1. Linear Programming MGMT E-5050 THE ALLOCATION PROBLEM
MEETING MULTIPLE AND
DIVERSE GOALS WITHIN
THE CONTEXT OF LIMITED
RESOURCES
Applied Management Science for Decision Making, 2e © 2014 Pearson Learning Solutions Philip A. Vaccaro , PhD

2. History Linear programming was conceptually developed
before World War II by the
Soviet mathematician
Andrei Nikolaevich
Kolmogorov
( 1903 – 1987 )
Андрей Николаевич Κолмогоров

3. History Leonid Vitalyevich Kantorovich, another Soviet mathematician,
won the Nobel Prize in
Economics for advancing the
concepts of optimal planning.
( 1912 – 1986 )
Лєонид Канторович

4. History In 1947, George Bernard Dantzig developed the solution
procedure known as the
simplex algorithm, while
working on Air Force
logistics problems.
( 1914 – 2005 )

5. Linear Programming Models
Goal
Programming
Graphical Method
Integer
Dynamic
Transportation
Algorithm
and
Assignment
Non-Linear
Simplex
Method
ALL DEVELOP AN OPTIMAL SOLUTION

6. Application Examples PRODUCT MIX PROBLEM
determining the optimal number of several
products to make in order to maximize total
profit or minimize total cost within the
context of limited resources.
BLENDING PROBLEM
determining the lowest-cost mixture of food
groups that will meet a set of minimum
nutritional requirements.

7. Application Examples FINANCIAL PROBLEM
determining the allocation of funds among
various investment classes such as stocks,
bonds, real estate, and commodities so
as to maximize total returns over time.
PROMOTION PROBLEM
determining the allocation of marketing budgets
over various media such as billboards, radio,
television, newspapers, and magazines
so as to maximize total exposure.

8. Problem Statement A firm makes two kinds of clocks: regular and alarm.
3 resources are required to produce these clocks:
Available daily labor hours - 1,600
Available daily processing hours - 1,800
Available daily alarm assemblies - 350

9. Problem Statement
Continued

10. Model Development Let X1 = the number of regular clocks produced
Let X2 = the number of alarm clocks
IF THERE WERE A 3rd and 4th TYPE CLOCK,
THEY WOULD BE DESIGNATED AS X3 AND X4
RESPECTIVELY

11. The Objective Function
controllable, decision, or real variables
Maximize Z = 3X1 + 8X2
contribution margins or
objective function coefficients
total daily profit

12. Labor Resource Constraint
real variables
2X1 + 4X2 =< 1,600 hours
usage coefficients
right-hand side (RHS)

13. Processing Resource Constraint
real variables
6X1 + 2X2 =< 1,800 hours
usage coefficients
right-hand side (RHS)

14. Alarm Assemblies Resource Constraint
real variables
0X1 + 1X2 =< 350 units
usage coefficients
right-hand side (RHS)

15. Non-Negativity Constraints
X1 => 0
X2 => 0 or
X1 , X2 => 0
THE FIRM MUST PRODUCE NOTHING
OR SOMETHING OF EACH PRODUCT.
NEGATIVE VALUES FOR PRODUCT
DO NOT EXIST !

16. The Model Maximize Z = 3X1 + 8X2 Subject to:
2X1 + 4X2 =< 1,600 ( labor hours )
6X1 + 2X2 =< 1,800 ( processing hours )
X2 =< ( alarm assemblies )
X1 =>
X2 =>
THE MODEL WILL FIND
VALUES FOR THE
CONTROLLABLE VARIABLES ( X1 and X2 )
THAT WILL MAXIMIZE THE
OBJECTIVE FUNCTION ( Z )
SUBJECT TO THE THREE
RESOURCE CONSTRAINTS
AND NON-NEGATIVITY CONSTRAINTS

17. The Graphical Method of Linear Programming
Graph the resource constraints.
Identify the feasible solution region.
Compute the values of X1 and X2 at each
corner point of the feasible region.
Select the corner point with the maximum
profit. ( the optimal solution )
Four Steps:

18. Plotting the 1st Resource Constraint
2X1 + 4X2 =< 1,600 labor hours
assume that labor hours are the only resource
needed to produce X1 and X2 clocks.
if we only made X1 s, we could make 800 units.
If we only made X2 s , we could make 400 units.
therefore, the coordinates for plotting the labor
hour constraint are X1 = 800 and X2 = 400.

