Solving Linear Programs (LP Section 2)

1
Solving Linear Programs (LP Section 2)

• Chapter 2 (Revisited)
• Graphical Method
• Computer Solutions
• Using LINDO
• Interpreting Computer Output for LP Section 1
  problems

2
Solving Linear Programs

• The goal of a linear program is to find the
  values of your xj variables that provide the
  largest (or smallest) possible objective value
  when substituted into the objective function.
• The Graphical Method can be used to solve the LP
  for these values when you have one or two
  decision variables. This technique helps to
  visualize key LP concepts.
• Computer Programs are generally used to solve all
  LP problems.
• We will learn to use the graphical method to
  solve small problems and gain insight into some
  theory behind solving LPs. Nevertheless, we will
  focus on using a computer software to solve our
  LP models.

3
Solving LPs using the Graphical MethodChapter 2

• We begin the graphical solution process by
  graphing each linear constraint of the LP model
  on the same graph.
• Refer to the PAR, INC example from Section 1 of
  your LP notes.
• Maximize 10 x1 9 x2
• Subject to
• 7/10 x1 1 x2 630
• 1/2 x1 5/6 x2 600
• 1 x1 2/3 x2 708
• 1/10 x1 1/4 x2 135
• x1 0, x2 0

4
Graphical Method PAR, INC

• Constraint 1 (C1)
• 7/10 x1 1 x2 630 Always graph
  constraints as an equality.
• In order to graph a line we need two points that
  fall on the line. Arbitrarily, let x1 0. Then
  7/10 (0) x2 630. Solving this equation for x2
  gives x2 630. Therefore one point on this line
  is (0, 630).
• Arbitrarily, let x2 0. Then 7/10 x1 (0)
  630. Solving this equation for x1 gives x1
  6300/7900. Therefore a 2nd point on this line
  is (900, 0). Connect the points (0, 630) and
  (900,0) to draw this line.
• Determine the Feasible Side of the Line
• In LP, on one side of the line all points are
  feasible (i.e. they satisfy the constraint) and
  on the other side all points are not feasible.
  For instance take the point (0,0). The point x1
  0 and x2 0 satisfy the above constraint 7/10
  (0) (0) is less than 630. Thus all points
  that are on the side of the line containing this
  point will also satisfy this constraint. We put
  a red arrow to indicate the feasible side of the
  line.

5
Graphical Method PAR, INC
x2
C1
x1
6
Graphical Method PAR, INC

• Constraint 2 (C2)
• 1/2 x1 5/6 x2 600
• In order to graph a line we need two points that
  fall on the line. Arbitrarily, let x1 0. Then
  1/2 (0) 5/6 x2 600. Solving this equation for
  x2 gives x2 720. Therefore one point on this
  line is (0, 720).
• Arbitrarily, let x2 0. Then x1 1200.
  Therefore a 2nd point on this line is (1200, 0).
  Connect the points (0, 720) and (1200,0) to draw
  this line.
• Determine the Feasible Side of the Line
• Use the point (0,0) as a test point. Since x1
  0 and x2 0 satisfies the above constraint 1/2
  (0) 5/6 (0) is less than 600. Thus all
  points on the side containing the point (0,0)
  will satisfy this constraint. We put a red arrow
  to indicate the feasible side of the line.

7
Graphical Method PAR, INC
C2
x2
C1
x1
8
Graphical Method PAR, INC

• Constraint 3 (C3)
• 1 x1 2/3 x2 708
• In order to graph a line we need two points that
  fall on the line. Arbitrarily, let x1 0. Then
  x2 1062. Therefore one point on this line is
  (0, 1062).
• Arbitrarily, let x2 0. Then x1 708. Therefore
  a 2nd point on this line is (708, 0). Connect
  the points (0, 1062) and (708,0) to draw this
  line.
• Determine the Feasible Side of the Line
• Using our test point point (0,0), we find that
  x1 0 and x2 0 satisfies the above constraint 1
  (0) 2/3 (0) is less than 708. Thus all
  points on the side containing the point (0,0)
  will satisfy this constraint. We put a red arrow
  to indicate the feasible side of the line.

