Introduction to Operations Research and Linear Programming

Introduction• Our world is filled with limited resources.

• Oil• Land

• Human• Time

• Business• Resource (Budget)

• Manufacturing • m/c , number of worker

• Restaurant• Space available for seating

Introduction

• How best to used the limited resources available ?

• How to allocate the resource in such a way as to maximize profits or minimize costs ?

Introduction

• Mathematical Programming(MP) is a field of management science or operations research that fines most efficient way of using limited resources/ to achieve the objectives of a business.

Optimization

Characteristics of Optimization Problems

• One or more decisions• Restrictions or constraints

e.g. Determining the number of products to manufacturea limited amount of raw materialsa limited amount of labor

• Objective– The production manager will choose the mix of

products that maximizes profits– Minimizing the total transportation cost

Expressing optimization problems mathematically

• Decision variablesX1 , X2 , X3 , … , Xn

e.g. the quantities of different products Index n = the number of product types

• Constraints– a less than or equal to constraint : f(X1 , X2 , X3 , … , Xn) < b– a greater than or equal to constraint : f(X1 , X2 , X3 , … , Xn) > b– an equal to constraint : f(X1 , X2 , X3 , … , Xn) = b

• Objective – MAX(or MIN) : f(X1 , X2 , X3 , … , Xn)

Mathematical formulation of an optimization problem

MAX(or MIN) : f(X1 , X2 , X3 , … , Xn)

Subject to: f(X1 , X2 , X3 , … , Xn) < b1

….f(X1 , X2 , X3 , … , Xn) > bk

….f(X1 , X2 , X3 , … , Xn) = bm

note : n variables , m constraints

Mathematical programming techniques

The function in model : f(x) -Linear-Nonlinear

Decision variable -Fractional value e.g. 2.33-Integer value e.g. 1,2,3,4..-Binary e.g. 0,1

Test : Linear functions?1.

2.

3.

5032 321 xxx

602 21 xx

75)3/1(4 21 xx

9.0323

321

321 xxx

xxx

4.

5.

4573 221 xx

Problem• Blue Ridge Hot Tubs manufactures and sells two models of hot tubs :

Aqua Spa and the Hydro-Lux. – Howie Jones, the owner and manager of the company, needs to

decide how many of each type of hot tubs to produce– 200 pumps available – Howie expects to have 1,566 production labor hours and

2,880 feet of tubing available.– Aqua-Spa requires 9 hours of labor and 12 feet of tubing– Hydro-Lux require s 6 hours of labor and 16 feet of tubingAssuming that all hot tubs can be sold To maximize profits, how many Aqua-Spas and Hydra-Luxs should be

produce?

Formulating LP Models1. Understand the problem2. Indentify the decision variable3. State the objective as a linear combination of decision variables

Max : 350x1 + 300x2

4. State the constraints as linear combinations of the decision variable 4.1 Pumps available 4.2 labor available4.3 Tubing available

5. Identify any upper or lower bounds on the decision variable x1 > 0

x2 > 0

Max : 350x1 + 300x2

Subject to:

x1 + x2 < 200 , Pumps available

9x1 + 6x2 < 1,566 , Labors available

12x1 + 16x2 < 2,880 , Tubing available

x1 > 0

x2 > 0

General Form of an LP model

MAX(or MIN) : C1X1+ C2X2+ , … , CnXn

Subject to: a11X1+ a12X2+ , … , a1nXn < b1

ak1X1+ ak2X2+ , … , aknXn > bk

am1X1+ am2X2+ , … , amnXn = bm

Notations

a11X1+ a12X2+ , … , a1nXn < b1

X1+ X2+ , … ,+ Xn = b1

X11 + X21+ X31 = b1

X12 + X22+ X32 = b2

11

1 bxa i

n

ii

11

bxn

ii

ji

ij bx

3

1j,

Graphical Method:Solving LP Problems

Graphical Method

x1+x2 ≤ 200 (1)

Graphical Method

9x1+ 6x2 ≤ 1,566 (2)

x1+x2 ≤ 200 (1)

Note : X1 = 0 , x2 = 1566/6 = 261X2 = 0 , x1 = 1566/9 =174

Graphical Method

9x1+ 6x2 ≤ 1,566 (2)

x1+x2 ≤ 200 (1)

12x1+ 16x2 ≤ 2,880 (3)

