
primal problem
Minimization with  Constraints
 In the last section we developed the simplex method to
solve linear programming problems that satisfy three
conditions:
1. The objective function is to be maximized.
2. All the variables involved are nonnegative.
3. Each linear constraint may be written so that the
expression involving the variables is less than or equal to
a nonnegative constant.
 We will now see how the simplex method can be used to
solve minimization problems that meet the second and
third conditions listed above.
Example
 Solve the following linear programming problem:
Minimize C  2 x  3 y
subject to
5x  4 y  32
x  2 y  10
x, y  0
 This problem involves the minimization of the objective
function and so is not a standard maximization problem.
 Note, however, that all the other conditions for a standard
maximization hold true.
Example
 We can use the simplex method to solve this problem by
converting the objective function from minimizing C to its
equivalent of maximizing P = – C.
 Thus, the restated linear programming problem is
Maximize P  2 x  3 y
subject to
5x  4 y  32
x  2 y  10
x, y  0
 This problem can now be solved using the simplex method
as discussed in section 6.4.
Example
Solution
Step 1. Set up the initial simplex tableau.
✦ Turn the constraints into equations adding to them the
slack variables u and v. Also rearrange the objective
function and place it below the constraints:
5x  4 y  u
 32
x  2y
v
 10
2 x  3 y
P 0
✦ Write the coefficients of the system in a tableau:
xMaximize
y
uP  2vx  3 y P
Constant
5
4
1
0
0
32
1
2
0
1
0
10
–2
–3
0
0
1
0
Example
Solution
x
y
u
v
P
Constant
5
4
1
0
0
32
1
2
0
1
0
10
–2
–3
0
0
1
0
Step 2. Determine whether the optimal solution has been
reached.
✦ Since there are negative entries in the last row of the
tableau, the initial solution is not optimal.
Example
Solution
xMaximize
y
uP  2vx  3 y P
Constant
5
4
1
0
0
32
1
2
0
1
0
10
–2
–3
0
0
1
0
Step 3. Perform the pivot operation.
✦ Since the entry – 3 is the most negative entry to the left
of the vertical line in the last row of the tableau, the
second column in the tableau is the pivot column.
Example
Solution
xMaximize
y
uP  2vx  3 y P
Constant
Ratio
5
4
1
0
0
32
32
4
8
1
2
0
1
0
10
10
2
5
–2
–3
0
0
1
0
Step 3. Perform the pivot operation.
✦ Divide each positive number of the pivot column into the
corresponding entry in the column of constants and
compare the ratios thus obtained.
✦ We see that the ratio 10/2 = 5 is less than the ratio
32/4 = 8, so row 2 is the pivot row.
Example
Solution
xMaximize
y
uP  2vx  3 y P
Constant
5
4
1
0
0
32
1
2
0
1
0
10
–2
–3
0
0
1
0
Step 3. Perform the pivot operation.
✦ The entry 2 lying in the pivot column and the pivot row
is the pivot element.
Example
Solution
xMaximize
y
uP  2vx  3 y P
1
2
R2
Constant
5
4
1
0
0
32
1
2
0
1
0
10
–2
–3
0
0
1
0
Step 3. Perform the pivot operation.
✦ Convert the pivot element into a 1.
Example
Solution
xMaximize
y
uP  2vx  3 y P
1
2
R2
Constant
5
4
1
0
0
32
1/2
1
0
1/2
0
5
–2
–3
0
0
1
0
Step 3. Perform the pivot operation.
✦ Convert the pivot element into a 1.
Example
Solution
xMaximize
y
uP  2vx  3 y P
Constant
R1  4 R2
5
4
1
0
0
32
R3  3R2
1/2
1
0
1/2
0
5
–2
–3
0
0
1
0
Step 3. Perform the pivot operation.
✦ Use elementary row operations to convert the pivot
column into a unit column.
Example
Solution
xMaximize
y
uP  2vx  3 y P
Constant
R1  4 R2
3
0
1
–2
0
12
R3  3R2
1/2
1
0
1/2
0
5
–1/2
0
0
3/2
1
15
Step 3. Perform the pivot operation.
✦ Use elementary row operations to convert the pivot
column into a unit column.
Example
Solution
xMaximize
y
uP  2vx  3 y P
Constant
3
0
1
–2
0
12
1/2
1
0
1/2
0
5
–1/2
0
0
3/2
1
15
Step 3. Perform the pivot operation.
✦ This completes an iteration.
✦ The last row of the tableau contains a negative number,
so an optimal solution has not been reached.
✦ Therefore, we repeat the iteration step.
Example
Solution
xMaximize
y
uP  2vx  3 y P
Constant
3
0
1
–2
0
12
1/2
1
0
1/2
0
5
–1/2
0
0
3/2
1
15
Step 3. Perform the pivot operation.
✦ Since the entry –1/2 is the most negative entry to the left
of the vertical line in the last row of the tableau, the first
column in the tableau is now the pivot column.
Example
Solution
xMaximize
y
uP  2vx  3 y P
Constant
3
0
1
–2
0
12
1/2
1
0
1/2
0
5
–1/2
0
0
3/2
1
15
12
3
5
1/2
4
 10
Step 3. Perform the pivot operation.
✦ Divide each positive number of the pivot column into the
corresponding entry in the column of constants and
compare the ratios thus obtained.
✦ We see that the ratio 12/3 = 4 is less than the ratio
5/(1/2) = 10, so row 1 is now the pivot row.
