Chapter 8
Linear programming is used to allocate
resources, plan production, schedule
workers, plan investment portfolios and
formulate marketing (and military)
strategies. The versatility and economic
impact of linear programming in today’s
industrial world is truly awesome.--Eugene
Lawler
Linear Programming
1
What is a Linear Program?
 A linear program is a mathematical model
that indicates the goal and requirements of
an allocation problem.
 It has two or more non-negative variables.
 Its objective is expressed as a mathematical
function. The objective function plots as a
line on a two-dimensional graph.
 There are constraints that affect possible
levels of the variables. In two dimensions
these plot as lines and ordinarily define
areas in which the solution must lie.
2
Redwood Furniture
Problem Formulation
 Let XT and XC denote the number of tables and
chairs to be made. (Define variables)
 Maximize P = 6XT + 8XC (Objective function)
 Subject to: (Constraints)
30XT + 20XC < 300 (wood)
5XT + 10XC < 110 (labor)
 where XT and XC > 0 (non-negativity conditions)
 Letting XT represent the horizontal axis and XC the
vertical, the constraints and non-negativity
conditions define the feasible solution region.
3
Feasible Solution Region for
Redwood Furniture Problem
4
Graphing to Find Feasible
Solution Region
 For an inequality constraint (with < or >),
first plot as a line: 30XT + 20XC = 300.
 Get two points. Intercepts are easiest:
 Set XC = 0, solve for XT for horizontal intercept:
30XT + 20(0) = 300 => XT = 300/30 = 10
 Set XT = 0, solve for XC for vertical intercept:
30(0) + 20XC = 300 => XC = 300/20 = 15
 Above gets wood line. Do same for labor.
 Mark valid sides and shade feasible solution
region. Any point there satisfies all
constraints and non-negativity conditions.
5
Graphing to Find Feasible
Solution Region
 To establish valid side, pick a test point
(usually the origin). If that point satisfies
the constraint, all points on same side are
valid. Otherwise, all points on other side
are instead valid.
 Equality constraints have no valid side. The
solution must be on the line itself.
 Some constraint lines are horizontal or
vertical. These involve only one variable
and one intercept.
6
Finding Most Attractive Corner
 The optimal solution will always correspond to a
corner point of the feasible solution region.
 Because there can be many corners, the most
attractive corner is easiest to find visually.
 That is done by plotting two P lines for arbitrary
profit levels.
 Since the P lines will be parallel, just hold your
pencil at the same angle and role it in from the
smaller P’s line toward the bigger one’s That is
the direction of improvement.
 Continue rolling until only one point lies beneath
the pencil. That is the most attractive corner.
(Problems can have two most attractive corners.)
7
Most Attractive Corner for
Redwood Furniture Problem
8
Finding the Optimal Solution
 The coordinates of the most attractive
corner provide the optimal levels.
 Because reading from graph may be
inaccurate, it is best to solve algebraically.
 Simultaneously solving the wood and labor
equations, the optimal solution is:
XT = 4 tables
XC = 9 chairs
P = 6(4) + 8(9) = 96 dollars
 Note: supply the computed level of the
objective in reporting the optimal solution.
9
Advice for Solving Linear
Programs
 The most attractive corner need not be
where the two lines cross. Verify by
 doubling the table profit. Then the P lines will
be steeper, and (XT = 10, XC = 0) would be best.
 doubling instead the chair profit. The P lines
will be flatter, and (XT = 0, XC = 11) is best.
 Problems can have more than 2 constraints.
 The objective function can involve negative
coefficients.
10
 Therefore, the better Ps may not lie to the right.
 Use 2 lines to guarantee getting right direction.
Minimizing Cost:
Feed-Mix Problem
 Let XB and XS denote pounds of buckwheat
and sunflower in mixture.
Minimize C=.18XB + .10XS
Subject to: .04XB + .06XS > 480 (fat)
.12XB + .10XS > 1,200 (prot.)
