Linear Programming Example

1. LP Quizzes 4 to 8
Professor Ed Dansereau

2. Decision Variables
The two decision variables are the number of social media and newspaper ads.
Decision variables represent what management control. We can decide how
many of each to run.
X1 = Newspaper
X2 = Social Media
All other expressions (OF and Constraints are defined in terms of decision
variables).

3. Objective Function
Objective: Maximize Profit
Objective Function: How much profit from each of our decision variables.
$4 profit from Newspapers
$2 profit from Social Media
So
Zmax = 4X1 + 2X2

4. Constraints
Constraints are limitations, we can only dream of having unlimited resources
such as time and money.
For Budget Constraint
We have a maximum of 240 dollars, we could use 200 dollars but can not
spend 250. Our limitation is less than or equal to 240 dollars.
Each newspaper ad consumes $3 and each social media $1
Budget → 3X1 + 1X1 <= 240

5. Constraints continued
Work-Hour constraint
A maximum of 100 hours are available. Each Newspaper consumes 1 hour and
each social media 2.5 hours.
X1 + 2.5X2 <= 100

6. Graph
Each constraint is Linear - a straight line. To graph we take advantage of the
straight lines and that the axis of any variable equals zero. The horizontal axis
is X1 and the Vertical is X2.
Budget
X1 → set X2 = 0, X1 + 2.5(0) <= 100, solve for X1, X1 = 100
X2 → set X1 = 0, 1(0) + 2.5X2 = 100, X2 = 40
Work Hours

7. Graph

8. Feasible Region
Feasible Region - the area limited by the
constraints.
Since our constraints are less than or equal to,
the feasible region is area below both lines.
Note that for most of the graph the binding
constraint is the blue line, after the point (76.9,
9.2) the binding constraint is the red line. All LP
problems assume greater than zero, so the axis
represents the other boundaries (you would not
do a project if your profits were negative).
Feasible Region

9. Extreme Points
Extreme Points are the points at the edges of
the feasible region. With two constraints there
are four extreme points, there can be more if
there are additional constraints.
Our extreme points are (the labels are arbitrary)
A: (0,0)
B: (0,40)
C: (76.9, 9.2)
Extreme Points

10. Optimal Solution
Any point within the feasible area is a solution
and will provide some measure of profit or at
least break-even
The Optimal Solution is the point that provides
the maximum profit given our constraints. In
Linear Programming it will always be one of the
extreme points regardless if the objective is to
maximizing profit or minimizing costs.
Put Coordinates of extreme points into
Objective Function to determine Maximum
Profit or Minimum Cost (Max profit in our
example).
Zmax = 4X1 + 2X2
Za (0,0) = 4(0)+ 2(0) = 0
Zb (0,40) = 4(0) +2(40) =40
Zc (76.9, 9.2) = 4(76.9) + 2(9.2) = 326Optimal Solution

11. Binding and Slack
A BINDING constraints limits the optimal
solution.
Slack is extra capacity.
For example the Budget constraint has slack up
until it intersects with the Work Hour constraint.
After that point, the Budget constraint is
binding.

12. Finding the intersection of two lines
There are three ways to determine the
intersection of two lines:
1. Estimate, take your best guess by looking
at graph, for our example I would guess
(77,10).
2. Simultaneous Solution
3. Mathematical Method

13. Simultaneous Solution
1. Write out the two linear lines, the constraint equations
2. Solve for one variable in terms of the other for first equation
3. Substitute solution into second equation, which will result a solution for
one variable
4. Solve for second variable

14. Simultaneous Solution
1. Write out two linear lines
a. Work Hours → X1 + 2.5X2 <= 100
b. Budget → 3X1 + X2 <= 240
2. Solve for one variable in terms of the other for first equation
a. X1 + 2.5X2 = 100, so X1 = 100 - 2.5X2
b. Solve for any variable in either equation. I choose Work Hours and the variable X1 because
the math was easy. With a coefficient of 1, (240-2.5X2)/1 = (240-2.5X2). You do not even
need a calculator for that!

15. Simultaneous Solution
3. Substitute solution into second equation, which will result a solution for one
variable
a. Budget → 3X1 + X2 <= 240
b. Work Hours → X1 = 100 - 2.5X2
c. Sub work hours into Budget → 3(100-2.5X2) + X2 = 240
i. 300 - 7.5X2 +x2 =240
ii. -6.5X2 = 240 - 300

16. Simultaneous Solution
4. Solve for second variable
X2 = 9.27
3X1 + X2 = 240 (either equation will work)
3X1 + 9.27 = 240
X1 = (240 - 9.27) / 3
X1 = 76.92, solution for second variable

17. Multiplication Method
In this method we try to eliminate one variable but multiplying by an inverse
(negative number).
1. Look at your two linear lines (constraints) and choose a variable to
eliminate. I look for a number that is easy to multiple.
2. Multiple the other equation by the inverse.
3. Add two equations together and solve for first variable.
4. Solve for second variable

18. Multiplication Method
1. Look at your two linear lines (constraints) and choose a variable to
eliminate
a. Budget → 3X1 + 1X2 <= 240
b. Hours → 1X1 + 2.5X2 <= 100
c. The X1 in budget is three times the X1 in hours, so I choose to multiple the Work Hours
constraint by inverse, or simply -3.
2. Multiple the other equation by the inverse
a. Hours → -3(1X1 +2.5X2 = 100),

19. Multiplication Method
3. Add two equations together and solve for first variable.
3X1 + 1X2 = 240
+ -3X1 - 7.5X2 = -300
= 0X1 - 6.5X2 = -60
X2 = (-60)/(-6.5)
X2 = 9.23 (Does this number look familiar?)

20. Multiplication Method
4. Solve for second variable
X2 = 9.23
Budget → 3X1 + 1X2 <= 240, you may choose either equation
3X1 + 9.23 = 240
X1 = (240 -9.23) / 3
X1 = 76.92

21. Finding the intersection of two lines
Both the Simultaneous Solution and Mathematical Method produce the same
solution for X1 and X2 (76.92, 9.23).
These coordinates are the point on the graph where the two lines intersect.
Yes, you need to learn both methods.
We will use the Mathematical Method in the next section.