19. The Labor Constraint PlotX2
THE LINEAR INEQUALITY
MUST BE CONVERTED TO
A LINEAR EQUALITY BEFORE
IT CAN BE PLOTTED
1000
900
800
700
600
500
400
300
200
100
2X1 + 4X2 = 1,600 labor hours X1

20. Plotting the 2nd Resource Constraint
6X1 + 2X2 =< 1,800 process hours
assume that process hours are the only resource
needed to produce X1 and X2 clocks.
if we only made X1 s , we could make 300 units.
if we only made X2 s , we could make 900 units.
therefore, the coordinates for plotting the pro-
cess hour constraint are X1 = 300 and X2 = 900.

21. The Process Constraint PlotX2
THE LINEAR INEQUALITY
MUST BE CONVERTED TO
A LINEAR EQUALITY BEFORE
IT CAN BE PLOTTED
1000
900
800
700
600
500
400
300
200
100
6X1 + 2X2 = 1,800 process hours X1

22. Plotting the 3rd Resource Constraint
0X1 + 1X2 =< 350 alarm assemblies
assume that alarm assemblies are the only
resource needed to produce X2 clocks.
if we only made X2 s , we could make 350 units.
therefore, the coordinates for plotting the alarm
assemblies constraint are X1 = 0 and X2 = 350. *
ALARM
ASSEMBLIES
ARE NOT
USED IN
MAKING
REGULAR
CLOCKS

23. The Alarm Assemblies Constraint PlotX2
THE LINEAR INEQUALITY
MUST BE CONVERTED TO
A LINEAR EQUALITY BEFORE
IT CAN BE PLOTTED
1000
900
800
700
600
500
400
300
200
100
0X1 + 1X2 = 350 alarm assemblies X1

24. The Fully-Plotted Graph
CORNER POINTS A,B,C,D,E DEFINE THE FEASIBLE REGION X2
1000
900
800
700
600
500
400
300
200
100
PROCESSING HOURS
ALARM ASSEMBLIES B C D
FEASIBLE
REGION
LABOR HOURS A E X1

25. The Corner Point Evaluations
Objective Function = Maximize Z = 3X1 + 8X2
THE PROFITS
AT EACH
CORNER
POINT
VIA THE
OBJECTIVE
FUNCTION
Point A Z = 3 ( 0) + 8( 0) = $0.00
Point B Z = 3 ( 0) + 8(350) = $2,800.00
Point C Z = 3(100) + 8(350) = $3,100.00
Point D Z = 3(200) + 8(300) = $3,000.00
Point E Z = 3(300) + 8( 0) = $900.00

26. Corner Point “C” Coordinates
AT THIS INTERSECTION,
THE VALUES OF X1 AND X2
MUST BE IDENTICAL
IN BOTH CONSTRAINTS.
SINCE X2 = 350 ,
WE SOLVE FOR X1:
2X1 + 4 (350) = 1,600
2X1 + 1,400 = 1,600
2X1 = 200
X1 = 100
THE TWO CONSTRAINT
LINES THAT INTERSECT
TO FORM POINT “C” ARE:
2X1 + 4X2 = 1,600
0X1 + 1X2 =

27. Corner Point “D” Coordinates
AT THIS INTERSECTION,
THE VALUES OF X1 AND X2
MUST BE IDENTICAL
IN BOTH CONSTRAINTS.
WE SOLVE FOR X2
BY CANCELING
OUT X1 IN BOTH
EQUATIONS
(CONSTRAINTS)
AS FOLLOWS
THE TWO CONSTRAINT
LINES THAT INTERSECT
TO FORM POINT “D” ARE:
2X1 + 4X2 = 1,600
6X1 + 2X2 = 1,800

28. Corner Point “D” Coordinates
3 ( 2X1 + 4X2 = 1,600 )
6X1 + 2X2 = 1,800
6X1 + 12X2 = 4,800
- 6X X2 = - 1,800
10X2 = 3,000
X2 =

