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3QA3/2DA3

Management Science for Business

Lecture 02

C01: June 23, 2021

C02: June 24, 2021

Instructors:

Seyyed Hossein Alavi Zdravko Dimitrov

Linear Programming (LP)

Intended Learning Outcomes

 At the end of this course, you will be able to

 define mathematical programming and linear programming

concepts.

 solve simple 2 variable questions with ‘graphical method’.

 model LP questions on Excel and solve by using an add-in called

Solver.

Mathematical Programming

 Many business problems are concerned with resource allocation

 Resource, to manufacture products or provide services:

 Machinery,

 Labor,

 Time,

 Space, and

 Raw Materials

 Resource allocation

✓ Decision makers would like to explore all possible decisions or

alternatives to identify the best, or optimal, decision

✓To achieve this, the mathematical programming is commonly used

 Programming: setting up and solving a problem mathematically

Linear Programming

 Mathematical Programming

— A mathematical representation of business decision problems

 Linear Programming (LP) is a powerful form of mathematical

programming

 Widely used to construct a set of mathematical relationships to model real-

world problems, when

➢Objective along with all relationships in the problem can be linearly

described

 LP problems deal with optimization!

 Computationally, the LP models can be polynomially solved to

optimality.

What is Optimization?

 The process of finding the optimum (Min/Max) for an objective

function by changing decision variables with respect to constraints.

 Put yourself in the position of manager for:

 Fortinos, Walmart, Loblaws, …

 Supplier selection problem:

 Decision variables:  Which suppliers are selected

 Number of products ordered from the selected suppliers

 Constraints:  No orders can exceed a supplier’s production capacity

 Orders cannot exceed vehicles loading capacity

 Objective:

 Minimize the total cost of purchasing and transportation 6

Optimization Example

 Maximize the profit

 Minimize the cost

 Minimizing the # of cashiers  Objective function: MIN # of cashiers  Decision variables: # of cashiers  Constraints: # of cashiers > C , sum of salary < Budget

 Maximizing the variety and # of items

 Minimizing the distance between aisles

 Minimizing the # of ON lights 7

More Examples

 Crew Scheduling

 Machine Scheduling

 Location/Allocation problems

 Covering problems

 Transportation/Routing problems

 Inventory management

Linear Programming

 Linear programming (LP) is constructing a set of mathematical

relationships to model real-world problems.

 It includes:

 Objective function

 Constraints

 Decision variables

 The objective function and all the constraints must be

expressed in terms of linear equations or inequalities.

LP Example

— Company manufactures two products A and B

— A requires 9 hours of labor and 5 units of material

— B requires 5 hours of labor and 7 units of material

— The company has 800 hours of labor and 200 units of material

— Profit: product A: $150/unit product B: $130/unit

 Decision Variables:

 How many A to produce? (𝑥1)

 How many B to produce? (𝑥2)

LP Example - Objective

— Company sells two products A and B

— A requires 9 hours of labor and 5 units of material

— B requires 5 hours of labor and 7 units of material

— The company has 800 hours of labor and 200 units of material

— Profit: product A: $150/unit product B: $130/unit

 Objective:

 Maximize the profit

𝑀𝑎𝑥 150𝑥1 + 130𝑥2

Decision Variables:

𝑥1: How many A to produce? 𝑥2: How many B to produce?

LP Example - Constraints

— Company sells two products A and B — A requires 9 hours of labor and 5 units of material — B requires 5 hours of labor and 7 units of material — The company has 800 hours of labor and 200 units of material — Profit: product A: $150/unit product B: $130/unit

 Constraints:  Hours of labor

9𝑥1 + 5𝑥2 ≤ 800

 Units of material

5𝑥1 + 7𝑥2 ≤ 200

 Non-negativity constraints

𝑥1≥ 0, 𝑥2≥ 0

Decision Variables:

𝑥1: How many A to produce? 𝑥2: How many B to produce?

