Managerial Decision Making
DSCN 205
Introduction to Modeling and Decision Analysis
General Form of an Optimization Problem

MAX (or MIN):
f0(X1, X2, …, Xn)

Subject to:
f1(X1, X2, …, Xn) <= b1
:

fk(X1, X2, …, Xn) >= bk

:

fm(X1, X2, …, Xn) = bm


Two general categories of optimization problems are:


Linear problems (Chapters 2-7)
Non-linear problems (Chapter 8)
Linear Programming (LP) Problem

MAX (or MIN):
c1X1 + c2X2 + … + cnXn

Subject to:
a11X1 + a12X2 + … + a1nXn <= b1




:
ak1X1 + ak2X2 + … + aknXn >= bk
:
am1X1 + am2X2 + … + amnXn = bm
An Example LP Problem

Blue Ridge Hot Tubs produces two types of hot tubs: AquaSpas & Hydro-Luxes. The production mix for the next week
should be determined. The profit should be maximized.
Pumps
Labor
Tubing
Unit Profit

Aqua-Spa
1
9 hours
12 feet
$350
Hydro-Lux
1
6 hours
16 feet
$300
There are 200 pumps, 1566 hours of labor, and 2880 feet of
tubing available for the next week.
LP Model for Blue Ridge Hot Tubs

Decisions:



Objective:


X1 = number of Aqua-Spa tubs to produce
X2 = number of Hydro-Lux tubs to produce
Max
350X1 + 300X2
(profit in dollars)
Constraints:




Subject to X1 + X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1, X2 >= 0
(pumps)
(hours of labor)
(feet of tubing)
Plotting the First Constraint
X2
250
(0, 200)
200
boundary line of pump constraint
X1 + X2 = 200
150
100
50
(200, 0)
0
0
50
100
150
200
250
X1
Plotting the Second Constraint
X2
(0, 261)
250
boundary line of labor constraint
9X1 + 6X2 = 1566
200
150
100
50
(174, 0)
0
0
50
100
150
200
250
X1
Plotting the Third Constraint
X2
250
(0, 180)
200
150
boundary line of tubing constraint
12X1 + 16X2 = 2880
100
Feasible Region
50
(240, 0)
0
0
50
100
150
200
250
X1
Plotting A Level Curve of the
Objective Function
X2
250
200
(0, 116.67)
objective function
150
350X1 + 300X2 = 35000
100
(100, 0)
50
0
0
50
100
150
200
250
X1
A Second Level Curve of the
Objective Function
X2
250
(0, 175)
200
objective function
350X1 + 300X2 = 35000
objective function
350X1 + 300X2 = 52500
150
100
(150, 0)
50
0
0
50
100
150
200
250
X1
Using A Level Curve to Locate
the Optimal Solution
X2
250
objective function
350X1 + 300X2 = 35000
200
150
optimal solution
100
objective function
350X1 + 300X2 = 52500
50
0
0
50
100
150
200
250
X1
Calculating the Optimal Solution
The optimal solution occurs where the “pumps” and “labor”
constraints intersect.
This occurs where:
X1 + X2 = 200
(1)
and
9X1 + 6X2 = 1566
(2)
From (1) we have, X2 = 200 -X1
(3)
Substituting (3) for X2 in (2) we have,
9X1 + 6 (200 -X1) = 1566
which reduces to X1 = 122
So the optimal solution is,
X1=122, X2=200-X1=78
Total Profit = $350*122 + $300*78 = $66,100
Enumerating The Corner (Extreme)
Points
X2
250
obj. value = $54,000
(0, 180)
200
obj. value = $64,000
150
(80, 120)
obj. value = $66,100
(122, 78)
100
50
obj. value = $60,900
(174, 0)
obj. value = $0
(0, 0)
0
0
50
100
150
200
250
X1
Summary of Graphical Solution
to LP Problems
1. Plot the boundary line of each constraint
2. Identify the feasible region
3. Locate the optimal solution by either:
a. Plotting level curves
b. Enumerating the extreme points
Binding and Redundant Constraint

A constraint is binding if it prevents from achieving a better objective value.
If removed, the solution may improve.

