Ch 7 LINEAR PROGRAMMING

1
Ch 7 LINEAR PROGRAMMING

• Linear programming1 is a mathematical technique
  that enables a decision maker to arrive at the
  optimal solution to problems involving the
  allocation of scarce resources.
• Typically, many economic and technical problems
  involve maximization or minimization of a certain
  objective subject to some restrictions.
• 1During World War II US army began to formulate
  certain linear optimization problems. Their
  solutions were called plans or programs. Today
  important application areas include airline crew
  scheduling, shipping or telecommunication
  networks, oil refining and blending, and stock
  and bond portfolio selection.

2
The History of Linear Programming
Leonid Kantorovich
George Dantzig
3

• Programming problems, in general, are concerned
  with the use or allocation of scarce resources -
  labor, materials, machines, and capital - in the
  best possible manner so that costs are
  minimized or profits maximized.
• In using the term best possible it is implied
  that some choice or set of alternative courses of
  actions is available for making the decision.

4
Typical Applications of Linear Programming

• 1. A manufacturer wants to develop a production
  schedule and inventory policy that will satisfy
  sales demand in future periods and same time
  minimize the total production and inventory cost.
• 2. A financial analyst must select an investment
  portfolio from a variety of stock and bond
  investment alternatives. He would like to
  establish the portfolio that maximizes the return
  on investment.

5
Typical Applications of Linear Programming
continued

• 3. A marketing manager wants to determine how
  best to allocate a fixed advertising budget among
  alternative advertising media such as radio, TV,
  newspaper, and magazines. The goal is to maximize
  advertising effectiveness.
• 4. A company has warehouses in a number of
  locations throughout the country. For a set of
  customer demands for its products, the company
  would like to determine how much each warehouse
  should ship to each customer so that the total
  transportation costs are minimized.

6
Constructing Linear Programming Models

• Next we list what is required in order to
  construct a linear programming model
• 1. Objective Function. There must be an objective
  (or goal or target) the firm or organization
  wants to achieve. For example, maximize dollar
  profits, minimize dollar cost, maximize total
  number of expected potential customers, minimize
  total time used, and so forth.

7
Constructing Linear Programming Models continued

• 2. Restrictions and Decisions. There must be
  alternative courses of action or decisions, one
  of which will achieve the objective.
• 3. Linear Objective Function and Linear
  Constraints. We must be able to express the
  decision problem incorporating the objective and
  restrictions on the decisions using only linear
  equations and linear inequalities. i.e., we must
  be able to state the problem as a linear
  programming model.

8
Economic Significance of Linearity

• The simplifying assumption of linearity causes
  some problems
• 1. In profit-maximizing production problem
  linearity of the objective function implies
  constant profit rate per unit as output
  increases this means
• a) that the selling price is constant and

9

• b) that average variable cost is constant
• the law of diminishing returns does not
  influence the production process, and input
  prices are constant
• ? perfect competition in output and input markets
• 2. Linear resource constraints imply constant
  combination of inputs
• this means constant returns to scale.

10
Summary of the Economic Implications of the
Linearity Assumption
Linear objective function
Constant gross profit per unit (GP P - AVC
constant)
Price (P) is constant
Constant input prices
Constant returns to variable inputs
Firm is a price taker in the output market
11
The three basic steps in constructing a linear
programming model

• Step I
• Identify the unknown variables to be determined
  (decision variables), and represent them in terms
  of algebraic symbols.

12

• Step II
• Identify all the restrictions or constraints in
  the problem and express them as linear equations
  or inequalities which are linear functions of the
  unknown variables.

13

• Step III
• Identify the objective or criterion and
  represent it as a linear function of the decision
  variables, which is to be maximized or minimized.

14
Example 1 Product-Mix Problem

• The Handy-Dandy Company wishes to schedule the
  production of a kitchen appliance which requires
  two resources labor and material. The company
  is considering three different models of this
  appliance and its engineering department has
  furnished the following data

15
The supply of raw materials is restricted to 200
pounds per day. The daily availability of
manpower is 150 hours. Formulate a linear
programming model to determine the daily
production rate of the various models of
appliances in order to maximize the total profit.
16

• Step I
• Identify the Decision Variables. The unknown
  activities to be determined are the daily rate of
  production for the three models (A, B, C) in
  order to maximize the total profit. Representing
  them by algebraic symbols,
• xA daily production of model A
• xB daily production of model B
• xC daily production of model C

17

• Step II
• Identify the Constraints. In this problem the
  constraints are the limited availability of the
  two resources (labor and material).
• Model A requires 7 hours of labor for each unit,
  and its production quantity is xA. Hence, the
  requirement of manpower for model A alone will be
  7xA hours (assuming a linear relationship).

