Transcript Document

Chapter 7
LINEAR PROGRAMMING
7.1 GRAPHING LINEAR
INEQUALITIES IN 2 VARIABLES
Terms:
– Boundary
– Half-plane
– Feasible region
1. Example 1: 3x – 2y  6
2. Example 2: y < – 3x + 12
x < 2y
7.1 GRAPHING LINEAR
INEQUALITIES IN 2 VARIABLES
1. Graph the boundary line. Decide
whether the line is part of the solution
(use solid line for  and , dashed line
for > and <)
2. Solve the inequality for y: shade the
region above the line if y > mx + b;
shade the region below the line
if y < mx + b.
7.1 GRAPHING LINEAR
INEQUALITIES IN 2 VARIABLES
Applications:
A company makes platic plate and cups, both of
which require time on 2 machines. Producing a
unit of plates requires 1h on machine A and 2h on
machine B, while producing a unit of cups
requires 3h on machine A and 1h on machine B.
Each machine is operated for at most 15h per
day. Write a system of inequalities expressing
these conditions, and graph the feasibleregion.
7.2 LINEAR PROGRAMMING:
THE GRAPHICAL METHOD
• Terms: Linear programming, objective
function, constraints
• EXAMPLE 1:
Find the maximum and minimum values of
the objective function z = 2x + 5y, with the
following constraints:
3x + 2y  6
– 2x + 4y  8
x + y  1,
x  0, y  0
CORNER POINT THEOREM
• If the feasible region is bounded, then the
objective function has both a maximum and
a minimum value, and each occurs at one or
more corner points.
• If the feasible region is unbounded, then the
objective function may not have a maximum
or minimum. But if a maximum or minimum
value exists, it will occur at one or more
corner points.
SOLVING LINEAR
PROGRAMMING PROBLEM
1. Write the objective function and all
necessary constraints.
2. Graph the feasible region.
3. Determine the cordinates of each corner
points.
4. If the feasible region is bounded, the
solution is given by the corner point
producing the optimum value of the
objective function.
SOLVING LINEAR
PROGRAMMING PROBLEM
5. If the feasible region is unbounded in the first
quadrant and both coefficients of the objective
function are positive, then the minimum value
of the objective function occurs at a corner
point and there is no maximum value.
Example: find max and min of z = 5x + 2y with the
following constraints: 3 y  2 x  0

 y  8 x  52


 y  2x  2
 x  3
7.3 APPLICATIONS OF LINEAR
PROGRAMMING
• EXAMPLE 1 (P. 340)
Raising geese and pigs
• EXAMPLE 2 (P. 341)
Purchasing filing cabinets
• EXAMPLE 3 (P. 343)
Buying food for animals
7.4 THE SIMPLEX METHOD:
MAXIMIZATION
STANDARD MAXIMUM FORM
A Linear Programming Problem Is In
Standard Maximum Form If:
1. The Objective Function Is To Be Maximized;
2. All Variables Are Nonnegative.
3. All Constraints Involve .
4. The Constants On The Right Side In The
Constraints Are All Nonnegative (b  0).
Example 1 (p. 392).
TERMS
•
•
•
•
•
•
•
Slack Variable
Initial Simplex Tableau
Indicators
Pivot, Pivot Column, Pivot Row
Pivoting
Basic Variable
Non-basic Variable
SIMPLEX METHOD
1. Determine The Objective Function.
2. Write All Necessary Constraints.
3. Convert Each Constraint Into An
Equation By Adding Slack Variables.
4. Set Up The Initial Simplex Tableau.
5. Locate The Most Negative Indicator. If
There Are Two Such Indicators,
Choose One. This Indicator
Determines The Pivot Column.
SIMPLEX METHOD (cont.)
6. Use The Positive Entries In The Pivot
Column To Form The Quotients
Necessary For Determining The Pivot.
If There Are No Positive Entries In The
Pivot Column, No Maximum Solution
Exists. If 2 Quotients Are Equally The
Smallest, Let Either Determines The
Pivot.
SIMPLEX METHOD (cont.)
7. Multiply every entry in the pivot row
by the reciprocal of the pivot to
change the pivot to 1. The use row
operations to change all other entries
in the pivot column to 0 by adding
suitable multiplies of the pivot to the
other rows.
SIMPLEX METHOD (cont.)
8. If the indicators are all positive or 0, this is
the final tableau. If not, go back to step 5
above and repeat the process until a
tableau with no negative indicators is
obtained.
9. Determine the basic and non-basic
variables and read the solution from the
final tableau. The maximum value of the
objective function is the number in the
lower right corner of the final tableau.
7.5 MAXIMIZATION
APPLICATIONS
EXAMPLE 1 (P. 405):
A farmer has 110 acres of available land he
wishes to plant with a mixture of potatoes, corn
and cabbage. It costs him $400 to produce an acre
of potatoes, $160 to produce an acre of corn and
$280 to produce an acre of cabbage. He has a
maximum of $20000 to spend. He makes a profit
of $120 per acre of potatoes, $40 per acre of corn,
and $60 per acre of cabbage. How many acres of
each crop should he plant to maximize his profit?
7.5 MAXIMIZATION APPLICATIONS
EXAMPLE 2 (P. 407)
Ana has $96000 to buy TV advertising time. Ads
cost $400 per minute on a local cable channel,
$4000 per minute on a regional channel, and
$12000 per minute on a national channel. The TV
stations can provide at most 30 minutes of
advertising time, with a maximum of 6 minutes on
the national channel. At any given time during the
evening, approx. 100000 people watch the local
channel, 200000 the regional channel, and 600000
the national channel. To get maximum exposure,
how much time should Ana buy from each station?
7.5 MAXIMIZATION APPLICATIONS
EXAMPLE 3 (P. 408)
A chemical plant makes 3 products – glaze,
solvent and clay – each of which brings in different
revenue per truckload. Production is limited, first
by the number of air pollution units the plant is
allowed to produce each day and second by the
time available in the evaporation tank. The plant
manager wants to maximize the daily revenue.
Using information not given here, he sets up an
initial simplex tableau and uses the simplex
method to produce the following final simplex
tableau:
7.5 MAXIMIZATION APPLICATIONS
EXAMPLE 3 (P. 408)
 10  25 0 1  1 60
 3

