INTRODUCTION TO OPERATIONS
RESEARCH
Linear Programming
LINEAR PROGRAMMING REVIEW

Linear Programming provides methods for allocating
limited resources among competing activities in an
optimal way.

Any problem whose model fits the format for the
linear programming model is a linear programming
problem.

Wyndor Glass Co. example


Two variables – Graphical method
Maximize profit
RADIATION THERAPY EXAMPLE

Mary diagnosed with cancer of the bladder →
needs radiation therapy

Radiation therapy



Involves using an external beam to pass radiation through the
patient’s body
Damages both cancerous and healthy tissue
Goal of therapy design is to select the number, direction and
intensity of beams to generate best possible dose distribution

Doctors have already selected the number (2) and direction
of the beams to be used

Goal: Optimize intensity (measured in kilorads) referred to
as the dose
RADIATION THERAPY
Area
Healthy Anatomy
Critical Tissues
Tumor Region
Tumor Center
Fraction of Entry
Dose
Absorbed by
Area (Average)
Beam1 Beam2
0.4
0.5
0.3
0.1
0.5
0.5
0.6
0.4
Restriction on Total
Average Dosage,
Kilorads
minimize
≤ 2.7
= 6.0
≥ 6.0
RADIATION THERAPY
Graph the equations
to determine
relationships
Minimize
Z = 0.4x1 + 0.5x2
Subject to:
0.3x1 + 0.1x2 ≤ 2.7
0.5x1 + 0.5x2 = 6
0.6x1 + 0.4x2 ≥ 18
x1 ≥ 0, x2 ≥ 0
MIXTURE PROBLEM


In order to ensure optimal health (and thus accurate test results), a
lab technician needs to feed the rabbits a daily diet containing a
minimum of 24 grams (g) of fat, 36 g of carbohydrates, and 4 g of
protein. But the rabbits should be fed no more than five ounces of
food a day.
Rather than order rabbit food that is custom-blended, it is cheaper to
order Food X and Food Y, and blend them for an optimal mix.



Food X contains 8 g of fat, 12 g of carbohydrates, and 2 g of protein per ounce,
and costs $0.20 per ounce.
Food Y contains 12 g of fat, 12 g of carbohydrates, and 1 g of protein per
ounce, at a cost of $0.30 per ounce.
What is the optimal blend?
MIXTURE PROBLEM
Daily Amount
Fat
Carbohydrates
Protein
Food Type
X
Y
8
12
12
12
2
1
maximum weight of the food is
five ounces:
X+Y≤5
Daily Requirements
(grams)
≥ 24
≥ 36
≥4
Minimize the cost:
Z = 0.2X + 0.3Y
MIXTURE PROBLEM
Graph the equations
to determine
relationships
Minimize
Z = 0.2x + 0.3y
Subject to:
fat:
8x + 12y ≥ 24
carbs:
12x + 12y ≥ 36
protein:
2x + 1y ≥ 4
weight:
x+y≤5
x ≥ 0, y ≥ 0
MIXTURE PROBLEM
When you test the corners at:
(0, 4), (0, 5), (3, 0), (5, 0), and (1, 2)
you get a minimum cost of sixty cents per daily
serving, using three ounces of Food X only.
Only need to buy Food X
INVESTMENT EXAMPLE
You have $12,000 to invest, and three different funds from
which to choose.
Municipal bond:
CDs:
High-risk acct:



7% return
8% return
12% return (expected)
To minimize risk, you decide not to invest any more than
$2,000 in the high-risk account.
For tax reasons, you need to invest at least three times as
much in the municipal bonds as in the bank CDs.
Assuming the year-end yields are as expected, what are the
optimal investment amounts?
INVESTMENT EXAMPLE
Bonds (in thousands):
CDs (in thousands):
High Risk:

x
y
z
Um... now what? I have three variables for a
two-dimensional linear plot
 Use
the "how much is left" concept
 Since $12,000 is invested, then the high risk
account can be represented as

z = 12 – x – y
INVESTMENT EXAMPLE
Constraints:
Amounts are non-negative:
x≥0
y≥0
z≥0 

Z = 0.07x + 0.08y + 0.12z
12 – x – y ≥ 0
y ≤ –x + 12
High risk has upper limit
z≤2 

12 – x – y ≤ 2
y ≤ –x + 10
Taxes:
3y ≤ x 
Objective to maximize the
return:
y ≤ 1/3 x

Z = 1.44 - 0.05x – 0.04y
INVESTMENT EXAMPLE
When you test the corner points at (9, 3), (12, 0), (10, 0),
and (7.5, 2.5), you should get an optimal return of $965
when you invest $7,500 in municipal bonds, $2,500 in CDs,
and the remaining $2,000 in the high-risk account.
MANUFACTURING EXAMPLE
Machine data
Machine \ Product
A
B
C
D
Total processing time
Product data
Unit processing times
(min)
P
Q
20
10
12
28
15
6
10
15
57
59
Revenue per unit
Material cost per unit
Profit per unit
Maximum sales
Availability
(min)
R
10
16
16
0
42
2400
2400
2400
2400
P
Q
R
$90
$45
$45
100
100
$40
$60
40
$70
$20
$50
60
PRODUCT STRUCTURE
Machines A,B,C,D
Available t ime:
2400 min/week
Operating expenses:
$6000/week
P urchased
part :
$5/unit
P
Revenue:
$90/unit
Max sales:
100 units/week
Q
Revenue:
$100/unit
Max sales:
40 units/week
D
15 min/unit
D
10 min/unit
R
Revenue:
$70/unit
Max sales:
60 units/week
C
16 min/unit
C
9 min/unit
C
6 min/unit
B
16 min/unit
A
20 min/unit
B
12 min/unit
A
10 min/unit
RM1
$20/unit
RM2
$20/unit
RM3
$20/unit
Component 1
Component 2
Component 3
LP FORMULATION
max
s.t.
45 xP
20 xP
12 xP
15 xP
10 xP
+
+
+
+
+
60 xQ
10 xQ
28 xQ
6 xQ
15 xQ
Objective Function




1800
1440
2040
2400
xP  100, xQ  40
Are we done?
xP ≥ 0, xQ ≥ 0
Structural
constraints
demand
nonnegativity
Are the LP assumptions
valid for this problem?
= 81.82
Optimal solution x *
P
x *Q = 16.36
FEASIBLE REGION
P
240
Max Q
200
D
160
120
Max P
80
A
40
0
B
0
40
C
80 120 160 200 240 280 320
360
Q
OPTIMAL SOLUTION
P
Optimal solut ion = (16.36, 81.82)
120
Max Q
100
80
A
60
40
Z = $4664
20
Z = $3600
B
0
0
10
20
30
40
50
60
Q
DISCUSSION OF RESULTS

Optimal objective value is $4664 but when we
subtract the weekly operating expenses of $3000 we
obtain a weekly profit of $1664.

Machines A & B are being used at maximum level
and are bottlenecks.

There is slack production capacity in Machines C & D.