.10XB + .15XS > 1,500 (rough.)
where
XB, XS > 0
 The optimal solution is:
XB = 3,750 pounds
XS = 7,500 pounds
C = .18(3,750) + .10(7,500) = 1,425 dollars
11
Minimizing Cost:
Feed-Mix Problem
12
Other Constraint Types
 Resources: amount used < available level.
 Requirements: quan. > minimum (< max.).
 XT > 5 (demand)
XC < 5 (capacity)
 Mixture: product > (or <) multiple of other.
 XC > 4XT (at least 4 chairs per table made)
 XB < .5XS (buckw. not exceed 1/2 wt of sunfl.)
 Transform before plotting:
 XC - 4XT > 0
XB - .5XS < 0
 Equality: XT + XC = 10 (exactly 10 items)
13
Special Problem Types
 Infeasible Problems: These arise from
contradictions among the constraints. No
solution possible until conflict is resolved.
 Ties for optimal solution: Multiple optimal
solutions can exist. Any linear combination
of two optimal corners is also optimal.
 Unbounded problems: Feasible solution
regions may be open-ended, and the
direction of improvement coincides.
 Mathematically, any profit is possible.
 Generally nonsensical, possibly due to a
missing constraint. Fix and solve again.
14
Graphing Linear Programs
Using Spreadsheets
The Redwood Furniture Company
Maximize
P = 6XT + 8XC
Subject to
30XT + 20XC
5XT + 10XC
where
XT, XC
15
(objective)
< 300 (wood)
< 110 (labor)
>0
First Step
The Formulas
The first step is to solve the
objective function and each constraint
for one of the variables. In this case,
solving for XC gives
16
XC = (P - 6XT)/8
(objective)
XC = (300 - 30XT)/20
(wood)
XC = (110 - 5XT)/10
(labor)
These formulas are entered on the
following spreadsheet.
Second Step
The Spreadsheet (Figure 8-18)
A
1
2
3 Profit
4 Wood
5 Labor
6
7
8 Tables, XT
9
0
10
1
11
2
12
3
13
4
14
5
15
6
16
7
17
8
18
9
19
10
B
C
D
E
F
G
H
I
J
Redwood Furniture Company
48
300
110
Wood
15
13.5
12
10.5
9
7.5
6
4.5
3
1.5
0
Chairs, XC
Labor
11
10.5
10
9.5
9
8.5
8
7.5
7
6.5
6
Profit
6
5.25
4.5
3.75
3
2.25
1.5
0.75
0
-0.75
-1.5
8
9
10
11
12
13
14
15
16
17
18
19
A
B
C
D
Tables, XT
Wood
Labor
Profit
0
=($B$4-30*A9)/20 =($B$5-5*A9)/10 =($B$3-6*A9)/8
1
=($B$4-30*A10)/20 =($B$5-5*A10)/10 =($B$3-6*A10)/8
2
=($B$4-30*A11)/20 =($B$5-5*A11)/10 =($B$3-6*A11)/8
3
=($B$4-30*A12)/20 =($B$5-5*A12)/10 =($B$3-6*A12)/8
4
=($B$4-30*A13)/20 =($B$5-5*A13)/10 =($B$3-6*A13)/8
5
=($B$4-30*A14)/20 =($B$5-5*A14)/10 =($B$3-6*A14)/8
6
=($B$4-30*A15)/20 =($B$5-5*A15)/10 =($B$3-6*A15)/8
7
=($B$4-30*A16)/20 =($B$5-5*A16)/10 =($B$3-6*A16)/8
8
=($B$4-30*A17)/20 =($B$5-5*A17)/10 =($B$3-6*A17)/8
9
=($B$4-30*A18)/20 =($B$5-5*A18)/10 =($B$3-6*A18)/8
10
=($B$4-30*A19)/20 =($B$5-5*A19)/10 =($B$3-6*A19)/8
For example,
Cell B9: XC = ($B$4 - 30XT)/20 = ($B$4-30*A9)/20
Cell B10: XC = ($B$5 - 5XT)/10 = ($B$5-5*A9)/10
Cell B11: XC = ($B$3 - 30XT)/20 = ($B$3-30*A9)/20
17
K
Third Step
Graphing with the Chart Wizard
Highlight cells B8:D19 and click on the
chart icon.