LP Example – The Model

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 150𝑥1 + 130𝑥2 Subject to

9𝑥1 + 5𝑥2 ≤ 800

5𝑥1 + 7𝑥2 ≤ 200

𝑥1 ≥ 0, 𝑥2 ≥ 0

Objective Function

Constraints

RHS

(Right-hand side)

Decision Variables

Non-negativity

Constraints

LP Example – The Model

 Note: Solution may result in fractional values for the decision

variables!

 View them as work-in-process inventory, carried forward to the next

period

 Round off the fractional value, if necessary

 Restrict the decision variables to be integers

→ integer programming problem → computational complexity will increase 14

Decision Variables:

𝑥1: Number of A to be manufactured 𝑥2: Number of B to be manufactured

LP Assumptions

1. Certainty  All parameters are known.

2. Proportionality  If the level of any activity is multiplied by a constant factor, then the contribution of

that activity in the objective function or constraints will be multiplied by the same factor.

3. Additivity  Total of all activities equals the sum of individual activities.

4. Divisibility  Non-integer variables are acceptable.

and of course “ Linearity “ !

Input data values,

coefficients used in the

• objective function, and

• constraints,

are known with certainty.

e.g.

Production of 1 unit of an

item uses 3 hours of a

particular resource,

→ making 10 units of that

item uses 30 hours of the

resource.

e.g.

One product worth $8 per unit is produced,

and

the other product worth $3 per unit is also

produced,

→ the total profit will be 8 plus 3.

LP Example – Why Linear?

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 150𝑥1 + 130𝑥2 Subject to

9𝑥1 + 5𝑥2 ≤ 800

5𝑥1 + 7𝑥2 ≤ 200 𝑥1 ≥ 0, 𝑥2 ≥ 0

• There is no non-linear term, such as

• 𝑥1𝑥2, 𝐿𝑜𝑔 𝑥, 𝑒 𝑥, sin 𝑥

• Exercise:

• 9𝑥1 + 𝑙𝑜𝑔𝑥2 ≤ 3

• 𝑥1 ≤ 2

𝑥2 • 𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 𝑥1

2 + 3𝑥2 4

Solving LP

 Graphical Solving  For problems with at most 2 variables

 Very limited

 Good illustration

 Algorithms:  SIMPLEX

 Interior-point methods  …

 Computer packages  CPLEX  GAMS

 AMPL

 Excel Solver

Graphical Solution

1. Draw all constraint to identify Feasible Region.

2. Draw the objective function contour lines (level, or iso line).

3. Identify the direction where we obtain a better objective value.

Move the contour line (in that direction) to find the last corner point (of the feasible region) that contour lines pass.

The point is called the Optimal Solution.

4. Find the value of the objective function for the optimal solution found.

The region which consists of all points, that

simultaneously satisfy all constraints in the problem.

Ex1: 1. Draw constraints

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 3𝑥 + 5𝑦 Subject to

𝑥 ≤ 4 2𝑦 ≤ 12

3𝑥 + 2𝑦 ≤ 18 𝑥 ≥ 0, 𝑦 ≥ 0

3 regular constraints and 1 pair of non-negativity constraint.

1. Draw all constraint to identify Feasible Region.

 There are two variables:

 We can plot each of the problem’s constrains on a two-dimension graph.

 We plot one variable, x, on the horizontal axis of the graph, and

the other variable, y, on the vertical axis. 19

Ex1: 1. Draw constraints

𝑥 ≥ 0, 𝑦 ≥ 0 y

10 –

–

8 –

–

6 –

–

4 –

–

2 –

–

0 –| | | | | | | | | | | | 0 2 4 6 8 10

Feasible

Region

Ex1: 1. Draw constraints

𝑥 ≤ 4, 2𝑦 ≤ 12

Convert inequality to a linear equation: 𝑥 = 4, 𝑦 = 6

10 –

–

8 –

–

6 –

–

4 –

–

2 –

–

0 –| | | | | | | | | | | | 0 2 4 6 8 10

y = 6

x = 4

Feasible

Region

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 3𝑥 + 5𝑦 Subject to

𝑥 ≤ 4 2𝑦 ≤ 12

3𝑥 + 2𝑦 ≤ 18 𝑥 ≥ 0, 𝑦 ≥ 0

 To draw a constraint which contains both variables, we need:  2 points for the line, and one identifying point