Note : X1 = 0 , x2 = 2880/16 = 180X2 = 0 , x1 = 2880/12 = 240

Graphical Method

MAX 350x1 + 300x2

Using level curves

• MAX 350x1 + 300x2

1. Set a number of objective e.g. Obj = 35,000

2. Finding points (x1,x2) which has obj = 35,000 x1 = 100 , x2 = 0

x1 = 0 , x2 = 116.67

Graphical Method

9x1+ 6x2 ≤ 1,566 (2)

x1+x2 ≤ 200 (1)

12x1+ 16x2 ≤ 2,880 (3)

MAX 350x1 + 300x2

Finding the optimal solution

9x1+ 6x2 = 1,566 (2)x1+x2 = 200 (1)

9x1+ 6(200 –x1) = 1,5663x1+ 1,200 = 1,5663x1 = 366Optimal Solution = x1 = 122X2 = 78

Special Conditions in LP Models

9x1+ 6x2 ≤ 1,566 (2)

x1+x2 ≤ 200 (1)

12x1+ 16x2 ≤ 2,880 (3)

MAX 450x1 + 300x2

1. Alternate Optimal Solutions - having more than one optimal solution

Special Conditions in LP Models

9x1+ 6x2 ≤ 1,566 (2)

x1+x2 ≤ 225 (1)

12x1+ 16x2 ≤ 2,880 (3)

MAX 350x1 + 300x2

2. Redundant Constraints A constraint that plays no role in determining the feasible region of the problem.

Special Conditions in LP Models

-x1+ 2x2 ≤ 400 (2)

x1+ x2 ≥ 400 (1)

x1 ≥ 0

MAX x1 + x2

3. Unbounded Solutions The objective function can be made infinitely large.

x2 ≥ 0

Special Conditions in LP Models

4. Infeasibility

x1+ x2 ≤ 150 (1)

x1+ x2 ≥ 200 (2)

x1 ≥ 0

MAX x1 + x2

x2 ≥ 0 ?

– No ways to satisfy all of the constraints

Using Excel Solver

Using Excel Solver

Model Excel SolverObjective Set Target CellDecision Variable Changing CellsConstraints Subject to the constraints

Using Excel Solver

9x1+ 6x2 ≤ 1,566 (2)

x1+x2 ≤ 200 (1)

12x1+ 16x2 ≤ 2,880 (3)

MAX 350x1 + 300x2

Parameters Profit /unit of each product , Pumps Req /unit, no. Pumps available ,Labor Req /unit, no. labor available ,Tubing Req /unit, no. Tubing available

Decision Variables

No. Aqua-Spas produceNo. Hydra-Luxs produce

Objective Maximize Profit

Constraint Pumps available , labor available , Tubing available

Using Excel Solver

Using Excel Solver

Modeling LP Problems

1. Make vs. Buy Decisions - The Electro-Poly corporation received a $750,000 order for various quantities of 3 types of slip rings.- Each slip ring requires a certain amount of time to wire and hardness.

Model1 Model2 Model3

Number ordered 3,000 2,000 900

Hours of wiring required per unit 2 1.5 3

Hours of harnessing required per unit 1 2 1

The company has only 10,000 hours of wiring capacity.The company has only 5,000 hours of harnessing capacity.

Make vs. Buy Decisions (Continue)

The company can sub contract to one of its competitors.

Determine the number of slip rings to make and the number to buy in order to fill the customer order at the least possible cost

Cost($) Model1 Model2 Model3

Cost to make 50 83 130

Cost to buy 61 97 145

Make vs. Buy Decisions (Continue)• Defining the decision variables

mi = number of model i slip rings to make in-house

bi = number of model i slip rings to buy from competitor

• Defining the objective function– To minimize the total cost

Min : 50m1 + 83m2 + 130m3 + 61b1+ 97b2 + 145b3

Subject to:

2m1 + 1.5m2 + 3m3 < 10,000 , wiring constraint

1m1 + 2m2 + m3 < 5,000 , harness constraint

m1 + b1 = 3,000 , Required order of model1

m2 + b2 = 2,000 , Required order of model2

m3 + b3 = 900 , Required order of model3

m1 , m2 ,m3 , b1 , b2 ,b3 , > 0

Make vs. Buy Decisions

Parameters Cost to buy, Cost to make, Number ordered, Hours of wiring , Hours of harnessing

Decision Variable - number of model i slip rings to make in-house- number of model i slip rings to buy from competitor