Sometimes there is a constraint that could be removed without changing
the feasible region. Such constraint is called redundant.

A constraint remains redundant provided that all other constraints remain
the same. Don’t rush with removing redundant constraints.

Example 1


Which constraints are binding and which are non-binding in the Blue Ridge
problem? Are there any redundant constraints in the Blue Ridge problem?
Example 2

15
Would a new constraint X2 <= 190 be redundant?
Special Conditions in LP Models
A number of anomalies can occur in LP problems:
•
•
•
•
Alternate Optimal Solutions
Redundant Constraints
Unbounded Solutions
Infeasibility
Special Conditions in LP Models
MAX: 450X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1 >= 0
X2 >= 0
Example of Alternate Optimal Solutions
X2
250
objective function level curve
450X1 + 300X2 = 78300
200
150
100
alternate optimal solutions
50
0
0
50
100
150
200
250
X1
Multiple (Alternate) Optimal Solutions

Sometimes more than one solution is optimal.

In 2-variable problems, multiple optimal solutions occur
when the objective level line is parallel to one of the
binding constraint lines; all points of a feasible segment on
that constraint line are optimal.

Any of the multiple optimal solutions could be adopted.
Secondary objectives could be used to choose.
Unbounded Solution
MAX : X1 + X2
S.T. : X1 + X2 >= 400
-X1 + 2X2 <= 400
X1 >= 0
X2 >= 0
DCSN20200 section 4
October 3
Example of an Unbounded Solution
X2
1000
objective function
X1 + X2 = 600
800
-X1 + 2X2 = 400
objective function
X1 + X2 = 800
600
400
200
X1 + X2 = 400
0
0
200
400
600
800
1000
X1
Unbounded Problem

A problem is unbounded if its objective can be made arbitrarily “good”.



Min – we can decrease objective value to –infinity while maintaining feasibility.
Max – we can increase objective value to +infinity while maintaining feasibility.
Possible reasons:



Missing constraints
Incorrect direction of optimization (min instead of max or vice versa)
Other mistakes in the model or assumptions

A problem can be unbounded only if its feasible region is unbounded.
However, an unbounded feasible region does not necessary imply that the
problem is unbounded.

Example 1 (unbounded problem)


Max 2X1 + X2
subject to
–X1 + X2 <= 0,
X1 >= 3,
Example 2 (unbounded feasible region, bounded problem)

Try minimization instead of maximization in Example 1
0 <= X2 <= 5
Infeasibility
MAX: X1 + X2
S.T.: X1 + X2 <= 150
X1 + X2 >= 200
X1 >= 0
X2 >= 0
Example of Infeasibility
X2
250
200
X1 + X2 = 200
feasible region for
second constraint
150
100
feasible region
for first
constraint
50
X1 + X2 = 150
0
0
50
100
150
200
250
X1
Infeasible Problem

A problem is infeasible if there is no solution that satisfies
all constraints.

Possible reasons:



Mistakes in the constraints of the model
Too restrictive business requirements
Example

Consider the Blue Ridge problem with an additional
requirement that at least 200 Aqua Spa tubs should be
produced to satisfy a previously signed contract.
Analysis of Feasible Region

A feasible region of an LP problem is always a convex polygon



Internal points
Boundary points
Corner points (special boundary points; intersections of at least 2
constraints in 2-variable problems)

In LP problems, optimal solution is always one or more of the
boundary points. If there is a single optimal solution, it is
always in a corner point (not true for non-linear problems).

Optimal corner point can be determined graphically, but exact
values (of X1, X2, and profit in the Blue Ridge example) should
always be calculated analytically.

If feasible region is bounded from all sides, optimal solution can
be found by comparing values of all corner points.