3𝑥 + 2𝑦 ≤ 18

1. Convert inequality to a linear equation:

3𝑥 + 2y = 18

2. Find two points:

✓ Normally, the two easiest points are the intersections between the linear

equation and two axes:

• If 𝑥 = 0, then 𝑦 = 9 ⇒ (0, 9)

• If 𝑦 = 0, then 𝑥 = 6 ⇒ (6, 0)

3. Draw a line through two points.22

Ex1: 1. Draw constraints

10 –

9 –

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –| | | | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 10 11

(x = 0, y = 9)

(x = 6, y = 0)

3x+2y = 18

x = 4

y = 6

3𝑥 + 2𝑦 ≤ 18

10 –

9 –

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –| | | | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 10 11

(x = 0, y = 9)

(x = 6, y = 0)

3x+2y = 18

x = 4

y = 6

3𝑥 + 2𝑦 ≤ 18

Use (0,0) as a specifying point:

3𝑥 + 2𝑦 = 0 ≤ 18

Therefore, (0,0) belongs to the

feasible region.

Ex1: 1. Draw constraints- Specifying Point

Ex1: 1. Draw constraints- Feasible Region

Any point in the feasible region is a “feasible solution”

Ex1: 2. Draw Objective Contours

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 3𝑥 + 5𝑦 Subject to

𝑥 ≤ 4 2𝑦 ≤ 12

3𝑥 + 2𝑦 ≤ 18 𝑥 ≥ 0, 𝑦 ≥ 0

2. Draw the objective function contour lines (level lines or iso

lines)

 Let 3𝑥 + 5𝑦 = 𝑍 and set 𝑍 to be an arbitrary value

 Only one line is enough

Ex1: 2. Draw Objective Contours

10 –

9 –

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –| | | | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 10 11

Feasible

Region

3x+5y = 15

3𝑥 + 5𝑦 = 𝑍

Ex1: 2. Draw Objective Contours

10 –

9 –

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –| | | | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 10 11

Feasible

Region

3x+5y = 15

3x+5y = 30

3𝑥 + 5𝑦 = 𝑍

Ex1: 2. Draw Objective Contours

10 –

9 –

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –| | | | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 10 11

Feasible

Region

3x+5y = 15

3x+5y = 30

3x+5y = 35

3𝑥 + 5𝑦 = 𝑍

Ex1: 3. Examine Corner Points

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 3𝑥 + 5𝑦 Subject to

𝑥 ≤ 4 2𝑦 ≤ 12

3𝑥 + 2𝑦 ≤ 18 𝑥 ≥ 0, 𝑦 ≥ 0

3. Find the last corner point that contour lines pass.

 Move the contour line to leave the feasible region.

 It is guaranteed that the best solution is one of the corner points.

 The last corner node is the Optimal Solution.

Ex1: 4. Optimal Solution

10 –

9 –

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –| | | | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 10 11

Feasible

Region

3x+5y = z

y = 6

3x+2y = 18

(𝑥, 𝑦) = 2,6 is the optimal solution

Objective function value = 3 × 𝑥 + 5 × 𝑦 = 36

(x = 2, y = 6)

Ex1: 4. Optimal Solution

(𝑥, 𝑦) = 2,6 is the optimal solution

Example 2: Flair Company

 Flair Furniture Company produces tables and chairs:

 Manager/Market restrictions:

 Make no more than 450 chairs (the on-hand inventory of chairs is high)

 Make at least 100 tables (the inventory of table is low)