29. Corner Point “D” Coordinates
WE SUBSTITUTE
X2 = 300
INTO THE
1st EQUATION
AND SOLVE
FOR X1:
2X1 + 4X2 = 1,600
2X1 + 4(300) = 1,600
2X1 + 1,200 = 1,600
2X1 = 400
X1 = 200

30. The Corner Point Evaluations

31. The Corner Point Evaluations
THE OPTIMAL SOLUTION IS
CORNER POINT
“C”

32. Optimal Solution X1 X2 = Manufacturing Decision Maximized Total Profit
Produce
350
Alarm
Clocks
Manufacturing
Decision
100
Regular
Total
Profit
$3,100.00
Maximized
Total Profit
500
Processing
Hours
Unused X2 =
Real (Decision)
Variable Values X1

33. Slack Variables ( S ) labor hours will be represented by “S1”
Every resource has its own unique slack
variable to represent it.
Its value is the difference between what was
consumed and what was originally available
for that particular resource.
In our example:
labor hours will be represented by “S1”
process hours will be represented by “S2”
alarm assemblies will be represented by “S3”
slack
variables
can be
computed
once the
optimal
solution
has been
found

34. Slack Variable S1 Computation
Given that
X1 = 100
regular
clocks
and
X2 = 350
alarm
clocks:
2X1 + 4X2 =< 1,600 labor hours
becomes:
2X1 + 4X2 + 1S1 = 1,600
(substituting)
2(100) + 4(350) + 1S1 = 1,600
, S1 = 1,600
1, S1 = 1,600
therefore:
S1 = 0
total consumed hours
originally available hours

35. Slack Variable S1 Interpretation
All labor hours are completely consumed in
the optimal product mix solution.
The labor hour constraint is binding, that is ,
we are prevented from producing more clocks
because labor hours are fully consumed.
Labor hours carry a positive shadow price ,
that is, we would be willing to buy additional
labor hours if they were available.

36. Slack Variable S2 Computation
Given that
X1 = 100
regular
clocks
and
X2 = 350
alarm
clocks:
6X1 + 2X2 =< 1,800 process hours
becomes:
6X1 + 2X2 + 1S2 = 1,800
(substituting)
6(100) + 2(350) + 1S2 = 1,800
S2 = 1,800
1, S2 = 1,800
therefore:
S2 = 500
total consumed hours
originally available hours

37. Slack Variable S2 Interpretation
There are 500 process hours left over in the
optimal product mix solution.
The process hour constraint is non-binding,
because it does not prevent us from producing
more clocks.
Process hours carry a zero shadow price, mean-
ing that we are not willing to buy additional hours
since we still have an excess.

38. Slack Variable S3 Computation
Given that
X1 = 100
regular
clocks
and
X2 = 350
alarm
clocks:
0X1 + 1X2 =< 350 alarm assemblies
becomes:
0X1 + 1X2 + 1S3 = 350
(substituting)
0(100) + 1(350) + 1S3 = 350
S3 = 350
S3 = 350
therefore:
S3 = 0
total consumed assemblies
originally available assemblies

39. Slack Variable S3 Interpretation
All alarm assemblies are completely con-
sumed in the optimal product mix solution.
The alarm assembly constraint is binding ,
that is, we are prevented from producing
more clocks because alarm assemblies are
fully consumed.
Alarm assemblies carry a positive shadow
price , that is, we would be willing to buy
additional alarm assemblies if they were
available.

40. The Shadow Price
The price we would pay for an additional unit of a resource
A ‘unit’ may be an hour of labor or processing
A ‘unit’ may be an ounce, pound, or ton of material
The shadow price may be zero or positive

41. Zero vs. Positive Shadow Prices
Resources > 0 in the
optimal solution have
zero shadow prices
because we have an
excess.
Resources = 0 in the
optimal solution have
positive shadow prices
because we would like
to buy more of them.