4
1
0
.
1
24


 7
13 0 0 .4 96
The 3 variables represent the number of truckloads of
glaze, solvent and clay. The 1st slack variable comes from
the air pollution constraint and the 2nd slack variable from
the time constraint on the evaporation tank. The revenue
function is given in hundreds of dollars.
a) What is the optimal solution?
b) Interpret the solution.
7.6 THE SIMPLEX METHOD:
DUALITY AND MINIMIZATION
Minimization problem:
1. The objective function is to be minimized.
2. All variables are nonnegative.
3. All constraints involve .
4. All the coefficients of the objective function
are nonnegative.
Note:
In minimization problem:
•
•
Variables are denoted y1, y2…
Objective function is denoted w.
TRANSPOSE OF A MATRIX
• Is the matrix obtained from the initial
matrix by exchanging its rows and
columns.
• Example: transpose of matrix
 2 1 5 
6
8 0


 3 7  1
is matrix
 2 6  3
 1 8 7 


 5 0  1
Example of minimization problem:
Minimize
w = 8y1 + 16y2
Constraints: y1 + 5y2  9
2y1 + 2y2  10
y1  0, y2  0.
Dual problem:
Maximize
z = 9x1 + 10 x2
Constraints x1 + 2x2  8
5x1 + 2x2  16
x1  0, x2  0
SUMMARY
Minimization problem
M variables
N constraints
Coefficients of objective
function
Constants
Dual problem
N variables
M constraints
Constants
Coefficients of objective function
THE SOLUTION OF MAXIMIZING PROBLEM PRODUCES THE SOLUTION
OF ASSOCIATED MINIMIZING PROBLEM, AND VICE-VERSA.
THEOREM OF DUALITY
The objective function w of a minimizing
linear programming problem takes on a
minimum value if and only if the objective
function z of the corresponding dual
maximizing problem takes on a maximum
value. The maximum value of z equals the
minimum value of w.
• Example 4 (p. 414)
• Example 5 (p. 415, 416)
SOLVING MINIMUM PROBLEMS WITH DUALS
1. Find the dual standard maximum problem.
2. Solve the maximum problem using the
simplex method.
3. The minimum value of the objective
function w is the maximum value of the
objective function z.
4. The optimal solution is given by the entries
in the bottom row of the columns
corresponding to the slack variables.
• Example 6 (p. 417)
• Example 7 (p. 418)
• FURTHER USES OF THE DUAL (p. 418)
7.7 THE SIMPLEX METHOD:
NONSTANDARD PROBLEMS
• Surplus variable
• Example:
2x1 – x2 + 5x3  12
 2x1 – x2 + 5x3 – x4 = 12
• Basic variable
Variable whose column has one entry 1 and the rest
0s.
• Basic solution
Solution obtained by setting all nonbasic variables
equal to 0 and solving for the basic variables.
• Basic feasible solution: all basic solution are
nonnegative
• Examples 1, 2 (p. 423, 424)
Stage 1: FINDING A BASIC FEASIBLE
SOLUTION
1. If any basic variable has a negative value,
locate the –1 in that variable’s column and
note the row it is in.
2. In the row determined in Step 1, choose a
positive entry (other than the one at the far
right) and note the column it is in. This is
the pivot column.
3. Use the positive entries in the pivot column
(except in the objective row) to form
quotients and select the pivot.
FINDING A BASIC FEASIBLE SOLUTION
4. Pivot as usual, which results in the pivot
column’s having one entry 1 and the rest
0s.
5. Repeat Steps 1-4 until every basic
variable is nonnegative, so that the basic
solution given by the tableau is feasible.
If it ever becomes impossible to
continue, the the problem has no feasible
solution.
Example 3 (p. 426)
Stage 2: Use simplex method
The same technique as in Standard
Maximization Problem
Example 4 (p. 426)
Example 5 (p. 428)
Example 6 (p. 430)
SOLVING NONSTANDARD PROBLEMS
1. Replace each equation constraint by an equivalent
pair of inequality constraints.
2. If necessary, write each constraints with a positive
constant
3. Convert the problem to a maximum problem by letting
z = – w.
4. Add slack variables and subtract surplus variables as
needed to convert the constraints into equations.
5. Write the initial simplex tableau.
6. Find a basic feasible solution for the problem, if one
exists (Stage I).
7. When a basic feasible solution is found, use the
simplex method to solve the problem (Stage II).