Per Table Per Chair

Availability

Profit ($) 7 5

Carpentry Department

(labor hours) 3 4 2,400

Painting Department

(labor hours) 2 1 1,000

EX2: Flair Company LP Model

 Decision Variables:

 # of tables produced : T

 # of chairs produced : C

 Objective function:

 Maximize the profit

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 7𝑇 + 5𝐶

EX2: Flair Company LP Model

 Constraints:

 Carpentry hour (the number of labor hours available at the department)

3𝑇 + 4𝐶 ≤ 2,400

 Painting hour (the number of labor hours available at the department)

2𝑇 + 1𝐶 ≤ 1,000

 Maximum chair

𝐶 ≤ 450

 Minimum table

𝑇 ≥ 100

 Non-negativity 𝑇 ≥ 0 𝐶 ≥ 0

EX2: Flair Company LP Model

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 7𝑇 + 5𝐶 Subject to

3𝑇 + 4𝐶 ≤ 2,400

2𝑇 + 1𝐶 ≤ 1,000

𝐶 ≤ 450

𝑇 ≥ 100

𝑇 ≥ 0

𝐶 ≥ 0

EX2: Flair Graphical LP Solution

N u

m b

e r

o f C

h a

ir s (

C )

Number of Tables (T)

1,000 –

–

800 –

–

600 –

–

400 –

–

200 –

–

0 –| | | | | | | | | | | | 0 200 400 600 800 1,000

Carpentry Constraint Line 3𝑇 + 4𝐶 = 2,400

EX2: Flair Graphical LP Solution

N u

m b

e r

o f C

h a

ir s (

C )

Number of Tables (T)

1,000 –

–

800 –

–

600 –

–

400 –

–

200 –

–

0 –| | | | | | | | | | | | 0 200 400 600 800 1,000

Region Satisfying 3𝑇 + 4𝐶 ≤ 2,400

(𝑇 = 300,𝐶 = 200)

(𝑇 = 0,𝐶 = 0)

EX2: Flair Graphical LP Solution

N u

m b

e r

o f C

h a

ir s (

C )

Number of Tables (T)

1,000 –

–

800 –

–

600 –

–

400 –

–

200 –

–

0 –| | | | | | | | | | | | 0 200 400 600 800 1,000

Carpentry Constraint 3𝑇 + 4𝐶 ≤ 2,400

Painting Constraint 2𝑇 + 1𝐶 ≤ 1,000

EX2: Flair Graphical LP Solution

N u

m b

e r

o f C

h a

ir s (

C )

Number of Tables (T)

1,000 –

–

800 –

–

600 –

–

400 –

–

200 –

–

0 –| | | | | | | | | | | | 0 200 400 600 800 1,000

Painting Constraint 2𝑇 + 1𝐶 ≤ 1,000

Carpentry Constraint 3𝑇 + 4𝐶 ≤ 2,400

Maximum Chairs Allowed Constraint 𝐶 ≤ 450

Feasible

Region

Infeasible Solution

Minimum Tables Required Constraint 𝑇 ≥ 100

EX2: Flair Graphical LP Solution

2 3

| | | | | | | | | |

0 200 400 600 800 1,000

800 –

–

600 –

–

400 –

–

200 –

–

0 –

N u

m b

e r

o f C

h a

ir s (

C )

Number of Tables (T)

Painting Constraint

Carpentry Constraint

Optimal Level Profit Line

Optimal Corner Point Solution

𝑀𝑎𝑥𝑖𝑚𝑖𝑧𝑒 7𝑇 + 5𝐶 → 7𝑇 + 5𝐶 = 𝑍

EX2: Flair Graphical LP Solution

2 3

| | | | | | | | | |

0 200 400 600 800 1,000

800 –

–

600 –

–

400 –

–

200 –

–

0 –

N u

m b

e r

o f C

h a

ir s (

C )

Number of Tables (T)