18

• Similarly, models B and C will require 3xB and
  6xC hours, respectively. Thus, the total
  requirement of labor will be
• 7xA 3xB 6xC, which should not exceed the
  available 150 hours. So the labor constraint
  becomes
• 7xA 3xB 6xC ? 150

19

• Similarly, the raw material constraint is given
  by
• 4xA 4xB 5xC ? 200
• In addition, we restrict the variables to have
  non-negative values. This is called the
  non-negativity constraint, which the variables
  must satisfy
• xA, xB and xC ? 0.

20

• Step III
• Identifying the Objective. The objective is to
  maximize the total profit from the sales.
  Assuming that perfect market exists for the
  product such that all that is produced can be
  sold, the total profit from sales becomes
• Z 4xA 2xB 3XC.

21
Thus, the linear programming model for our
product mix problem is

• Find numbers xA, xB, xC which will maximize
• Z 4xA 2xB 3XC
• subject to the constrains
• 7xA 3xB 6xC ? 150
• 4xA 4xB 5xC ? 200
• xA ? 0, xB ? 0, xC ? 0

22
Exercise 1 Formulating an LP-Problem

• Advertising Media Selection
• An advertising company wishes to plan an
  advertising campaign in three different media
  television, radio, and magazines. The purpose of
  the advertising program is to reach as many
  potential customers as possible. Result of a
  market study are given below
•   

23
Solving Linear Programming Problems

• Graphical Technique
• First graph the constraints
• the solution set of the system is that region
  (or set of ordered pairs), which satisfies ALL
  the constraints. This region is called the
  feasible set

24
Solving Linear Programming Problems Graphical
Technique continued

• Locate all the corner points of the graph
• the coordinates of the corners will be
  determined algebraically
• It is important to note that the optima is
  obtained at the boundary of the solution set and
  furthermore at the corner points.
• For linear programs, it can be shown that the
  optima will always be obtained at corner points.

25
Solving Linear Programming Problems Graphical
Technique continued

• Determine the optimal value
• test all the corner points to see which yields
  the optimum value for the objective function

Objective function
Feasible set
Optimum
26
Example 2

• Suppose a company produces two types of widgets,
  manual and electric. Each requires in its
  manufacture the use of three machines A, B, and
  C. A manual widget requires the use of the
  machine A for 2 hours, machine B for 1 hour, and
  machine C for 1 hour. An electric widget requires
  1 hour on A, 2 hours on B, and 1 hour on C.
  Furthermore, suppose the maximum numbers of hours
  available per month for the use of machines A, B,
  and C are 180, 160, and 100, respectively. The
  profit on a manual widget is 4 and on electric
  widget it is 6. See the table below for a
  summary of data. If the company can sell all the
  widgets it can produce, how many of each type
  should it make in order to maximize the monthly
  profit?

27
Example 2 continued

• Step I Identify decision variables
• x number of manual widgets
• y number of electric widgets
• Step II Identify constraints
• 2x y ? 180
• x 2y ? 160
• x y ? 100
• x ? 0
• y ? 0

28
Example 2 continued

• Step III Define objective function
• max P 4x 6y
• Solving P for y gives
• y -2/3 P/3.
• This defines a so-called family of parallel
  lines, isoprofit lines.
• Each line gives all possible combinations of x
  and y that yield the same profit.

29
Example 2 continued
30
Example 2 continued
31
Exercise 2 Solving an LP-problem

• A California vintner has available 660 lbs of
  Cabernet Sauvignon (CS) grapes, 1860 lbs of Pinot
  Noir (PN) grapes, and 2100 lbs of Barbera (B)
  grapes. The vintner makes a Pinot Noir (PN) wine,
  which contains 20 CS, 60 PN, and 20 B grapes
  and sells 3 a bottle, and a Barbera (B) wine,
  which contains 10 CS, 20 PN, and 70 B grapes
  and sells for 2 a bottle. Assuming each bottle
  of wine requires 3 lbs of grapes, determine how
  many bottles of each type of wine should be
  produced to maximize income.