9
Graphical Method PAR, INC
C3
x2
C1
C2
x1
10
Graphical Method PAR, INC

• Constraint 4 (C4)
• 1/10 x1 1/4 x2 135
• Arbitrarily, let x1 0. Then x2 540.
  Therefore one point on this line is (0, 540).
• Arbitrarily, let x2 0. Then x1 1350.
  Therefore a 2nd point on this line is (1350, 0).
  Connect the points (0, 540) and (1350,0) to draw
  this line.
• Determine the Feasible Side of the Line
• Using our test point (0,0), we find that x1 0
  and x2 0 satisfies the above constraint 1 (0)
  2/3 (0) is less than 708. Thus all points on
  the side containing the point (0,0) will satisfy
  this constraint. We put a red arrow to indicate
  the feasible side of the line.

11
Graphical Method PAR, INC
C3
x2
C1
C4
C2
x1
12
Graphical Method PAR, INC

• Nonnegativity Constraints
• These constraints simply restrict us to the 1st
  quadrant of our coordinates (i.e. only 0 or
  positive values of x1 and x2 are considered
  feasible on our graph when we highlight our final
  feasible region).

13
Graphical Method PAR, INC
C3
x2
C1
C4
C2
x1
14
Graphical Method PAR, INC

• After graphing each of the constraints and their
  feasible sides individually, we must determine
  all points on the coordinates which satisfy all
  constraints. These points make up the feasible
  region.

15
PAR, INC GRAPHICAL METHODFEASIBLE REGION
C3
C1
x2
C4
x1
16
FEASIBLE REGION

• Once the feasible region is drawn, we must
  determine the point in this region that maximizes
  (or minimizes) the objective function.

17
Extreme Points and the Optimal Solution

• The corners or vertices of the feasible region
  are referred to as the extreme points.
• Fundamental Theorem of Linear Programming
• An optimal solution to an LP problem can be
  found at an extreme point of the feasible region.
• Optimal point occurs on the objective function
  line corresponding to the optimal objective
  function value
• When looking for the optimal solution, you do not
  have to evaluate all feasible solution points.
• You have to consider only the extreme points of
  the feasible region.

18
PAR, INC GRAPHICAL METHODFEASIBLE REGION AND
EXTREME POINTS
x2
x1
19
Finding the Optimal Point

• How to find the optimal point
• Randomly draw objective function line.
• Push the line in the direction of decrease (if
  minimization problem) or the direction of
  increase (if maximization problem)
• The last point that you encounter in the feasible
  region is the optimal point

20
Drawing an Objective Function Line

• Randomly select a point in the feasible region.
• Substitute the coordinates of this point into the
  objective function to obtain an objective
  function value.
• Plot the line obtained by setting the original
  objective function equal to the objective
  function value obtained in 2.
• PAR, INC Example
• Step 1 The point (200,0) is in the feasible
  region.
• Step 2 PAR INC OBJECTIVE FUNCTION 10 x1 9 x2
  . (200,0) gives 10 (200) 9 (0) 2000.
• Step 3 Plot the line 10 x1 9 x2 2000.
• Connect the points (0, 222.22) and
  (200, 0)

21
PAR, INC GRAPHICAL METHODFEASIBLE REGION AND
EXTREME POINTS
x2
Obj. Func. line
x1
22
FINDING THE DIRECTION OF INCREASE (For
Maximization problems)

• Maximization problem Finding the direction of
  increase
• Let v1 the coefficient of x1 in the objective
  function.
• Let v2 the coefficient of x2 in the objective
  function.
• Plot the point (v1, v2) on the graph.
• Draw an arrow from the origin (the point x10,
  x20) to the point (v1,v2). This is the
  direction of increase. Push the objective
  function in this direction until you encounter
  the last point in the feasible region. This point
  is optimal.

23
FINDING THE DIRECTION OF DECREASE (For
Minimization problems)

• Minimization problem Finding the direction of
  decrease
• Let v1 - (the coefficient of x1 in the objective
  function).
• Let v2 - (the coefficient of x2 in the objective
  function).
• Plot the point (v1, v2) on the graph.
• Draw an arrow from the origin (the point x10,
  x20) to the point (v1,v2). This is the
  direction of decrease. Push the objective
  function in this direction until you encounter
  the last point in the feasible region. This
  point is optimal.