Painting Constraint

3𝑇 + 4𝐶 ≤ 2,400

Carpentry Constraint

2𝑇 + 1𝐶 ≤ 1,000

Optimal Level Profit Line

Optimal Corner Point Solution

EX2: Flair Graphical LP Solution

 Optimal solution:

 Intersection of carpentry constraint and painting constraint:

 Objective function value:

7𝑇 + 5𝐶 = 7 × 320 + 5 × 360 = $4,040

൝ 3𝑇 + 4𝐶 = 2,400

2𝑇 + 1𝐶 = 1,000 ⇒ 𝑇 = 320,𝐶 = 360

EX2: Flair Graphical LP Solution

Point 1 (T = 100, C = 0)

Profit = $7 x 100 + $5 x 0 = $700

Point 2 (T = 100, C = 450)

Profit = $7 x 100 + $5 x 450 = $2,950

Point 3 (T = 200, C = 450)

Profit = $7 x 200 + $5 x 450 = $3,650

Point 4 (T = 320, C = 360)

Profit = $7 x 320 + $5 x 360 = $4,040

Point 5 (T = 500, C = 0)

Profit = $7 x 500 + $5 x 0 = $3,500

Self Exercise

Maximize 𝑍 = 30𝑥 + 20𝑦

Subject to

2𝑥 + 𝑦 ≤ 1,000 𝑥 + 𝑦 ≤ 800

𝑥 ≤ 350

𝑥 ≥ 0

𝑦 ≥ 0

1. Draw the feasible region for this LP model and identify all the corner points

2. Draw at least one objective contour line and find the direction where the

value of 𝑍 increases

3. Find the corner point that maximizes the value of 𝑍 and report the point’s

coordinates and the optimal value of 𝑍.

Self Exercise

1,000 –

–

800 –

–

600 –

–

400 –

–

200 –

–

0 –| | | | | | | | | | | | 0 200 400 600 800 1,000

Feasible

Region

30x+20y = 12,000

30x+20y = 18,000

2𝑥 + 𝑦 = 1,000

𝑥 + 𝑦 = 800

➔ 𝑥∗=200, 𝑦∗ = 600

Special situations in LP

1) Alternate Optimal Solution

2) Redundant Constraints

3) Unbounded Solutions

4) Infeasibility

1- Alternate Optimal Solution

 When there is more than one corner point that maximizes or

minimizes the objective function

 Happens when the objective contour line is parallel to one of

the constraints (same slope)

𝑀𝑎𝑥 3𝑥1 + 4𝑥2 Subject to

𝑥2 ≤ 2 6𝑥1 + 8𝑥2 ≤ 24

𝑥1 ≤ 3

𝑥1 ≥ 0

𝑥2 ≥ 0

6 = 4

𝑆𝑙𝑜𝑝𝑒𝑠 = − 3

1- Alternate Optimal Solution

2 3

Feasible

Region

| | | | | | | | | |

0 200 400 600 800 1,000

800 –

–

600 –

–

400 –

–

200 –

–

0 –

Level Profit Line Is Parallel to Painting Constraint

Optimal Solution Consists of All

Points Between Corner

Points 4 and 5

Z = 7𝑇 + 5𝐶

Z = 6𝑇 + 8𝐶 3𝑇 + 4𝐶 ≤ 2,400

2- Redundant Constraint

 A redundant constraint plays NO role in identifying the feasible region!

𝑀𝑎𝑥 𝑥1 + 𝑥2 Subject to

𝑥1 + 𝑥2 ≤ 4 𝑥2 ≤ 3

𝑥1 + 𝑥2 ≤ 6

𝑥1 ≤ 5 𝑥1 ≥ 0 𝑥2 ≥ 0

Redundant Constraints

 𝑥1 ≤ 4 − 𝑥2 ≤ 4

2- Redundant Constraint

Carpentry Constraint is Redundant

Painting Constraint is Redundant

F e a s ib

le R

e g io

N u

m b

e r

o f C

h a

ir s (

C )

Number of Tables (T)