Example
Solution
x
y
u
v
P
Constant
3
0
1
–2
0
12
1/2
1
0
1/2
0
5
–1/2
0
0
3/2
1
15
Step 3. Perform the pivot operation.
✦ The entry 3 lying in the pivot column and the pivot row
is the pivot element.
Example
Solution
1
3
R1
x
y
u
v
P
Constant
3
0
1
–2
0
12
1/2
1
0
1/2
0
5
–1/2
0
0
3/2
1
15
Step 3. Perform the pivot operation.
✦ Convert the pivot element into a 1.
Example
Solution
1
3
R1
x
y
u
v
P
Constant
1
0
1/3
–2/3
0
4
1/2
1
0
1/2
0
5
–1/2
0
0
3/2
1
15
Step 3. Perform the pivot operation.
✦ Convert the pivot element into a 1.
Example
Solution
x
y
u
v
P
Constant
R2  12 R1
1
0
1/3
–2/3
0
4
R3  12 R1
1/2
1
0
1/2
0
5
–1/2
0
0
3/2
1
15
Step 3. Perform the pivot operation.
✦ Use elementary row operations to convert the pivot
column into a unit column.
Example
Solution
x
y
u
v
P
Constant
R2  12 R1
1
0
1/3
–2/3
0
4
R3  12 R1
0
1
–1/6
5/6
0
3
0
0
1/6
7/6
1
17
Step 3. Perform the pivot operation.
✦ Use elementary row operations to convert the pivot
column into a unit column.
Example
Solution
x
y
u
v
P
Constant
1
0
1/3
–2/3
0
4
0
1
–1/6
5/6
0
3
0
0
1/6
7/6
1
17
Step 3. Perform the pivot operation.
✦ The last row of the tableau contains no negative
numbers, so an optimal solution has been reached.
Example
Solution
x
y
u
v
P
Constant
1
0
1/3
–2/3
0
4
0
1
–1/6
5/6
0
3
0
0
1/6
7/6
1
17
Step 4. Determine the optimal solution.
✦ Locate the basic variables in the final tableau.
In this case, the basic variables are x, y, and P.
 The optimal value for x is 4.
 The optimal value for y is 3.
 The optimal value for P is 17, which means that
the minimized value for C is –17.
The Dual Problem
 Another special class of linear programming problems we
encounter in practical applications is characterized by the
following conditions:
1. The objective function is to be minimized.
2. All the variables involved are nonnegative.
3. All other linear constraints may be written so that the
expression involving the variables is greater than or
equal to a nonnegative constant.
 Such problems are called standard minimization
problems.
The Dual Problem
 In solving this kind of linear programming problem, it
helps to note that each maximization problem is associated
with a minimization problem, and vice versa.
 The given problem is called the primal problem, and the
related problem is called the dual problem.
Example
 Write the dual problem associated with this problem:
Minimize
subject to
C  6x  8 y
40 x  10 y  2400
10 x  15 y  2100
5 x  15 y  1500
x, y  0
Primal
Problem
 We first write down a tableau for the primal problem:
x
y
Constant
40
10
2400
10
15
2100
5
15
1500
6
8
Example
x
y
Constant
40
10
2400
10
15
2100
5
15
1500
6
8
 Next, we interchange the columns and rows of the tableau
and head the three columns of the resulting array with the
three variables u, v, and w, obtaining
u
v
w
Constant
40
10
5
6
10
15
15
8
2400
2100
1500
Example
u
v
w
Constant
40
10
5
6
10
15
15
8
2400
2100
1500
 Consider the resulting tableau as if it were the initial
simplex tableau for a standard maximization problem.
 From it we can reconstruct the required dual problem:
Maximize
P  2400u  2100v  1500w
subject to
40u  10v  5w  6
10u  15v  15w  8
u, v , w  0
Dual
Problem
Theorem 1
The Fundamental Theorem of Duality
 A primal problem has a solution if and only if the
corresponding dual problem has a solution.
 Furthermore, if a solution exists, then:
a. The objective functions of both the primal and
the dual problem attain the same optimal value.
b. The optimal solution to the primal problem
appears under the slack variables in the last row
of the final simplex tableau associated with the
dual problem.
Example
 Complete the solution of the problem from our last example:
Maximize
P  2400u  2100v  1500w
subject to
40u  10v  5w  6
10u  15v  15w  8
u, v , w  0
Dual
Problem
Example
Solution
 The dual problem associated with the given primal
problem is a standard maximization problem.
 Thus, we can proceed with the simplex method.
 First, we introduce to the system of equations the slack
variables x and y, and restate the inequalities as equations,
obtaining
40u 
10v 
5w  x
10u  15v  15w
2400u  2100v  1500w
6
y
8
P0
Example
Solution
 Next, we transcribe the coefficients of the system of
equations
40u  10v 
5w  x
6
10u  15v  15w
y
8
2400u  2100v  1500w
P0
into an initial simplex tableau:
u
v
w
x
y
P
Constant
40
10
5
1
0
0
6
10
15
15
0
1
0
8
–2400
–2100
–1500
0
0
1
0
Example
Solution
 Continue with the simplex iterative method until a final
tableau is obtained with the solution for the problem:
u
v
w
x
y
P
Constant
1
0
–3/20
3/100
–1/50
0
1/50
0
1
11/10
–1/50
2/25
0
13/25
0
0
450
30
120
1
1140
Solution for the
primal problem
 The fundamental theorem of duality tells us that the
solution to the primal problem is x = 30 and y = 120, with a
minimum value for C of 1140.
End of
Chapter

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