24
PAR INC GRAPHICAL METHOD

• PAR INC objective function is
• Maximize 10 x1 9 x2
• Thus v110 and v29. I can draw an arrow from
  the origin in the direction of the point (10,9).
  This is the direction of increase.

25
PAR, INC GRAPHICAL METHODFEASIBLE REGION AND
EXTREME POINTS
x2
Direction of Increase
This is the optimal point.
Obj. Func. line
x1
26
PAR, INC GRAPHICAL METHODFEASIBLE REGION AND
EXTREME POINTS
x2
Slide 15 shows that the optimal point is formed
by the intersection of Constraint 1 and
Constraint 3.
x1
27
PAR INC GRAPHICAL SOLUTION

• Constraint 1 7/10 x1 1 x2 630
• Constraint 3 1 x1 2/3 x2 708
• Find the point where these two lines intersect.
• Solve the first equation for x2
• x2 630-7/10 x1
• Substitute the above equation for x2 in the 2nd
  equation.
• 1 x1 2/3(630-7/10 x1)708
• Solving the above equation for x1 gives, x1
  540. Thus
• x2 630-7/10 (540) 252.
• The optimal point denoted x1540, x2252.

28
PAR INC GRAPHICAL SOLUTION

• If we substitute the optimal point x1540,
  x2252 into the objective function, 10 (540) 9
  (252) 7668.
• Thus PAR INC should produce 540 Standard bags and
  252 Deluxe bags to receive a maximum profit of
  7668.

29
Discussion about the Graphical Method

• Refer to Additional Graphical Method Problems
  link in LP Section 2 folder and/ or your textbook
  for more practice solving problems via the
  graphical method.
• You will have one homework (containing 2
  problems) dedicated to this approach. YOU WILL
  NOT HAVE TO SOLVE A PROBLEM USING THIS TECHNIQUE
  ON AN EXAM.
• This approach to solving Linear Programming
  models allows us to visualize the process that
  computer software use to soft LP models.
• Computer software use the knowledge that the
  optimal point is an extreme point of the feasible
  region (There are characteristics of extreme
  points that allow the computer to recognize
  them). The software will intelligently search
  these extreme points to determine the optimal
  solution to an LP model.

30
Computer Solutions

• Computer programs designed to solve LP problems
  are now widely available.
• Most large LP problems can be solved with just a
  few minutes of computer time.
• Small LP problems usually require only a few
  seconds.
• We will use LINDO to solve our LP problems.

31
Using LINDO to Solve

• Look in the LP Section 2 folder and click on the
  link Downloading Free LINDO Software and follow
  the instructions.
• The Par Inc file located in this folder is a MS
  Word file which explains how to enter (and solve)
  a program in LINDO.
• Enter the other examples presented in this
  chapter into LINDO.

32
Interpretation of Computer Output

• We will discuss the following output
• objective function value
• values of the decision variables
• reduced costs
• slack/surplus

33
Reduced Cost

• The reduced cost for a decision variable whose
  value is 0 in the optimal solution is the amount
  the variable's objective function coefficient
  would have to improve (increase for maximization
  problems, decrease for minimization problems)
  before this variable could assume a positive
  value.
• The reduced cost for a decision variable with a
  positive value is 0.
• Example
• Consider the following objective function
• Min 2 x1 5 x2 4 x3
• Suppose the optimal value of x1 is zero, with a
  reduced cost of 1.2. Since this is a
• minimization problem, this tells us that the
  current coefficient of x1 , which is 2,
• must be decreased by 1.2 in order for the optimal
  value of x1 to be nonzero. Thus
• if the objective function coefficient of x1 was
  0.8 (or less), resolving the LP would
• yield a nonzero value of x1.

34
SLACK/SURLUS

• The slack for less than or equal to constraints
  is the difference between the right hand side of
  an equation and the value of the left hand side
  after substituting the optimal values of the
  decision variables.
• The slack represents the amount of unused units
  of the right hand side resources.
• The surplus for greater than or equal to
  constraints is the difference between the right
  hand side of an equation and the value of the
  left hand side after substituting the optimal
  values of the decision variables.
• The surplus represents the number of units in
  which the optimal solution causes the constraint
  to exceed the right hand side lower limit.

