 Chapter 7 LINEAR PROGRAMMING.

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Chapter 7 LINEAR PROGRAMMING

7.1 GRAPHING LINEAR INEQUALITIES IN 2 VARIABLES
Terms: Boundary Half-plane Feasible region Example 1: 3x – 2y  6 Example 2: y < – 3x + 12 x < 2y

7.1 GRAPHING LINEAR INEQUALITIES IN 2 VARIABLES
Graph the boundary line. Decide whether the line is part of the solution (use solid line for  and , dashed line for > and <) 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
Write the objective function and all necessary constraints. Graph the feasible region. Determine the cordinates of each corner points. 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
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:

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: The Objective Function Is To Be Maximized; All Variables Are Nonnegative. All Constraints Involve . 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 Determine The Objective Function.
Write All Necessary Constraints. Convert Each Constraint Into An Equation By Adding Slack Variables. Set Up The Initial Simplex Tableau. Locate The Most Negative Indicator. If There Are Two Such Indicators, Choose One. This Indicator Determines The Pivot Column.

SIMPLEX METHOD (cont.) 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.) 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.) 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. 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 people watch the local channel, the regional channel, and 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) 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. What is the optimal solution? Interpret the solution.

7.6 THE SIMPLEX METHOD: DUALITY AND MINIMIZATION
Minimization problem: The objective function is to be minimized. All variables are nonnegative. All constraints involve . 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 is matrix

Example of minimization problem:
Minimize w = 8y1 + 16y2 Constraints: y1 + 5y2  9 2y1 + 2y2  10 y1  0, y2  0. Dual problem: Maximize z = 9x x2 Constraints x1 + 2x2  8 5x1 + 2x2  16 x1  0, x2  0

SUMMARY Minimization problem Dual problem M variables N variables
N constraints M constraints Coefficients of objective function Constants 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
Find the dual standard maximum problem. Solve the maximum problem using the simplex method. The minimum value of the objective function w is the maximum value of the objective function z. 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
If any basic variable has a negative value, locate the –1 in that variable’s column and note the row it is in. 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. 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
Pivot as usual, which results in the pivot column’s having one entry 1 and the rest 0s. 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
Replace each equation constraint by an equivalent pair of inequality constraints. If necessary, write each constraints with a positive constant Convert the problem to a maximum problem by letting z = – w. Add slack variables and subtract surplus variables as needed to convert the constraints into equations. Write the initial simplex tableau. Find a basic feasible solution for the problem, if one exists (Stage I). When a basic feasible solution is found, use the simplex method to solve the problem (Stage II).