Step 1 - Chart Type
Step 2 - Chart Source Data
Step 3 - Chart Options
Step 4 - Chart Location
18
Chart Wizard
Chart Type
Select Line
as the Chart
type and pick
the first
Chart subtype (Line)
and click
Next.
19
Chart Wizard
Sources of Data, Series Tab
20
Enter the
horizontal axis
values by clicking
on the Series tab
and entering the
range of numbers
to be on the
horizontal axis,
cells A:9:A19, in
the Category (X)
axis labels line.
Alternately, click
in the Category
(X) axis labels line
and then highlight
cells A9:A19.
Click Next.
Chart Wizard
Chart Options
In the Chart title line type Redwood Furniture
Company, in the Category (X) axis put Tables, T, and in
the Value (Y) axis line write Chairs, C. Click Next.
21
Chart Wizard
Chart Location
Click on Finish and the Chart shown next appears.
22
Step Four
The Final Graph (Figure 8-19)
16
14
12
Chairs, XC
The final
graph
(after
making
formatting
changes).
Redwood Furniture Company
(P = 48)
10
8
Wood
6
4
Labor
Profit
2
0
-2 0
-4
1
2
3
4
5
6
Tables, XT
23
7
8
9
10
The Graph with P = 96
(Figure 8-20)
Redwood Furniture Company
(P = 96)
16
14
12
Chairs, XC
Increasing
the number
in cell B3
moves the
objective
function line
up and to the
right. This
graph show
the objective
function for
P = 96.
10
Wood
8
Labor
6
Profit
4
2
0
0
1
2
3
4
5
6
Tables, X T
24
7
8
9
10
The Graph with P = 96 and 80 Hours
of Labor (Figure 8-21)
25
Redwood Furniture Company
(P = 96 and 80 hours of labor)
16
14
12
Chairs, XC
To see what
happens when the
amount of wood or
labor vary, change
the numbers in
cells B4 (for wood)
or B5 (for labor)
and the
corresponding line
will move. This
graph show the
result when 80 is
entered in cell B5
(and P = 96).
10
Wood
8
Labor
6
Profit
4
2
0
0
1
2
3
4
5
6
Tables, XT
7
8
9
10
Drawing Horizontal
and Vertical Lines
Drawing two types of lines with
Excel require special attention:
horizontal and vertical lines. The
constraint Y = 3 is a horizontal line
and the constraint X = 7 is a vertical
line. Figure 9-21 shows what an
Excel spreadsheet looks like for these
two constraints.
26
Spreadsheet for
Horizontal and Vertical Lines (Figure 8-22)
The vertical
line equation
has an Yintercept of
100,000 and
a slope of (100,000/7).
Thus, it is
not exactly
vertical but it
is sufficiently
close to
vertical for
our
purposes.
A
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
B
C
D
E
F
Graphing Horizontal and Vertical Lines
X
0
1
2
3
4
5
6
7
8
9
10
Y=3
3
3
3
3
3
3
3
3
3
3
3
X=7
100000
85714
71429
57143
42857
28571
14286
0
-14286
-28571
-42857
4
5
6
7
8
9
10
11
12
13
14
C
=100000-(100000/7)*A4
=100000-(100000/7)*A5
=100000-(100000/7)*A6
=100000-(100000/7)*A7
=100000-(100000/7)*A8
=100000-(100000/7)*A9
=100000-(100000/7)*A10
=100000-(100000/7)*A11
=100000-(100000/7)*A12
=100000-(100000/7)*A13
=100000-(100000/7)*A14
Vertical Line Equation: Y = 100,000 – (100,000/7)X
27
Graphing Horizonal
and Vertical Lines (Figure 8-23)
To change the
position of the
vertical line,
change the 7 in
the
denominator of
all the formulas
in column C to
the desired
number.
Plotting Horizontal and
Vertical Lines
10
8
Y
6
Y=3
X=7
4
2
0
0
2
6
4
X
28
8
10