Quantitative Analysis for
Management
Goa Institute of Management
PGDM-FT 2023-2024
27-09-2023
Deepti Mohan | Goa Institute of Management
Problem A (System of linear equations)
• An animal feed company
must produce 200 kg of a
mixture consisting of
ingredients A and B daily.
Ingredient A costs Rs. 3 per
kg, and ingredient B costs
Rs. 8 per kg. If the total cost
of the mixture is Rs. 1100,
determine how much of
each ingredient should be
used.
27-09-2023
𝑥𝐴 + 𝑥𝐵 = 200
3𝑥𝐴 + 8𝑥𝐵 = 1100
Deepti Mohan | Goa Institute of Management
Problem A (System of linear equations)
• An animal feed company
must produce 200 kg of a
mixture consisting of
ingredients A and B daily.
Ingredient A costs Rs. 3 per
kg, and ingredient B costs
Rs. 8 per kg. If the total cost
of the mixture is Rs. 1100,
determine how much of
each ingredient should be
used.
27-09-2023
𝑥𝐴 + 𝑥𝐵 = 200
3𝑥𝐴 + 8𝑥𝐵 = 1100
Deepti Mohan | Goa Institute of Management
Problem B (LPP)
• An animal feed company must produce 200 kg
of a mixture consisting of ingredients A, and B
daily. A costs Rs. 3 per kg and B Rs. 8 per kg.
Not more than 80 kg A can be used and at
least 60 kg of B must be used. Find how much
of each ingredient should be used if the
company wants to minimize the cost.
Maximize 3𝑋1 + 8𝑋1
s.t.
𝑋1 + 𝑋2 =200
𝑋1 ≤ 80
𝑋2 ≥ 60
𝑋1 , 𝑋2 ≥ 0
27-09-2023
Deepti Mohan | Goa Institute of Management
Problem B (LPP)
• An animal feed company must produce
200 kg of a mixture consisting of
ingredients A, and B daily. A costs Rs. 3 per
kg and B Rs. 8 per kg. Not more than 80 kg
A can be used and at least 60 kg of B must
be used. Find how much of each ingredient
should be used if the company wants to
minimize the cost.
Maximize 3𝑋1 + 8𝑋1
s.t.
𝑋1 + 𝑋2 =200
𝑋1 ≤ 80
𝑋2 ≥ 60
𝑋1 , 𝑋2 ≥ 0
27-09-2023
Deepti Mohan | Goa Institute of Management
Plotting an inequality
•𝑥≤3
• 𝑥+𝑦 ≤5
• 𝑥−𝑦 ≥3
• 2𝑥 − 3𝑦 ≤ −3
27-09-2023
Tip: Put x=0 and y=0
and see if the
inequality is satisfied. If
yes, origin(0,0) is part
of the inequality, hence
it is the half plane that
includes the origin. If
no, it is the half plane
that does not include
the origin.
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
Maximize
Z = 10x1 + 15x2
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
Maximize
Z = 10x1 + 15x2
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
Maximize
Z = 10x1 + 15x2
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
Maximize
Z = 10x1 + 15x2
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
Maximize
A bounded feasible region is
composed of:
1) Interior points
2) Boundary points (excluding
corner points)
3) Corner points
Z = 10x1 + 15x2
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
Maximize
A bounded feasible region is
composed of:
1) Interior points
2) Boundary points
(excluding corner points)
3) Corner points
Z = 10x1 + 15x2
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
A bounded feasible region is
composed of:
1) Interior points
2) Boundary points
(excluding corner points)
3) Corner points
Maximize
Z = 10x1 + 15x2
Subject to
2x1 + x2 ≤ 26
Irrespective of the objective
function, either E or F has better
objective function value than that
of I and either G or H has a better
objective function value than that
of I. Interior points are always
dominated by boundary points.
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
A bounded feasible region is
composed of:
1) Interior points
2) Boundary points
(excluding corner points)
3) Corner points
Maximize
Z = 10x1 + 15x2
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
A boundary point which is
not a corner point is
dominated by one of the
two corner points on its
either side.
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
LPP – key ideas
• Every point inside the feasible region is dominated by a boundary
point.
• Every boundary point is dominated by a corner point (At best, a
boundary point can only be as good as a corner point).
• Hence optimal solution is one of the corner points.