42. Shadow Price Implications
If we pay less than the
shadow price, total profit
( Z ) will increase
If we pay the shadow price
exactly, total profit ( Z )
will not change
If we pay more than the
will decrease

43. Iso-Profit Lines From the Greek ( ίσώ ) meaning “equal”.
Any combination of the two products produced
along this line will yield the same total profit *
They are identified by dashed lines.
Primarily used today to quickly determine if a
certain level of profit (or cost) can be achieved
before proceeding further.
COST IN A MINIMIZATION PROBLEM

44. The Clock Production Graph
CORNER POINTS A,B,C,D,E
DEFINE THE FEASIBLE REGION X2
PROCESSING HOURS
1000
900
800
700
600
500
400
300
200
100
LABOR HOURS
ALARM ASSEMBLIES
UNFEASIBLE
REGION B C D
FEASIBLE
REGION A E X1

45. Iso-Profit Lines $1,200.00 = 3X1 + 8X2 400 X1 clocks and 0 X2 clocks
If we wanted to determine if an arbitrary profit of
$1, were achievable, we set the objective
function equal to $1, :
CONTINUED
$1, = 3X1 + 8X2
We find that we could make :
400 X1 clocks and 0 X2 clocks or
0 X1 clocks and 150 X2 clocks
in order to achieve a profit of $1,200.00

46. The Clock Production Graph
CORNER POINTS A,B,C,D,E
DEFINE THE FEASIBLE REGION X2 A
PROFIT OF
$1,200.00 IS
ACHIEVABLE
PROCESSING HOURS
1000
900
800
700
600
500
400
300
200
100
LABOR HOURS
ALARM ASSEMBLIES
UNFEASIBLE
REGION B C D
$1, Iso-Profit Line
FEASIBLE
REGION A E X1

47. Iso-Profit Lines $2,400.00 = 3X1 + 8X2 800 X1 clocks and 0 X2 clocks
If we wanted to determine if an arbitrary profit of
$2, were achievable, we set the objective
function equal to $2, :
$2, = 3X1 + 8X2
CONTINUED
We find that we could make :
800 X1 clocks and 0 X2 clocks or
0 X1 clocks and 300 X2 clocks
in order to achieve a profit of $2,400.00

48. The Clock Production Graph
CORNER POINTS A,B,C,D,E
DEFINE THE FEASIBLE REGION X2 A
PROFIT OF
$2,400.00 IS
ACHIEVABLE
PROCESSING HOURS
1000
900
800
700
600
500
400
300
200
100
LABOR HOURS
ALARM ASSEMBLIES B C
$2, Iso-Profit Line D
FEASIBLE
REGION A E X1

49. Iso-Profit Lines $3,000.00 = 3X1 + 8X2 1,000 X1 clocks and 0 X2 clocks
If we wanted to determine if an arbitrary profit of
$3, were achievable, we set the objective
function equal to $3, :
$3, = 3X1 + 8X2
We find that we could make :
1,000 X1 clocks and 0 X2 clocks or
0 X1 clocks and 375 X2 clocks
in order to achieve a profit of $3,000.00

50. The Clock Production Graph
THE $3, ISO-PROFIT LINE IS JUST BELOW
CORNER POINTS “C” AND “D” X2 A
PROFIT OF
$3,000.00 IS
ACHIEVABLE
PROCESSING HOURS
1000
900
800
700
600
500
400
300
200
100
LABOR HOURS
ALARM ASSEMBLIES
UNFEASIBLE
REGION B C
$3, Iso-Profit Line D
FEASIBLE
REGION A E X1

51. Iso-Profit Lines $3,100.00 = 3X1 + 8X2 1,033 X1 clocks and 0 X2 clocks
If we wanted to determine if an arbitrary profit of
$3, were achievable, we set the objective
function equal to $3, :
CONTINUED
$3, = 3X1 + 8X2
We find that we could make :
1,033 X1 clocks and 0 X2 clocks or
0 X1 clocks and 387 X2 clocks
in order to achieve a profit of $3,100.00

52. The Clock Production Graph
THE $3, ISO-PROFIT LINE CROSSES THROUGH
CORNER POINT “C” ONLY X2 A
PROFIT OF
$3,100.00 IS
ACHIEVABLE
PROCESSING HOURS
1000
900
800
700
600
500
400
300
200
100
LABOR HOURS
ALARM ASSEMBLIES
UNFEASIBLE
REGION B
$3, Iso-Profit Line C D
FEASIBLE
REGION A E X1