35
Par INC LINDO Solution

• LP OPTIMUM FOUND AT STEP 2
• OBJECTIVE FUNCTION VALUE
• 1) 7667.942
• VARIABLE VALUE REDUCED COST
• X1 539.984253 0.000000
• X2 252.011032 0.000000
• ROW SLACK OR DUAL PRICES
• SURPLUS
• 2) 0.000000
  4.374566
• 3) 120.007088
  0.000000
• 4) 0.000000
  6.937804
• 5) 17.998819
  0.000000
• 6) 539.984253
  0.000000
• 7) 252.011032
  0.000000

Constraint 1
Constraint 2
Constraint 3
Constraint 4
X1gt0 Constraint
X2gt0 Constraint
36
Par INC LINDO Solution

• RANGES IN WHICH THE BASIS IS UNCHANGED
• OBJ
  COEFFICIENT RANGES
• VARIABLE CURRENT ALLOWABLE
  ALLOWABLE
• COEF INCREASE
  DECREASE
• X1 10.000000 3.499325
  3.700000
• X2 9.000000 5.285714
  2.333000
• RIGHTHAND SIDE
  RANGES
• ROW CURRENT ALLOWABLE
  ALLOWABLE
• RHS INCREASE
  DECREASE
• 2 630.000000 52.358864
  134.400009
• 3 600.000000 INFINITY
  120.007088
• 4 708.000000 192.000000
  127.986000
• 5 135.000000 INFINITY
  17.998819
• 6 0.000000 539.984253
  INFINITY
• 7 0.000000 252.011032
  INFINITY

37
PAR INC

• Recall that we rounded values (instead of
  entering fractions) when entering our model.
  Therefore, our LINDO output will be slightly
  different from the actual solution.
• From the spreadsheet we see that the maximum that
  PAR can achieve while meeting the constraints is
  7667.942 or 7668.
• The optimal solution is to produce X1
  539.984253 or 540 standard bags, X2252.011032
  or 252 deluxe bags.
• The reduced costs for both decision variables is
  zero since their optimal values are nonzero.
• Row 1 represents the objective function.
• Row 2 represents constraint 1, row 3 represents
  constraint 2, row 4 represents constraint 3, etc
• Refer to the order in which you entered the model
  into LINDO.

38
PAR INC (Slack/Surplus)

• Constraint 1 is a less than or equal to
  constraint (0.7 x1 x2 630)
• Plugging the optimal values into the left hand
  side of constraint 1 gives
• 0.7(540) 1 (252)2 630
• Since the left hand side (with optimal values
  substituted) equals the left hand side, the slack
  is 0.
• Constraint 2 is a less than or equal to
  constraint (0.5 x1 0.8333 x2 600)
• Plugging the optimal values into the left hand
  side of constraint 2 gives
• 0.5 (540) 0.8333(252) 480
• The difference between this value and 600 is
  120. The spreadsheet lists 120 as the slack for
  this constraint.
• Recall from LP Section 1that Constraint 2
  represents the amount of sewing hours available
  for use. This slack indicates that the optimal
  solution leaves 120 sewing hours unused.
  Suggesting that less workers may be used to
  achieve the optima profit since so many hours
  will be wasted.

39
PAR INC (Slack/Surplus)

• The output also reveals that 18 hours of
  Inspection and packing will also go unused.
• The second output sheet gives information about
  sensitivity analysis. We will discuss this topic
  in the next section.

40
Floataway Tours LINDO Solution

• OBJECTIVE FUNCTION VALUE
• 1) 5040.000
• VARIABLE VALUE REDUCED COST
• X1 28.000000 0.000000
• X2 0.000000 2.000000
• X3 0.000000 12.000000
• X4 28.000000 0.000000
• ROW SLACK OR SURPLUS DUAL PRICES
• 2) 0.000000 0.012000
• 3) 6.000000 0.000000
• 4) 0.000000 -2.000000
• 5) 52.000000 0.000000
• 6) 28.000000 0.000000
• 7) 0.000000 0.000000
• 8) 0.000000 0.000000
• 9) 28.000000 0.000000