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 1
Maximize
Corner points
Z = 10x1 + 15x2
Objective Function
Value
10X1+ 15X2
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
(8,10)
10(8)+ 15(10) = 230
(13,0)
10(13)+ 15(0) = 130
(6,11)
10(6)+ 15(11) = 225
(0,5)
10(0)+ 15(5) = 75
(0,0)
10(0)+ 15(0) = 0
Graphical Method: Example - 1
Maximize
Coordinates
(X1,X2)
Point
Objective Function
Value
10X1+ 15X2
Z = 10x1 + 15x2
A
(8,10)
10(8)+ 15(10) = 230
B
(13,0)
10(13)+ 15(0) = 130
C
(6,11)
10(6)+ 15(11) = 225
D
(0,5)
10(0)+ 15(5) = 75
E
(0,0)
10(0)+ 15(0) = 0
Subject to
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
1x1 – 1x2 ≥ -5
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 2
Minimize
Z = 3x1 + 5x2
Subject to
-3x1 + 4x2 ≤ 12
2x1 + 3x2 ≥ 12
2x1 – 1x2 ≥ -2
x1 ≤ 4
x2 ≥ 2
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 2
Minimize
Z = 3x1 + 5x2
Subject to
-3x1 + 4x2 ≤ 12
2x1 + 3x2 ≥ 12
2x1 – 1x2 ≥ -2
x1 ≤ 4
x2 ≥ 2
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 2
Minimize
Z = 3x1 + 5x2
Subject to
-3x1 + 4x2 ≤ 12
2x1 + 3x2 ≥ 12
2x1 – 1x2 ≥ -2
x1 ≤ 4
x2 ≥ 2
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 2
Minimize
Z = 3x1 + 5x2
Coordinates
(X1,X2)
Point
Objective
Function Value
3X1+ 5X2
Subject to
-3x1 + 4x2 ≤ 12
A
(4/5,18/5)
3(4/5)+ 5(18/5) =
102/5=20.4
B
(4,6)
3(4)+ 5(6) = 42
C
(3/4,7/2)
3(3/4)+ 5(7/2) =
79/4=19.75
D
(3,2)
3(3)+ 5(2) = 19
E
(4,2)
3(4)+ 5(2) = 22
2x1 + 3x2 ≥ 12
2x1 – 1x2 ≥ -2
x1 ≤ 4
x2 ≥ 2
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 3
Maximize
Z = 6x1 + 8x2
Subject to
2x1 + 3x2 ≥ 15
4x1 + 2x2 ≥ 8
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 3
Maximize
Z = 6x1 + 8x2
Subject to
2x1 + 3x2 ≥ 15
4x1 + 2x2 ≥ 8
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 3
Maximize
Z = 6x1 + 8x2
Unbounded feasible region
Unbounded solution
Subject to
2x1 + 3x2 ≥ 15
4x1 + 2x2 ≥ 8
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 4
Minimize
Z = 6x1 + 8x2
Subject to
2x1 + 3x2 ≥ 15
4x1 + 2x2 ≥ 8
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 4
Minimize
Z = 6x1 + 8x2
Subject to
2x1 + 3x2 ≥ 15
4x1 + 2x2 ≥ 8
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 4
Minimize
Unbounded feasible region
Unique optimal solution
Z = 6x1 + 8x2
Subject to
2x1 + 3x2 ≥ 15
4x1 + 2x2 ≥ 8
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 5
Maximize
Z = 2000x1 + 4000x2
Subject to
1x1 - 2x2 ≥ 10
1x1 + 2x2 ≤ 30
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 5
Maximize
Z = 2000x1 + 4000x2
Subject to
1x1 - 2x2 ≥ 10
1x1 + 2x2 ≤ 30
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 5
Maximize
Coordinates
(X1,X2)
Z = 2000x1 + 4000x2
Subject to
1x1 - 2x2 ≥ 10
1x1 + 2x2 ≤ 30
(20,5)
2000(20)+
4000(5) =
60000
(10,0)
2000(10)+
4000(0) =
20000
(30,0)
2000(30)+
4000(0) =
60000
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Objective
Function Value
2000X1+
4000X2
Graphical Method: Example - 5
Maximize
Infinite number of optimum
solutions
Z = 2000x1 + 4000x2
Subject to
1x1 - 2x2 ≥ 10
1x1 + 2x2 ≤ 30
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 6
Maximize
Z = 20x1 + 30x2
Subject to
2x1 + x2 ≤ 40
4x1 - x2 ≤ 20
x1 ≥ 30
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 6
Maximize
Z = 20x1 + 30x2
Subject to
2x1 + x2 ≤ 40
4x1 - x2 ≤ 20
x1 ≥ 30
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 6
Maximize
No feasible solution
Z = 20x1 + 30x2
Subject to
2x1 + x2 ≤ 40
4x1 - x2 ≤ 20
x1 ≥ 30
