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Quantitative Methods - 2010

Linear Programming (LP)

1QM 2010 - Transcend

George Dantzig

Father of Linear Programming


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Preliminary

Programming in the OR sense means to find

the values for several variables. (Not to beconfused with computer coding).

LP is part of a larger field called MathematicalProgramming (MP)

Objective of LP is to optimize (maximize orminimize) a function of several variables, subject

to constraints on these variables.

Huge variety of very large, practical problems canbe expressed and solved as LP problems

QM 2010 - Transcend 2


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Motivating Scenario

(Finance)

An NBFC is considering investing in shares of 2companies, A and B.

Current price per share: A: Rs.250, B: Rs.160

Expected annual returns per share are: A: Rs. 45

B: Rs. 37

NBFC internal limit for this industry: Rs. 10 L SEBI limits: A 3000 shares; B 4000 shares

Question: How many shares to buy in A and B?



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Motivating Scenario

(Advertizing) A marketing manager is considering 2 different

media, A and B, for promotion of a product.

Cost per insertion in each is known Medium A: Rs. 25,000; Medium B: Rs. 36,000

Expected no. of prospects per insertion Medium A: 7 L; Medium B: 9 L

Ad budget: Rs.350,000

Ad availability Medium A: 4 insertions; Medium B: 6 insertions

Question: How many insertions in A, B?



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Motivating Scenario

(Distribution)

Product availability (no. of cartons) at our

warehouses in

Sangli: 700; Lonavla: 450; Palghar: 800

Requirements at our customers in

Vapi: 630; Nasik: 750; Bombay: 570

Transport costs from each to each are known.

Question: How much to send from which

warehouse to which customer?



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General LP Problem



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General Structure

We have

n decision variables: x1 --- xn An objective function to be optimized involving the

decision variables: Z = c1x1 + --- +cnxn A set of constraints involving the decision variables:

a11x1 + a12x2 - - - +a1nxn b1 a21x1 + a22x2 - - - +a2nxn b2

- - - - - - - - - -

am1x1 + am2x2 - - - +amnxn bm

For solving, we have to add non-negativity constraints:

All x 0



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Structure for 2 Decision Variables

We have

2 decision variables: x1 and x2 An objective function to be optimized involving the 2

decision variables: Z = c1x1 + c2x2 A set of constraints involving the decision variables:

a11x1 + a12x2 b1 a21x1 + a22x2 b2

- - - - - - -

am1x1 + am2x2 bm

For solving, we have to add non-negativity constraints:

All x 0



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Graphical Representation(Maximizing Z - For 2 Decision Variables)


a11x1 + a12x2 b1

a21

x1

+ a22

x2 b

2

a31x1 + a32x2 b3

x1

x2

Z = c1x1 + c2x2

Feasible Region

Solution is x1 = x1*, and x2 = x2*

(x1*, x2*)


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Graphical Representation(Minimizing Z - For 2 Decision Variables)


x1

x2

(x1*, x2*)

Solution is x1 = x1*, and x2 = x2*


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Formulating A (2 Variable) LP Problem

Identify your decision variables

Let x1 = no of shares of A; x2 = no of shares of B

Write objective function

Maximize Z = 45x1 + 37x2 Write constraints

250x1 + 180x2 10 L ---- investment constraint

X1 + x2 4500 ----- SEBI constraint

Add non-negativity constraints

X1, x2 0



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Plotting The Constraints

Estimate range of variables and draw axes

X1: 0 to 4500, x2: 0 to 6000

Take any 2 arbitrary x1 in this range, write

constraints as equations, solve for x2 (1) 250(1000) + 160(x2) = 10 L; x2 = 4687

250(3000) + 160(x2) = 10 L; x2 = 1563

(2) x1 = 3000 (3) x2 = 4000

Plot the constraints lines



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Feasible Region For NBFC Problem


x1

x2

250(x1) + 160 (x2) = 10 L

X1 = 3000

X2 = 4000

0 1000 2000 3000 6000

0

10

00

2000

3000

6000


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Objective Function

On Feasible region


x1

x2

250(x1) + 160 (x2) = 10 L

X1 = 3000

X2 = 4000

A

Z = 85,000


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Final Solution

Moving the Z line parallely up, we can see that the Zline thru point A will have a higher value than thru anyother point in the Feasible Region.

Calculate the coordinates of A:

Clearly, x2 of A = 4000 Put this value in the other line equation, to get x1 for A

250(x1) + 160(4000) = 10 L, so x1 = 3.60 L/250 = 1440

Find Z at A, by putting x1 = 1440, x2 = 4000 in Zequation Z = 45(1440) + 40(4000) = 224,800

Thus Max Z = 224,800, and solution is x1 = 1440, x2 =4000



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Illustrating Final Solution


x1

x2

250(x1) + 160 (x2) = 10 L

X1 = 3000

X2 = 4000

A

Z = 85,000

Z = 2,24,800


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Verification Of Final Solution

Point B has x1 = 3000 and x2 = 1563

Z at that point = 45(3000) + 40(1563) = 197,520

This less than the Z at A

All other points will have even less Z

So clearly Z at A is the maximum value, and

coordinates of A is the solution.



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Transportation Problem

(Special Case Of General LP Problem)



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General Structure

We have mxn table of m sources and n destinations. Table entries are transportation costs cij (per unit of some

item) from each destination to each source

Row totals Ai are availabilities at the sources

Column totals Rj are requirements at the destinations

We have mxn decision variables (how much to ship) : x11 --- xmn An objective function to be optimized involving the

decision variables: Z = c11x11 + --- +cmnxmn Constraints are shipments

Must not violate any availabilities

Must meet all requirements



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Transportation LP Table


Available

d1 d2 --- --- dn

s1 c11 c12 c1n A1s2 c21 A2

Sources --- ---

--- ---

sm cm1 cmn Am

Required R1 R2 --- --- Rm

Destinations


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Transportation LP Solution Table


Available

d1 d2 --- --- dn

s1 x11 x12 x1n A1

s2 x21 A2

Sources --- ---

--- ---

sm xm1 xmn Am

Required R1 R2 --- --- Rm

Destinations

X in rows must add up to Availability

X in columns must add up to Required

Some X can be zero.


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Try This Yourself

Cartons of milk powder are stored in 3 warehouses asfollows: Sangli: 700; Lonavla: 450; Palghar: 800

They are required at 3 customers as follows:

Vapi: 630; Nasik: 750; Bombay: 570 Transport costs per carton are as follows:

Suggest 3 distribution plans, and work out total cost foreach plan.


Vapi Nasik Bombay

Sangli 5 4 2

Lonavla 7 5 3

Palghar 2 8 6


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Assignment Problem

(Another Special Case Of General LP Problem)



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Assignment Problem Table


m1 m2 --- --- mmj1 c11 c12 c1n

j2 c21

Jobs ---

---

jm cm1 cmn

Machines


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Assignment Problem Solution Table


m1 m2 --- --- mm

j1 x11 x12 x1n

j2 x21

Jobs ---

---

jm xm1 xmn

Machines

There can be only one 1 in each row

and in each column.

All other entries must be 0


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End Of

Linear Programming


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