Objective Total Cost

Constraint wiring constraint harness constraint Required order of each model

A transportation problemSupply R/M Area Processing Plant Capacity

275,000 200,000

400,000 600,000

300,000 225,000

i1

i2

i3

j1

j2

j3

Distance : (mile) j1 j2 j3

i1 21 50 40

i2 35 30 22

I3 55 20 25

A transportation problem• How many product should be shipped from r/m area to the processing plant,

the trucking company charges a flat rate for every mile. Min the total distance ~ Min total cost of transportation• Defining the decision variable

xij = number of R/M to ship from node i to node j• Defining the objective function

Min : 21x11+50x12 +40x13+35 x21 +30x22 +22x23 +55x31 +20x32 +25x33

• Defining the constraints x11 + x21 + x31 < 200,000 capacity restriction for j1x12 + x22 + x32 < 600,000 capacity restriction for j2x13 + x23 + x33 < 225,000 capacity restriction for j3x11 + x12 + x13 = 275,000 supply restriction for i1x21 + x22 + x23 = 400,000 supply restriction for i2x31 + x32 + x33 = 300,000 supply restriction for i3 xij > 0 , for all i and j

A transportation problem

Parameters Distance from I to j, no supply , no capacity

Decision Variable number of R/M to ship from node i to node j

Objective Min the total distance

Constraint capacity restriction supply restriction

A transportation problem

A Blending problem

• Agri-Pro stocks bulks amounts of four types of feeds that can mix to meet a given customer’s specifications.

Nutrient(%) feed1 feed2 feed3 feed4Corn (i=1) 30% 5% 20% 10%Grain(i=2) 10% 30% 15% 10%Minerals(i=3) 20% 20% 20% 30%Cost per Pound($) 0.25 0.30 0.32 0.15

A Blending problem• Agri-Pro has just received an order from a local chicken farmer for 8,000

pounds of feed.• The farmers wants this feed to contain at least 20% corn, 15% grain , and

15% minerals. • What should Agri-Pro do to fill this order at minimum cost?• Defining the decision variable

xi = pounds of feed (j) to use in the mix • Defining the objective function

Min : 0.25x1 +0.30x2 +0.35x3+0.15 x4 , total cost• Defining the constraints

x1 + x2 + x3 + x4 = 8,0000.3x1 +0.05 x2 + 0.20x3 + 0.10x4 > 0.20 × 8,000 --- Corn0.1x1 +0.30 x2 + 0.15x3 + 0.10x4 > 0.15× 8,000 --- Grain0.2x1 +0.20 x2 + 0.20x3 + 0.30x4 > 0.15 × 8,000 --- Mineralx1 , x2 , x3 , x4 > 0

A Blending problem

Parameters Cost per Pound($), Nutrient(%) of each feed Order quantity , Required Nutrient(%)

Decision Variable pounds of feed (j) to use in the mix

Objective Minimize Cost

Constraint Satisfy the Required Nutrient(%)

A Blending problem

A production and inventory planning problem

• Upton Corporation is trying to plan its production and inventory levels for the next 6 months

• Given the size of Upton’s warehouse, a maximum of 6,000 units • The owner like to keep at least 1,500 units in inventory any months as

safety stock• The company wants to produce at no less than half of its maximum

production capacity each month.• Estimate that the cost of carrying a unit in any given month is equal to

1.5% of the unit production cost in the same month.

m1 m2 m3 m4 m5 m6

Unit production cost (Cm) 240 250 265 280 280 260

Unit Demanded(Dm) 1000 4500 6000 5500 3500 4000

Maximum Production(Maxp m) 4000 3500 4000 4500 4000 3500

A production and inventory planning problem

• The number of units carried in inventory using the averaging the beginning and ending inventory for each month

• Current inventory 2,750 units• To minimize production and inventory costs, the

company wants to identify the production and inventory plan.

• Defining the decision variable pm = the number of units to product in month m

bm = the beginning inventory for month m

A production and inventory planning problem

• Defining the objective functionMin : ,

Subject to

2/))(015.0( 1

6

1

6

1

mmmmm

mm bbcpC

mm Maxpp m,

m,

m,

mmmm Dpbb 1

m,

m,

mm Maxpp )2/1(000,6 mmm Dpb

500,1 mmm Dpb

A production and inventory planning problem

Parameters Maximum Production(Maxp),Unit production cost(Cm) , Unit Demanded(Dm)

Decision Variable the number of units to product in month m , pm

the beginning inventory for month m, bm

Objective Minimize Cost

Constraint Upper bound and Lower bound of the production and inventory , Balance constraint

A production and inventory planning problem