Chapter 3: Section 3-3 Solutions of Linear Programming ... · PDF fileChapter 3: Section 3-3...

Chapter 3: Section 3-3Solutions of Linear Programming Problems

D. S. MalikCreighton University, Omaha, NE

D. S. Malik Creighton University, Omaha, NE ()Chapter 3: Section 3-3 Solutions of Linear Programming Problems 1 / 21

Geometric Approach to Solve Linear Programming (LP)Problems

The objective of a linear programming problem is to maximize orminimize an objective function, subject to certain constraints.We have considered several examples illustrating how to formulate theconstraints and the objective function.In the constraints, the inequalities use either the symbol � or thesymbol �, i.e., none of the inequalities are strict.This is important because the �rst step in solving a linearprogramming problem using the geometric approach is to draw thegraphs of the inequalities and determine the region (solution set) thatsatis�es the linear inequalities.Next determine the corner points of the solution set.If an inequality is strict, then a corner point given by that inequalitymay not be in the solution set.Thus, in all of the linear programming problems that we consider, thelinear inequalities will not be strict.


Feasible Region

Feasible Region (Solution Set)

BoundedUnbounded


Bounded Feasible Region

Theorem(Bounded Region) Let F be the feasible region of a linear programmingproblem and F 6= ∅. Let f be the objective function of the linearprogramming problem. Suppose that F is bounded. Then(i) f attains its maximum value at a corner point of F ,(ii) f attains its minimum value at a corner point of F .


Procedure to �nd a solution of a linear programmingproblem when the feasible region is bounded.

1 Graph the linear inequalities and determine the feasible region(solution set).

2 Find the corner points of the feasible region.3 Evaluate the objective function at each of the corner points.4 If the problem is to maximize the objective function, then choose thelargest value of the objective function. The coordinates of the cornerpoint that gives the largest value is a solution of the linearprogramming problem. (If there is more than one such corner point,then choose one of those points.)

5 If the problem is to minimize the objective function, then choose thesmallest value of the objective function. The coordinates of thecorner point that gives the smallest value is a solution of the linearprogramming problem. (If there is more than one such corner point,then choose one of those points.)


ExampleConsider the following constraints of a linear programming problem:x � 3, x + y � 0, x � y � 0, �x + 2y � 5.

The region in the x-y plane that satis�es these inequalities is:

x

y

x + y = 0x - y = 0

A(-(5/3),(5/3))

B(3, 4)

C(3,3)

-x + 2y = 5

-5

x = 3

O(0,0)1

2

3

4

-1-2

0 1 2 3-1-2-3 4 5-4

Corner points of the feasible region are: O(0, 0), A�� 53 ,53

�, B(3, 4),

and C (3, 3).


Example

The corner points of the feasible region are: O(0, 0), A�� 53 ,53

�,

B(3, 4), and C (3, 3).

Let us minimize 6x � 3y subject to the given constraints.Because the feasible region is bounded, the minimum occurs at acorner point. Let us evaluate 6x � 3y at the corner points. Now

Corner Point Value of the Objective function: 6x � 3yO(0, 0) 6 � 0� 3 � 0 = 0A�� 53 ,53

�6 � (� 5

3 )� 3 �53 = �10� 5 = �15

B(3, 4) 6 � 3� 3 � 4 = 18� 12 = 6C (3, 3) 6 � 3� 3 � 3 = 18� 9 = 9

The minimum value of 6x � 3y is �15 and it occurs at A�� 53 ,53

�.


Example

The corner points of the feasible region are: O(0, 0), A�� 53 ,53

�,

B(3, 4), and C (3, 3).

Let us maximize 5x + 2y subject to the given constraints.

Because the feasible region is bounded, the maximum occurs at acorner point. Let us evaluate 5x + 2y at the corner points. Now

Corner Point Value of the Objective function: 5x + 2yO(0, 0) 5 � 0+ 2 � 0 = 0A�� 53 ,53

�5 � (� 5

3 ) + 2 �53 = �

253 +

103 = �

153

B(3, 4) 5 � 3+ 2 � 4 = 15+ 8 = 23C (3, 3) 5 � 3+ 2 � 3 = 15+ 6 = 21

The maximum value of 5x + 2y is 23 and it occurs at B(3, 4).


Exercise: Consider the following exercise described earlier.

A pet store specializes in cats and bunnies. Each cat costs $9 and eachbunny costs $6. The pro�t on each cat is $12 and on each bunny is $9.The store cannot house more than 30 animals and cannot spend morethan $216 to buy the pets. Under a special agreement the pet store musthouse at least 2 cats. How many pets of each type should be housed tomaximize the pro�t?

Let x be the number of cats and y be the numbers of bunnies.

The linear programming problem describing this problem is: Thelinear programming problem is:

maximize: 12x + 9ysubject to: x � 2, y � 0

x + y � 303x + 2y � 72


The graph of the inequalities x � 2, y � 0, x + y � 30,3x + 2y � 72 is:

x

y

0-10

10

20

30

3x + 2y = 72

x + y = 3040

10 20 30 40

C(12, 18)

B(2, 28)

A(2, 0)

x = 2

D(24, 0)

The corner points are A(2, 0), B(2, 28), C (12, 18), D(24, 0).


The corner points are A(2, 0), B(2, 28), C (12, 18), D(24, 0).

