Intro Formulation

8/12/2019 Intro Formulation

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Operations Research (Optimization Problem)
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Research into operations
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Allocating limited resources (funds, manpower, time, rawmaterial) among competing activities in the best possiblemanner.
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Search of optimal solution strategy in a given situation.
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Search of optimal solution strategy in a given situation.

Applicable to business operations

Operations Research at Taco BellOperational Cost Control at Kellogg
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Optimization Problem

What is an Optimization Problem all about??
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Minimizing or Maximizing some Objective subject to certainrestrictions.
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Why is an Optimization Problem required??
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All problems seek to maximize or minimize some quantity (theobjective function).
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The presence of restrictions or constraints, limits the degree to

which we can pursue our objective.
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The presence of restrictions or constraints, limits the degree to

which we can pursue our objective.

There must be alternative courses of action to choose from.
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Optimization Problem:(Mathematically)
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Return function: F(X), X Rn

Constraint functions: gi(X), i= 1, 2, . . . , n
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Linear Program:


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Linear Program:

All functions are linear = linear relationship
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Linear Program:


What do we mean by linear?


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Linear Program:


What do we mean by linear?

Linear:

Things are proportional

Double the production:Profit gets doubledRaw material consumption is doubled
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Example

Two Products (A and B)
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Example


Each requires two raw materials (P and Q)

One ton of A needs 2 tons of P, 4 tons of QOne ton of B needs 3 tons of P, 2 tons of Q
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Example




Raw material availability is limited 6 tons of P, 9 tons of Q
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Example





Each product yields different profit/ton

A: Rs. 400 per tonB: Rs. 500 per ton
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Example





Each product yields different profit/ton

A: Rs. 400 per tonB: Rs. 500 per ton

? Problem: How much of each product should beproduced so that Total Profit is maximized
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Elements of a Linear Programming Model


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Decision VariablesQuantities whose values the management wants to decide
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Decision VariablesQuantities whose values the management wants to decidex1: tons of product A to producex2: tons of product B to produce
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Decision VariablesQuantities whose values the management wants to decidex1: tons of product A to producex2: tons of product B to produce

Parameters (Problem Data)Quantities that cannot be changed by management
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Objective FunctionA mathematical function of decision variables, whose value

the management wishes to minimize or maximize (bycontrolling the values of decision variables)


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the management wishes to minimize or maximize (bycontrolling the values of decision variables)Maximize Z= 400x1+ 500x2
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Constraints

Restrictions under which the decision must be taken


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Constraints

Restrictions under which the decision must be takenRequirement Availability

2x1+ 3x2 6 (availability of P)4x1+ 2x2 9 (availability of Q)


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Constraints

Restrictions under which the decision must be takenRequirement Availability

2x1+ 3x2 6 (availability of P)4x1+ 2x2 9 (availability of Q)

Produced quantities can not be negative

x1, x2 0 (Non-negativity)

C


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Complete Model

Maximize 400x1+ 500x2Subject to 2x1+ 3x2 6

4x1+ 2x2 9x1, x2 0

The above is a mathematical formulation of the problem (LPP).

C l M d l
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Complete Model


4x1+ 2x2 9x1, x2 0


Study and Explore the mathematical structure.
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C l t M d l


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Complete Model


4x1+ 2x2 9x1, x2 0


Study and Explore the mathematical structure.

Learn how to formulate real life problems as LPs Learn techniques for solving it.

Li P i A ti


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Linear Programming Assumptions

Li ea P og a i g Ass tio s
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Problem data is known exactly (deterministic)

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All variables can assume any real value (continuous variables)

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All variables are non-negative



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All relationship between variables are linear:



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All relationship between variables are linear:Programming??

Not to be confused with computer programming!!!



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Not to be confused with computer programming!!!It is the systematic Planning of activities.



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Not to be confused with computer programming!!!It is the systematic Planning of activities.

Linear Programming: The problem of planning the activities tooptimize a linear objective function subject to linear constraints.

Formulation of a Linear Program
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Translating a verbal description of the problem into an equivalentmathematical description

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1 Identify Decision variables

2 Identify objective function

3 Identify constraints



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1 Identify Decision variables

2 Identify objective function

3 Identify constraints

More of an Art than a Science.

Takes practice and experience to formulate correctly and efficiently.