1,000 –

–

800 –

–

600 –

–

400 –

–

200 –

–

0 –| | | | | | | | | | | | 0 200 400 600 800 1,000

𝑇 ≤ 100𝑇 ≥ 100

3- Unbounded Solution

 Objective function can be infinitely large (for MAX) or infinitely small (for Min)

𝑀𝑎𝑥 𝑥1 + 𝑥2 𝑥1 + 𝑥2 ≥ 4

−𝑥1 + 𝑥2 ≤ 4

𝑥1 ≥ 0

𝑥2 ≥ 0

Note: It depends on the objective function.

Unbounded Solution VS. Unbounded Feasible Region

3- Unbounded Solution

Unbounded

Feasible Region

10 –

9 –

8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –| | | | | | | | | | | 0 1 2 3 4 5 6 7 8 9 10

4- Infeasibility

 An LP is infeasible if all constraints cannot be satisfied at the

same time

𝑀𝑎𝑥 𝑥1 + 𝑥2 𝑥1 + 𝑥2 ≤ 3

𝑥1 + 𝑥2 ≥ 5

𝑥1 ≥ 0

𝑥2 ≥ 0

No solution in this case!

4- Infeasibility

Region

Satisfying

Fourth

Constraint

Region

Satisfying

Three

Constraints

1,000 –

–

800 –

–

600 –

–

400 –

–

200 –

–

0 –| | | | | | | | | | | | 0 200 400 600 800 1,000

Two Regions

Do Not Overlap

Solving LP Using Excel Solver®

Modeling in Microsoft Excel

 Excel Solver Activation:

 Click ‘File’ and then the ‘Options’.

 On the ‘Excel Options’ dialog box, click the ‘Add-ins’ command.

 At the bottom, select “Excel Add-ins” for Manage and hit “Go…”.

 The ‘Add-ins’ dialog box will appear. Check ‘Solver Add-in’; click the ‘OK’.

 ‘Solver’ is now loaded and is available in the Excel ‘Data’ tab.

Flair Furniture Company - Solver

Maximize 7𝑇 + 5𝐶 Subject to

3𝑇 + 4𝐶 ≤ 2,400

2𝑇 + 1𝐶 ≤ 1,000

𝐶 ≤ 450

𝑇 ≥ 100

𝑇 ≥ 0

𝐶 ≥ 0

Carpentry hour

Painting hour

Maximum chairs

Minimum tables

Flair Furniture Company - Solver

59 Screenshot 2-1A Formula View of the Excel Layout for Flair Furniture – Page 41

Flair Furniture Company - Solver

Screenshot 2-1C Solver Add Constraint Window – Page 45

Flair Furniture Company - Solver

 Solver Options:

➢The values and check boxes

shown in the “All Methods” tabs

are good ones to use in tis

course.

Screenshot 2-1D Solver Options Window – Page 46

Flair Furniture Company - Solver

63 Screenshot 2-1E Excel Layout and Solver Solution for Flair

Furniture (Solver Results Window Also Shown) – Page 47

Solver Output for LP Situations1

1) Optimality  “Solver found a solution. All Constraints and optimality conditions are satisfied.”

2) Non-linearity  “The linearity conditions required by this LP Solver are not satisfied.”

3) Formulation Error  “Solver encountered an error value in the Objective Cell or a Constraint Cell.”

4) Unbounded Solutions  “The Objective Cell values do not converge”

5) Infeasibility  “Solver could not find a feasible solution”

64 1 - Table 2.2 – Page 48

Logical functions in the model:

IF(), SUMIF(), LOOKUP(), …

Exam Sample Questions

 Break-Even Point (quantity and dollar)

 Developing Mathematical Formulation

 Graphical Solution

 4 Special Situations of LP

 Solving LP with Solver®

Practice Problems

 CHAPTER 2

 Discussion Questions 3, 4, 7, 10, 11, 12

 Problems 13, 17, 27, 29, 43