53. The Clock Production Graph
THE $4, ISO-PROFIT LINE FALLS OUTSIDE THE FEASIBLE
REGION X2 A
$4,000.00
PROFIT
IS NOT
ACHIEVABLE
PROCESSING HOURS
1000
900
800
700
600
500
400
300
200
100
LABOR HOURS
ALARM ASSEMBLIES
UNFEASIBLE
REGION B C D
$4, Iso-Profit Line
FEASIBLE
REGION A E X1

54. Units of product produced
The Surplus Variable ( S )
Units of product produced
over and above the
minimum required in
the optimal solution.
The only way it can be
differentiated from a
slack variable is by
placing a negative (-)
sign in front of it.

55. EXCESS OF REGULAR CLOCKS,
The Surplus Variable
EXAMPLE
X1 => 60 regular clocks
MEANS PRODUCE AT LEAST
60 REGULAR CLOCKS
( A QUOTA CONSTRAINT )
REWRITTEN AS A LINEAR
EQUALITY WHERE S4 IS THE
EXCESS OF REGULAR CLOCKS,
IF ANY, PRODUCED IN THE
OPTIMAL SOLUTION
X1 – S4 = 60 regular clocks
SINCE X1 = 100 REGULAR
CLOCKS IN THE OPTIMAL
SOLUTION, THE EXCESS
PRODUCED IS 40 CLOCKS
= 40 regular clocks

56. QUOTA CONSTRAINTS ARE A NORMAL LUXURY OF MAKING WHAT THEY WANT
Management / market imposed operating
restrictions which, by their very nature,
have a high probability of degrading the
optimal solution, resulting in lower total
profits or higher total costs.
QUOTA CONSTRAINTS ARE A NORMAL
OCCURRENCE IN THE LINEAR PROGRAMMING PROBLEM BECAUSE MOST FIRMS DO NOT ENJOY THE
LUXURY OF MAKING WHAT THEY WANT

57. QUOTA CONSTRAINTS X1 = regular clocks X2 = alarm clocks X1 + X2 = 40
PRODUCE EXACTLY 40 CLOCKS BETWEEN
REGULAR AND ALARM CLOCKS
1st
EXAMPLE
X1 + X2 => 40
PRODUCE AT LEAST 40 CLOCKS BETWEEN
REGULAR AND ALARM CLOCKS
X1 + X2 =< 40
PRODUCE NO MORE THAN 40 CLOCKS
BETWEEN REGULAR AND ALARM CLOCKS

58. QUOTA CONSTRAINTS X2 X1 + X2 = 40 ALL COMBINATIONS OF X1 AND X2
The
Constraint
Plots X2
X1 + X2 = 40
ALL COMBINATIONS OF X1 AND X2
ARE ON THE CONSTRAINT LINE ITSELF X1 X2
X1 + X2 => 40
ALL COMBINATIONS OF X1 AND X2
ARE ON THE CONSTRAINT LINE AND
ABOVE IT X1 X2
X1 + X2 =< 40
ALL COMBINATIONS OF X1 AND X2
ARE ON THE CONSTRAINT LINE AND
BELOW IT X1

59. QUOTA CONSTRAINTS X1 = regular clocks X2 = alarm clocks X1 = 2X2
2nd
EXAMPLE
PRODUCE EXACTLY TWICE AS MANY
REGULAR CLOCKS AS ALARM CLOCKS
X1 = 2X2
PRODUCE AT LEAST TWICE AS MANY
REGULAR CLOCKS AS ALARM CLOCKS
X1 => 2X2
PRODUCE AT MOST, TWICE AS MANY
REGULAR CLOCKS AS ALARM CLOCKS
X1 =< 2X2