41
Floataway Tours LINDO Solution

• OBJ
  COEFFICIENT RANGES
• VARIABLE CURRENT ALLOWABLE
  ALLOWABLE
• COEF
  INCREASE DECREASE
• X1 70.000000
  44.999996 1.875001
• X2 80.000000
  2.000001 INFINITY
• X3 50.000000
  12.000001 INFINITY
• X4 110.000000
  INFINITY 16.363636
• RIGHTHAND SIDE RANGES
• ROW CURRENT ALLOWABLE
  ALLOWABLE
• RHS
  INCREASE DECREASE
• 2 420000.000000 INFINITY
  45000.000000
• 3 50.000000
  6.000000 INFINITY
• 4 0.000000
  70.000000 30.000000
• 5 200.000000
  52.000000 INFINITY
• 6 0.000000
  28.000000 INFINITY
• 7 0.000000
  0.000000 INFINITY
• 8 0.000000
  0.000000 INFINITY
• 9 0.000000
  28.000000 INFINITY

42
Floataway Tours LINDO Solution Discussion

• The maximum daily profit that they can achieve
  (while meeting constraints) is 5040.
• The optimal solution is x128 or purchase 28
  Speedhawks and x228 or purchase 28 Classys.
• The reduced costs for Speedhawks and Classys is 0
  since their optimal values are nonzero.
• The reduced cost for Silverbirds (x2) is 2. This
  means that the objective function coefficient for
  x2 (currently 80) must be improved by 2 in order
  for the optimal value of x2 to become nonzero
  (i.e. in order for it to become profitable to
  order Silverbirds). Since this is a maximization
  problem, improved means increased. Therefore if
  the daily expected profit for Silverbirds was 82
  opposed to 80, it would be profitable to order
  Silverbirds.
• Similarly if the objective function coefficient
  for Catmans (x3 ) was 62 (instead of 50) it would
  be profitable to order these type of boats.

43
Floataway Tours LINDO Solution Discussion

• The surplus for constraint 2 (the requirement
  that at least 50 boats be purchased) shows that
  they purchase 6 boats over the 50 minimum
  requirement.
• The surplus for constraint 4 ( the requirement
  that they have a seating capacity of at least
  200) shows that they can seat 52 passengers over
  the 200 minimum requirement.

44
Police Scheduling LINDO Solution

• OBJECTIVE FUNCTION VALUE
• 1) 19.00000
• VARIABLE VALUE REDUCED COST
• X1 0.000000 0.000000
• X2 6.000000 0.000000
• X3 4.000000 0.000000
• X4 3.000000 0.000000
• X5 1.000000 0.000000
• X6 5.000000 0.000000
• ROW SLACK OR SURPLUS DUAL PRICES
• 2) 0.000000 0.000000
• 3) 0.000000 -1.000000
• 4) 0.000000 0.000000
• 5) 0.000000 -1.000000
• 6) 0.000000 0.000000
• 7) 0.000000 -1.000000

45
Police Scheduling LINDO Solution

• OBJ
  COEFFICIENT RANGES
• VARIABLE CURRENT ALLOWABLE
  ALLOWABLE
• COEF
  INCREASE DECREASE
• X1 1.000000
  INFINITY 0.000000
• X2 1.000000
  0.000000 1.000000
• X3 1.000000
  0.000000 0.000000
• X4 1.000000
  0.000000 0.000000
• X5 1.000000
  0.000000 0.000000
• X6 1.000000
  0.000000 0.000000
• RIGHTHAND
  SIDE RANGES
• ROW CURRENT ALLOWABLE
  ALLOWABLE
• RHS
  INCREASE DECREASE
• 2 5.000000
  0.000000 3.000000
• 3 6.000000
  INFINITY 0.000000
• 4 10.000000
  0.000000 INFINITY
• 5 7.000000
  INFINITY 0.000000
• 6 4.000000
  0.000000 3.000000
• 7 6.000000
  3.000000 0.000000

46
Police Scheduling LINDO Solution

• The department needs 19 officers to meet the
  daily requirements.
• The optimal solution is to hire 0 officers to
  work the shift from 8AM 4PM (x10), 6 officers
  to work the shift from Noon 8PM (x26), 4
  officers to work the shift from 4PM Midnight
  (x34), 3 officers to work the shift from 8PM
  4AM (x43), 1 officers to work the shift from
  Midnight 8AM (x51), and 5 officers to work the
  shift from 4AM Noon (x65).
• The surplus for all constraints is 0. Thus each
  shift has exactly the number of required officers
  for each shift.

47
THE END
See your textbook for more examples and detailed
explanations of all topics discussed in these
notes.