ExampleLet us evaluate 12x + 9y at the corner points. Now

Corner Point Value of the Objective function: 12x + 9yA(2, 0) 12 � 2+ 9 � 0 = 24B(2, 28) 12 � 2+ 9 � 28 = 276C (12, 18) 12 � 12+ 9 � 18 = 306D(24, 0) 12 � 24+ 9 � 0 = 288

The maximum value of 12x + 9y is 306 and it occurs at C (12, 18).

To maximize the pro�t, the pet store should house 12 cats and 18bunnies.


Unbounded Feasible Set

Theorem(Unbounded Feasible Region) Let F be the feasible region of a linearprogramming problem and F 6= ∅. Let f be the objective function of thelinear programming problem. Suppose that F is unbounded.(i) If f attains its maximum value, then the maximum value occurs at acorner point of F ,(ii) If f attains its minimum value, then the minimum value occurs at acorner point of F ..


Unbounded Feasible Set

RemarkIf the feasible region is bounded, then the preceding theorem does notguarantee that an objective function will have a maximum or minimum. Itonly guarantees that, if the maximum or minimum occurs, then it wouldoccur at a corner point. So how do we know that a maximum or minimumwill occur? To give a general answer to this problem, we will state atheorem at the end of this section. However, there are some special cases.Let us consider some special cases that you will encounter frequently whensolving a real-world problem.


Suppose that the feasible region is restricted to the �rst quadrant andit extends inde�nitely in the �rst quadrant. Consider the objectfunction 2x + 3y . Because the objective function extends inde�nitelyin the �rst quadrant, the value of 2x + 3y increases without boundswhen the values of x and y increase in the feasible region. That is,2x + 3y has no �xed largest value. So 2x + 3y has no maximum inthe feasible region. In fact, if the objective function is ax + by , wherea � 0, b � 0, and both a and b are not zero, then the objectivefunction ax + by has no maximum.

Suppose that the feasible region is restricted to the �rst quadrant andit extends inde�nitely in the �rst quadrant. However, the objectivefunction, say 3x + 5y , is to be minimized. In this case, the minimumwill occur at a corner point, because the value of 3x + 5y cannot bemade arbitrarily small. Thus, in general, in this case, if the objectivefunction is ax + by , where a � 0, b � 0, and both a and b are notzero, then the objective function ax + by has the minimum, and theminimum will occur at a corner point.


Suppose that the feasible region is restricted to the third quadrantand it extends inde�nitely in the third quadrant. Suppose that theobjective function is ax + by , where a � 0, b � 0, and both a and bare not zero. Then ax + by has the maximum, and the maximum willoccur at a corner point. Furthermore, ax + by has no minimum.


Example

A hiker wants to take a snack mix of peanuts and raisins.The hiker wants 1000 calories and 120 grams of fat from the mix.Each gram of peanuts contains 8 calories and 0.4 grams of fat, andcosts 7 cents.Each gram of raisins contains 2 calories and 0.6 grams of fat, andcosts 4 cents.How many grams of each food should the hiker take so that the costof the snack is the minimum?Suppose that the hiker takes x grams of peanuts and y grams ofraisins.Then

Variable Calories Fat CostEach gram of peanuts x 8 0.4 7Each gram of raisins y 2 0.6 4Total (required) � 1000 � 120


The linear programming problem is:

minimize: 7x + 4y

subject tox � 0y � 08x + 2y � 10000.4x + 0.6y � 120.


x � 0, y � 0, 8x + 2y � 1000, 0.4x + 0.6y � 120.We graph these inequalities and obtain the following graph:

100

0 100 200 300 400 500

200

4x + y = 500

y

400

300

500

x600 700

A(0, 500)

C(300, 0)

B(90, 140)

600

2x + 3y = 600

The corner points are: A(0, 500), B(90, 140), C (300, 0)


The corner points are: A(0, 500), B(90, 140), C (300, 0)

Let us evaluate the objective function at the corner points. Thus,

Corner point Value of the objective function: 7x + 4yA(0, 500) 7 � 0+ 4 � 500 = 2000B(90, 140) 7 � 90+ 4 � 140 = 630+ 560 = 1190C (300, 0) 7 � 300+ 4 � 0 = 2100

From this table, the minimum value is 1190 and this value occurs atB(90, 140). Thus, the minimum occurs when x = 90 and y = 140.

Thus, the hiker must take 90 grams of peanuts and 140 grams ofraisins to minimize the cost.


Theorem(Solutions of Linear Programming Problems with UnboundedFeasible Sets). Let L be a linear programming problem and S be thefeasible region of L. Let f be the objective function of L. Suppose that S isunbounded.(i) Suppose that f is to be maximized. Let T be the set of all cornerpoints of S . Let A be a corner point in T that gives the maximum value off . Let l1 and l2 be the boundary lines of S such that l1 and l2 intersect atA. On each line l1 and l2, choose a point in the feasible region that is nota corner point and evaluate f at that point. If the value of f at that pointis less than or equal to the value of f at A, then the linear programmingproblem has a solution, and it is the largest value that is attained at acorner point. Otherwise the problem has no solution.


Theorem(ii) Suppose that f is to be minimized. Let T be the set of all cornerpoints of S . Let A be a corner point in T that gives the minimized valueof f . Let l1 and l2 be the boundary lines of S such that l1 and l2 intersectat A. On each line l1 and l2, choose a point in the feasible region that isnot a corner point and evaluate f at that point. If the value of f at thatpoint is greater than or equal to the value of f at A, then the linearprogramming problem has a solution, and it is the smallest value that isattained at a corner point. Otherwise the problem has no solution.


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