60. THE GENERAL CONSTRAINT
QUOTA CONSTRAINTS
X1 = 2X2 X1 => 2X2 X1 =< 2X2
ANY TWO PAIRS OF
X1 AND X2 COORDINATES
WILL ENABLE US TO PLOT
THE GENERAL CONSTRAINT
ARBITRARILY SELECT
VALUES FOR X1 AND X2
SUCH THAT THE LEFT
SIDE OF THE LINEAR
EQUALITY EQUALS THE
RIGHT SIDE.
X1 = 2X2
X1 = 2X2
4 = 2 ( 2 )
6 = 2 ( 3 )
8 = 2 ( 4 )
etc. X2 5 4 3 2 1 X1

61. QUOTA CONSTRAINTS X1 = regular clocks X2 = alarm clocks X1 = 1.75X2
3rd
EXAMPLE
PRODUCE EXACTLY 75% MORE
REGULAR CLOCKS THAN ALARM CLOCKS
X1 = 1.75X2
PRODUCE AT LEAST 75% MORE
REGULAR CLOCKS THAN ALARM CLOCKS
X1 => 1.75X2
PRODUCE AT MOST, 75% MORE
REGULAR CLOCKS THAN ALARM CLOCKS
X1 =< 1.75X2

62. THE GENERAL CONSTRAINT
QUOTA CONSTRAINTS
X1 = 1.75X2 X1 => 1.75X2 X1 =< 1.75X2
ANY TWO PAIRS OF
X1 AND X2 COORDINATES
WILL ENABLE US TO PLOT
THE GENERAL CONSTRAINT
ARBITRARILY SELECT
VALUES FOR X1 AND X2
SUCH THAT THE LEFT
SIDE OF THE LINEAR
EQUALITY EQUALS THE
RIGHT SIDE.
X1 = 1.75X2 X2
X1 = 1.75X2
7 = 1.75( 4)
14 = 1.75( 8)
21 = 1.75(12)
etc. 12 8 4 X1

63. Other Constraints X2 30 X1 = 20 on the line 20
4th Example X2
FEASIBLE REGION 30 20 10
X1 = on the line
X1 => 20 on and right of the line
X1 =< 20 on and left of the line X1

64. Other Constraints X2 30 20 10 X1 0 10 20 30 40 50 60 FEASIBLE REGION
X2 = on the line
X2 => 30 on and above
the line
X2 =< 30 on and below
the line X1

65. Graph Linear Programming
The objective function coefficients become costs.
The objective function becomes “Minimize Z”.
Usually more quota constraints are used.
MINIMIZATION X2
UNBOUNDED
FEASIBLE
REGION
“=“ and “=>”
The feasible region may
be unbounded. X1

66. Linear Programming Application
HUMAN RESOURCES
We have a choice of three different employee candidates: un-trained,
semi-trained, and highly-trained for new positions in our department.
The cost of training is $5.00 / hour, $8.50 / hour, and $10.50 / hour for
un-trained, semi-trained, and highly-trained respectfully.
An un-trained worker requires 28 hours of training for the painting
process and 35 hours for the packing process.
A semi-trained worker requires 23 hours of training for the painting
process and 30 hours of training for the packing process.
A highly-trained worker requires only 15 and 20 hours for painting
and packing process training.
We need at least 25 new employees. We only have 700 hours available
for training in the painting process and 775 hours available for training
in the packing process.

67. REQUIREMENT Formulate a linear programming model
that will minimize the total cost of train-
ing while hiring at least twenty-five new
employees.
Let X1 = the number of un-trained workers to hire
Let X2 = the number of semi-trained workers to hire
Let X3 = the number of highly-trained workers to hire

68. The Model Minimize Z = (63x$5.00)X1 + (53x$8.50)X2 + (35x$10.50)X3
SUBJECT TO:
28X1 + 23X2 + 15X3 =< 700 painting training hrs.
35X1 + 30X2 + 20X3 =< 775 packing training hrs.
1X1 + 1X2 + 1X3 => 25 new employees
X1, X2, X3 => 0

69. Linear Programming Application
ADVERTISING
The Salem Department of Tourism is developing a marketing strategy
for next year. They would like to maximize the number of tourists that
choose Salem for their vacation. They have narrowed their choices
down to three types of television ads: regional, statewide, or local.
The research team has determined that a regional ad will reach 100,000
people, a statewide ad will reach 70,000 people, and a local ad will
reach 20,000 people.
The cost per ad is $8, for regional, $6, for statewide, and
$ for local.
There is a one million ($1,000,000.00) dollar budget for advertising.
They would like to have at least twice as many regional ads as local
ads.