New Venture Finance
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Need to raise 50 lakhs

New Venture Finance
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Upto 20 lakhs in bonds @12%
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New Venture Finance


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Fin. Corp. A: 30 lakhs @18%Condition: Bond debt + loan from B loan from A

Fin. Corp. B: (@16%)Condition: loan amt. not to exceed the loan from A

New Venture Finance
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Decision variables:x1: bond debt, x2: loan from A, x3: loan from B

New Venture Finance
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Objective: Minimize 12x1+ 18x2+ 16x3

New Venture Finance
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Constraints:

x1+ x2+ x3 50 (Required Amount)

New Venture Finance


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Constraints:

x1

+ x2

+ x3

50 (Required Amount)x1 20 (Bond Debt)

New Venture Finance


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Constraints:

x1

+ x2

+ x3

50 (Required Amount)x1 20 (Bond Debt)x2 30 (Loan from A)

New Venture Finance
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Constraints:x1

+x2

+x3 50 (Required Amount)x1 20 (Bond Debt)

x2 30 (Loan from A)x1+ x3 x2 (Condition of Fin. Inst. A)

New Venture Finance
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Constraints:x1

+x2

+x3


x1+ x3 x2 (Condition of Fin. Inst. A)x3 x2 (Loan from B)

New Venture Finance
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Constraints:x1

+x2

+x3


x1+ x3 x2 (Condition of Fin. Inst. A)x3 x2 (Loan from B)

x1, x2, x3 0 (Non-negativity constraints)

A Job-Training Problem
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Trained Machinist:Work a machineTeachStay idle

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Decision variables:

x1: trained machinist teaching in Januaryx2: trained machinist idle in Januaryx3: trained machinist teaching in Februaryx4: trained machinist idle in Februaryx5: trained machinist teaching in Marchx6: trained machinist idle in March



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Decision variables:

x1: trained machinist teaching in Januaryx2: trained machinist idle in Januaryx3: trained machinist teaching in Februaryx4: trained machinist idle in Februaryx5: trained machinist teaching in Marchx6: trained machinist idle in March

Machining + Teaching + Idle = Trained machinists available

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Constraints:

100 + x1+ x2 = 130 (Jan)150 + x3+ x4 = 130 + 7x1 (Feb)200 + x5+ x6 = 130 + 7x1+ 7x3 (Mar)

130 + 7x1+ 7x3+ 7x5 = 250 (Apr)

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Constraints:


130 + 7x1+ 7x3+ 7x5 = 250 (Apr)

Objective:Minimize 400(10x1+ 10x3+ 10x5) + 700(x1+ x3+ x5)+500(x2+ x4+ x6)



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Constraints:


130 + 7x1+ 7x3+ 7x5 = 250 (Apr)

Objective:Minimize 400(10x1+ 10x3+ 10x5) + 700(x1+ x3+ x5)+500(x2+ x4+ x6) +700(100 + 150 + 200)



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Constraints:


130 + 7x1+ 7x3+ 7x5 = 250 (Apr)

Objective:Minimize 400(10x1+ 10x3+ 10x5) + 700(x1+ x3+ x5)+500(x2+ x4+ x6) +700(100 + 150 + 200)

Non-Negativity:x1, x2, x3, x4, x5, x6 0

Investment Planning
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Scheme A: x1 x4Scheme B: x5 x7Scheme C: x8Scheme D: x9


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Investment Planning


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Objective:Maximize 1.4x4+ 1.7x7+ 1.9x8+ 1.3x9

Constraints:

x1+ x5 60000 Year 1
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Investment Planning


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Constraints:

x1+ x5 60000 Year 1x2+ x6+ x8 60000 (x1+ x5) Year 2

x3+ x7 60000 (x1+ x5+ x2+ x6+ x8) + 1.4x1 Year 3

Investment Planning
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Constraints:

x1+ x5 60000 Year 1x2+ x6+ x8 60000 (x1+ x5) Year 2

x3+ x7 60000 (x1+ x5+ x2+ x6+ x8) + 1.4x1 Year 3

x4 60000 (x1+ x5+ x2+ x6+ x8+ x3+ x7)+1.4x1+ 1.4x2+ 1.7x5 Year 4
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Roster Planning

Decision variables: xij


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Decision variables: xijThe number of workers start working on day i(i= 1, . . . , 7) in shift

j(j = 1, 2, 3)

Roster Planning

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j(j = 1, 2, 3)Constraints:

7

i=1

3

j=1

xij 60 (No of Workers)

Roster Planning

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7

i=1

3

j=1


x1j +x7j +x6j +x5j D1j j = 1, 2, 3 (Monday)
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Roster Planning



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ij

The number of workers start working on day i(i= 1, . . . , 7) in shift


7

i=1

3

j=1


x1j +x7j +x6j +x5j D1j j = 1, 2, 3 (Monday)

x2j +x1j +x7j +x6j D2j j = 1, 2, 3 (Tuesday)x3j +x2j +x1j +x7j D3j j = 1, 2, 3 (Wednesday)x4j +x3j +x2j +x1j D4j j = 1, 2, 3 (Thursday)x5j +x4j +x3j +x2j D5j j = 1, 2, 3 (Friday)x6j +x5j +x4j +x3j D6j j = 1, 2, 3 (Saturday)x7j +x6j +x5j +x4j D7j j = 1, 2, 3 (Sunday)

xij 0

Dij is the number of workers required on day i and shift period j

Objective Function: Minimize7

i=1

3

j=1

xij
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Documents

Intro Formulation