32
Solving an LP-problem continued

• Algebraic Technique
• The graphical method solving linear programming
  problems can be used for problems with two
  variables (with some difficulty three)
• However, for problems were the number of
  variables might run into hundreds or thousands,
  algebraic techniques must be used
• The simplex method, with the aid of the computer,
  can solve these problems

33

• As with the graphical procedure, the simplex
  method finds the optimal corner-point solution of
  the set of feasible solutions.
• Regardless of the number of decision variables
  and regardless of the number of constraints, the
  simplex method uses the key property of a linear
  programming problem, which is

34

• A linear programming problem always has an
  optimal solution occurring at a corner-point
  solution
• Simplex method begins with a feasible solution
  and tests whether or not it is optimum.
• If not optimum, the method proceeds to a better
  solution.

35
Solution of a maximum-type linear programming
problem by the simplex algorithm involves the
following steps

• 1. Adding slack variables to convert the
  inequalities into equations
• In the case of less-than-or-equal-to
  constraints, slack variables are used to increase
  the left-hand side to equal the right-hand side
  limits of the constraint conditions.

36

• Example max Z 3x1 x2
• st. 2x1 x2 ? 8
• 2x1 3x2 ? 12
• adding slacks and rewriting the object row
• 2x1 x2 s1 8
• 2x1 3x2 s2 12
• -3x1 x2 Z 0
• where x1, x2, s1 and s2 are non-negative.

37
2x1 x2 s1 8 2x1 3x2
s2 12 -3x1 x2 Z 0

• 2. Setting up the initial simplex tableau
• Form an augmented coefficient matrix

38

• 3. Finding an initial feasible solution
• Simplex algorithm starts always from origin (if
  possible). In our case it means that the initial
  feasible solution is
• x1 0, x2 0, s1 8, s2 12.
• Simplex method proceeds from this corner point
  to an adjacent corner point. Such corner points
  are called basic feasible solutions.

39
4. Introducing basic and nonbasic variables as
the natural way to express basic feasible
solutions

• For any basic feasible solution (B.F.S.), the
  variables held zero are called nonbasic variables
  and the other are called basic variables.
• Moving form one B.F.S. to another means that one
  basic variable (we call it a departing or exiting
  variable) becomes nonbasic and a nonbasic
  variable (entering variable) becomes a basic
  variable.

40
(No Transcript)
41

• 5. Choosing the proper pivot element to advance
  the solution and maintain the non-negativity of
  all variables
• How to decide which variable to make basic and
  which nonbasic (in other words in which direction
  to move from the current B.F.S.)?

42
The natural direction is the one in which the
value of the objective function increases the
most

• If in Z 3x1 x2, x1 is allowed to become
  basic, x2 remains at 0 and Z 3x1 thus for each
  one-unit increase in x1, Z increases by 3 units.
  On the other hand, if x2 is allowed to become
  basic, Z will increase only by one unit if x2 is
  increased by one unit. This is one way to
  determine the pivot column.

43
Choosing the proper pivot element continued

• Another way to determine the pivot column is to
  examine the bottom row of the simplex tableau

indicators
entering variable
44

• The bottom row entries to the left of the
  vertical line are called indicators.
• We choose the column with the most negative
  indicator as the pivot column.
• Having chosen the pivot column, we must now
  determine the pivot row in order to know which
  element in our simplex tableau is the pivoting
  element.

45

• For that purpose we divide the right hand side
  entries of the simplex tableau by corresponding
  entries of the pivot column.
• Of the resulting quotients we choose the smallest
  (minimum quotient). So the pivot element is the
  intersection of the pivot column and pivot row.

46
The pivot element is the intersection of the
pivot column and pivot row
Quotients 8 2 4 (smaller) 12 2 6
departing variable
entering variable (most negative indicator)
47

• Since x1 and s2 will be the basic variables in
  our new B.F.S, it would be convenient to change
  our previous tableau by elementary row operations
  into a form where the values of x1, s2, and Z can
  be read off with ease just as in the initial
  basic solution.

48
To do this we want to find a matrix which is
equivalent to the tableau above but which has the
form
where the question marks represent numbers to be
determined.
49
6. Pivoting, which is done column wise

• We must transform the tableau to an equivalent
  matrix that has a 1 at the place of the pivot
  element (pivot entry) and 0s elsewhere in the
  column x1.

departing variable
entering variable
50
By elementary row operations, we have
R1 R2 R3
½ R1
-2R1 R2
3R1 R3
51

• 7. Continuing and recognizing when the algorithm
  terminates.
• General termination rule all values in the
  indicator row are non-negative.

52
We have a new simplex tableau

indicators
All values in the indicator row are non-negative.
Hence, we have found the optimal solution for the
problem. The solution is x1 4, x2 0 and Z 12