70. with their advertising
REQUIREMENT
Formulate a linear
programming
model that will
maximize the
number of people
they can reach
with their advertising
while having at
least twice as many
regional ads as local
ads.
Let X1 = the number of regional ads to place
Let X2 = the number of statewide ads to place
Let X3 = the number of local ads to place

71. The Model Maximize Z = 100,000 X1 + 70,000 X2 + 20,000 X3
EXPOSURE
SUBJECT TO:
ADVERTISING COSTS
BUDGET
AT LEAST TWICE AS MANY REGIONAL ADS

72. QM for WINDOWS Linear Programming

73. WE SELECT THE
LINEAR PROGRAMMING
MODULE

74. WE WANT TO SOLVE A NEW PROBLEM

75. THE DIALOGUE BOX

76. THERE ARE THREE RESOURCE CONSTRAINTS WE WANT TO MAXIMIZE TOTAL PROFIT
THERE ARE TWO
PRODUCTS
( 2 REAL VARIABLES )
WE WANT TO MAXIMIZE TOTAL PROFIT

77. THE DATA INPUT TABLE

78. 6X1 + 2X2 =< 1,800 process hours
THE MODEL
MAXIMIZE “Z” = $3.00 X1 + $8.00 X2
Subject to:
2X1 + 4X2 =< 1,600 labor hours
6X1 + 2X2 =< 1,800 process hours
0X1 + 1X2 =< 350 alarm assemblies

79. ( produce 100 regular clocks ) ( produce 350 alarm clocks )
THE SHADOW PRICES
Labor hours - $1.50
Process hours - $0.00
Alarm assemblies - $2.00
1st SOLUTION WINDOW
( produce 100 regular clocks )
( produce 350 alarm clocks )
( profit = $3, )

80. BASIC VARIABLES ARE THOSE VARIABLES > 0 NON-BASIC VARIABLES ( = 0 )
X1 = 100
X2 = 350
S2 = 500
NON-BASIC VARIABLES ( = 0 )
S1 = 0
S3 = 0

81. THIS IS THE SIMPLEX ALGORITHM SOLUTION TO OUR PROBLEM

82. THE GRAPH METHOD SOLUTION
THE OPTIMAL
SOLUTION
IS AT THIS
CORNER POINT
X1 = 100 , X2 = 350
THE GRAPH METHOD SOLUTION
*******
THE FEASIBLE REGION
SHOWN IN PURPLE

83. Linear Programming

87. Template

98. Linear Programming

99. Linear Programming Graph Method
Solved Problems
Linear Programming
Graph Method
COMPUTER-BASED
MANUAL

100. Solved Problem GRAPH METHOD OF LINEAR PROGRAMMING The Clothier Problem
A clothier makes coats and slacks. The two resources are wool cloth
and labor. The clothier has 150 square yards of wool and 200 hours of
labor available.
Each coat requires three (3) square yards of wool and ten (10) hours of
labor, while each pair of slacks requires five (5) square yards of wool
and four (4) hours of labor.
The profit for a coat is $50.00 and the profit for a pair of slacks is $40.00.
The clothier wants to determine the number of coats and slacks to make
so that profit will be maximized.
REQUIREMENT :
Formulate a linear programming model for this problem.
Solve the model using the graph method.
Solve the model using QM for WINDOWS.