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 7
Maximize
Z = 6x1 + x2
Subject to
2x1 + 3x2 ≤ 16
4x1 + 2x2 ≥ 16
2x1 + x2 =16
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 7
Maximize
Z = 6x1 + x2
Subject to
2x1 + 3x2 ≤ 16
4x1 + 2x2 ≥ 16
2x1 + x2 =16
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Graphical Method: Example - 7
Feasible region is
a point (only one
feasible solution)
Maximize
Z = 6x1 + x2
Subject to
2x1 + 3x2 ≤ 16
4x1 + 2x2 ≥ 16
2x1 + x2 =16
& x1, x2 ≥0
27-09-2023
Deepti Mohan | Goa Institute of Management
Convexity
▪ A set (or region) is convex if only if for any two points on the set, the
line segment joining those points lies entirely in the set.
▪ The feasible region of a linear programming problem is a convex set.
27-09-2023
Deepti Mohan | Goa Institute of Management
Convexity
▪ A set (or region) is convex if only if for any two points on the set, the
line segment joining those points lies entirely in the set.
▪ The feasible region of a linear programming problem is a convex set.
27-09-2023
Deepti Mohan | Goa Institute of Management
Convexity
▪ A set (or region) is convex if only if for any two points on the set, the
line segment joining those points lies entirely in the set.
▪ The feasible region of a linear programming problem is a convex set.
A non-convex
set
A convex set
27-09-2023
Deepti Mohan | Goa Institute of Management
LPP – key ideas
Feasible region: The region containing all the feasible solutions of
a linear programming problem.
A bounded feasible region is composed of:
1) Interior points
2) Boundary points
3) Corner points (also called extreme points)
• Every point inside the feasible region is dominated by a boundary
point. Every boundary point is dominated by a corner point (At best, a
boundary point can only be as good as a corner point). Hence optimal
solution is one of the corner (extreme) points.
• The feasible region of an LPP is a convex set.
27-09-2023
Deepti Mohan | Goa Institute of Management
Solve the problem
• The Healthy Juices Co. is renowned for producing organic juices and
smoothies. The company recently decided to explore full-scale
production and selling of Orange Juice and Mixed Berry Smoothie.
They have inaugurated a new production unit which can operate for
48 hours per week. Preparing a batch of Orange Juice consumes 2
hours, whereas a batch of Mixed Berry Smoothie requires 3 hours.
The profit contribution from selling each batch of Orange Juice is Rs.
4000, while a batch of Mixed Berry Smoothie contributes Rs. 8000 to
the profits. The sales team, after extensive market analysis, concluded
that a maximum of 15 batches of Orange Juice and 10 batches of
Mixed Berry Smoothie can be sold each week.
27-09-2023
Deepti Mohan | Goa Institute of Management
𝑥1 = Number of batches of Orange Juice produced and sold each week.
𝑥2 = Number of batches of Mixed Berry Smoothie produced and sold
each week.
𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑍 = 4000𝑥1 + 8000𝑥2
𝑆𝑢𝑏𝑗𝑒𝑐𝑡 𝑡𝑜:
2𝑥1 + 3𝑥2 ≤ 48[Time Constraint]
𝑥1 ≤ 15 [Orange Juice Demand Constraint]
𝑥2 ≤ 10[Smoothie Demand Constraint]
𝑥1 , 𝑥2 ≥ 0 [Non−negativity Constraint]
27-09-2023
Deepti Mohan | Goa Institute of Management
27-09-2023
Deepti Mohan | Goa Institute of Management
C (0, 10) 80000
H (9, 10) 116000
G (15, 6) 108000
E (15, 0) 60000
I (0, 0) 0
27-09-2023
Deepti Mohan | Goa Institute of Management
27-09-2023
Deepti Mohan | Goa Institute of Management