101. Solved Problem The Clothier Problem The Model
Let X1 = coats , Let X2 = slacks
Maximize Z = 50X1 + 40X2
subject to :
3X1 + 5X2 <= 150 square yards (wool)
10X1 + 4X2 <= 200 hours (labor)
X1 , X2 >= 0

102. The Clothier Problem Solved Problem Coordinates:
Wool: X1 = X1 = Labor: X1 = X1 = 0
X2 = X2 = X2 = X2 = 50
X2 SLACKS 50 40 30 20 10
10X1 + 4X2 = 200 hours, labor
3X1 + 5X2 = 150 sq. yards, wool B D
FEASIBLE REGION A C
X1 COATS

103. Corner Point “D” Coordinate Calculations
“10” ( 3X1 + 5X2 = 150)
“3” (10X1 + 4X2 = 200)
30X1 + 50X2 = 1,500
30X1 + 12X2 =
38X2 =
X2 = ≈ 23
Then: 3X1 + 5(23.68) = 150
3X = 150
3X1 = 31.6
X1 = ≈ 10
X1 = 10 coats
X2 = 23 slacks

104. Corner Point Summary and Solution
MAKE
10 COATS
AND
23 PAIRS
OF SLACKS
WITH TOTAL
PROFIT OF
$1,420.00
* OPTIMAL SOLUTION

107. Make 10 Coats
Make 23 Slacks
Total Profit = $1,420.00

109. FEASIBLE REGION

113. Template

125. Solved Problem GRAPH METHOD OF LINEAR PROGRAMMING
The Copperfield Mining Company
The Copperfield Mining Company owns two mines, each of which produces
three grades of ore: high, medium, and low. The company has a contract to
supply a smelting firm with at least twelve (12) tons of high-grade ore, eight
(8) tons of medium-grade ore, and twenty four (24) tons of low-grade ore.
Each mine produces a certain amount of each type of ore during each hour
that it is in operation.
Mine #1 produces 6, 2, and 4 tons respectively of high-, medium-, and low-
grade ore per hour.
Mine #2 produces 2, 2, and 12 tons respectively of high-, medium-, and low-
It costs Copperfield $ per hour to mine ore from mine #1, and $160.00
per hour to mine ore from mine #2.
The company wants to determine the number of hours it needs to operate
each mine so that its contractual obligations can be met at the lowest cost.

126. Solved Problem GRAPH METHOD OF LINEAR PROGRAMMING
The Copperfield Mining Company
REQUIREMENT :
Formulate a linear programming model for
this problem.
2. Solve this model using the graph method.
3. Solve this model using QM for WINDOWS.

127. Solved Problem GRAPH METHOD OF LINEAR PROGRAMMING
The Copperfield Mining Company
The Model:
Let X1 = mine #1 , Let X2 = mine #2
Minimize Z = 200X X2
subject to :
6X X2 >= 12 tons , high-grade ore
2X X2 >= 8 tons , medium-grade ore
4X1 + 12X2 >= 24 tons , low-grade ore
X1 , X2 >= 0
hourly operation cost
hourly
production
rates
supply demands

128. Solved Problem The Copperfield Mining Company
COORDINATES
High-grade ore: X1 = X1 = 0
X2 = X2 = 6
Medium-grade ore: X1 = X1 = 0
X2 = X2 = 4
Low-grade ore: X1 = X1 = 0
X2 = X2 = 2

129. Solved Problem The Copperfield Mining Company 8 Feasible Region 6
X2 ( Mine #2 ) 8 6 4 2
Feasible
Region
6X X2 = 12 tons , high-grade ore
2X X2 = 8 tons , medium-grade ore
4X1 + 12X2 = 24 tons , low-grade ore A B
Feasible
Region C D
( X1 Mine #1 )

130. Solved Problem The Copperfield Mining Company
Corner Point “B” Coordinate Calculations
6X1 + 2X2 = 12
2X1 + 2X2 = 8
4X1 = 4
X1 = 1
Then: 6(1) + 2X2 = 12
2X2 = 6
X2 = 3
Corner Point “C” Coordinate Calculations
4X X2 = 16
“2” (2X X2 = 8)
4X1 + 12X2 = 24
-8X2 = -8
X2 = 1
Then: 2X1 + 2(1) = 8
2X1 = 6
X1 = 3

131. Solved Problem Corner Point Summary and Solution
The Copperfield Mining Company

132. Solved Problem The Copperfield Mining Company solution extrapolation
total number
of tons
of each grade
produced
from
both mines

137. FEASIBLE REGION

141. Template

152. Linear Programming Graph Method
Solved Problems
Linear Programming
Graph Method
COMPUTER-BASED
MANUAL