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Operations Research – Massimo Paolucci – DIBRIS University of Genova

Linear Mathematical Programming (LP)

• A MP is LP if :– The objective function is linear

where

– The set X is defined by linear equality or inequality constraints

xcxf T=)(

],...,[ 1 nT ccc =

=

nx

xx

1

bxA ≤ where

=

mb

bb

1

=

mnm

n

aa

aaA

1

111


• An LP problem can be espressed as

• The quantitiesare two vectors and a matrix of constant coefficients representing the parameters of the problem

0

..)(max

≥≤

=

xbxA

tsxcxf T(LP)

Abc



• The set of feasible solutions of LP can be denoted as

• Types of LP problems:

}0,:{ ≥≤ℜ∈= xbxAxX n

Continuous LP (LP)}0,:{ ≥≤ℜ∈= xbxAxX n

}0,:{ ≥≤∈= xbxAxX nZ


Integer LP (IP)

}0,0,:Z,{ 21 ≥≥≤+∈ℜ∈= yxbyDxAyxX nn

Mixed Integer LP (MIP)


• How can a decision problem to be modeled as a LP problem?• What corresponds to the set X?• How can we find a solution to our decision problem?

• We see these aspects considering an example ...



LP – an example• A company producing paints wants to plan its daily production• Two types of paints are considered, a paint for interior (I) and a paint for

exterior (E)• Paint production uses two raw materials indicated with A and B• The daily availability of raw material is: A = 6 tons, B = 8 tons• The quantity of A and B consumed to produce one ton of paint E and I is

known:E I

A 1 2B 2 1Raw material

Paints

• We assume that all the paint produced is sold• The seling price per ton is 3K€ for E and 2K€ for I• Further information obtained by the company through a market survey:

– the daily demand of I paint never exceeds more than 1 ton that of paint E – the maximum daily demand of paint I is 2-ton

• Decision problem: determine the quantities to produce daily for the two paints in order to make the maximize the gain from sales


LP – problem formulation• MP formulation of a decision problem

• Variables: define the solution of the problem• Objective: quantifies the quality of a solution (as a function of the

variables)• Constraints: identify the set of allowed values for the variables

defining the feasible solution set (e.g., technological constraints, budget, operational constraints, etc.)

Mathematical formulation

Definition ofvariables

Definition ofobjective

Definition ofconstraints


LP – an example (cont.)• Problem variables

– Two variables representing the quantity (tons) of paint of the two types produced (and sold) daily:

• Production of paint for external: xE

• Production of paint for internal : xI– Continuous and non negative variables

• Problem objective function– Maximize the daily gain (K€) from selling the produced paint types

– It is a linear expressionIE xxZ 23 +=


• Problem constraints– Technological constraints: the use of raw materials cannot exceed the

material availability

– Constraints due to the market survey

– Variable positivity constraints (variable lower bounds)

812)(621)(

≤+≤+

IE

IExxBxxA

Raw material Availability

Consumption for unit of product

LP – an example (cont.)

21

≤≤−

I

EIx

xx

00 ≥≥ IE xx


• The complete problem formulation

• It is a LP problem

)6(0)5(0

)4(2)3(1)2(82)1(62

23max

≥≥

≤≤+−≤+≤+

+=

I

E

I

IE

IE

IE

IE

xx

xxxxxxx

xxZ

Constraints defining the set of feasible solutions

Objective function

LP – an example (cont.)


8

7

6

5

4

3

2

1

1 2 3 4 5 6-1-2 xE

xI (5)

(6)(1)

(2)(3)

(4)

X

)1(62 ≤+ IE xx

The (xE, xI) plane

)6(0)5(0

≥≥

I

Exx

Non negativity constraints

Intoducing the constraints

)2(82 ≤+ IE xx)3(1≤+− IE xx)4(2≤Ix

LP – Graphic solution


LP – Polyedra• The feasibility set X includes the inner points and the sides (in

case of strictly inequality the problem is non linear)• X is a Polyedron obtained from the intersection of the semi-

spaces defined by hyperplanes (lines in the example)• What’s the difference between LP and IP polyedra?

– A polyedron for an LP probleminclude ∞ feasible solutions

}:{ bxAxX n ≤∈= RX

x1

x2


• The polyhedron for an IP problem has a finite number of feasible solutions

Red dots on the sides and in the polyedron correspond to the integer feasible solutions

}:{ bxAxX n ≤∈= Z

Xx1

x2

LP - Polyedra


1 2 3 4 5 6

1

-1-2

2

3

4(2) (3)

(4)

(1)

(5)

(6)xE

xI

023 =+ IE xxIf Z=0

If Z=6

623 =+ IE xx

LP - Graphic solutionThe obj function is drawn in the plane (xE,xI) as a parametric line that depends on the value fixed for Z

IE xxZ 23 +=


1 2 3 4 5 6

1

-1-2

2

3

4(2) (3)

(4)

(1)

(5)

(6)xE

xI

A

F

E D

B

C

A=(0,0)B=(4,0)C=(10/3,4/3) D=(2,2)E=(1,2)F=(0,1)

Z=0 Z=6 Z=38/3

Optimal solution

LP - Graphic solution


1 2 3 4 5 6

1

-1-2

2

3

4(2) (3)

(4)

(1)

(5)

(6)xE

xI

A

F

E D

B

CC=(10/3,4/3)Optimal

solution

33823* =+= IE xxZ

Property:The optimal solution lies on thefrontier of polyedron X



• Linearity implies that optimal (finite) solutions are in the polyedron vertices• More than a single (finite) equivalent solutions (infinite number) if the

objective function is parallel to a constraint• Even in these cases the search for optimal solution is limited to polyedron

vertices

1

2

F

E D

B

C

xE

xI

1 2 3 4

X

A

All the points on DC are optimal



• In the paint production problem it can be interesting to know how the raw materials (available resources) are used:– Is there any advantage from an increase of the resource availability?

– Is it possible to decrease the resource availability without changing the optimal objective value?

• Another interesting point is to analyse how the optimal solution would change if the selling prices change

LP – Graphic solution and post-optimality


• Post-optimality ⇔ Sensitivity analysis w.r.t changes in resources availabilityAll the constraints are ≤ so they can be viewed as

Quantity of resource used ≤ Resource availability

– Constraints (1) and (2) are satisfied as equalities by the optimal solution– Then both raw materials A and B are used up to their availability– Constraints (1) and (2) are saturated and raw materials A and B are said

scarce resources

Saturated constraint Scarse resource

Non Saturated constraint Abundant resource

LP - Graphic solution and post-optimality


First case: increase the availability of a scarse resourceWhat is the maximum increment that is worth considering for resource A?

(1)1

2

A

F

E D

B

C

xE

xI

1 2 3 4

X

(2)

(4)K

z=38/3 z=13

Increasing resource Aavailability moves constraints (1) and so the optimal point

K=(3,2)



Increase beyond K=(3,2) is no profitable (why?)

(1)1

2

A

F

E D

B

C

xE

xI

1 2 3 4

X

(2)

(4)

Availability of A increases from 6 to 7

)'1(72 ≤+ IE xx

The new (1’)(1’)



Same behaviour for resource B

1

2

A

F

E D

B

C

xE

xI

X

(2)

(4)

(1)1 2 3 4 5 6

(6)L

z=38/3z=18

Increasing availability of B move (2)and optimal point

Beyond L=(6,0) no reasonfor further increase of B.

New availability for B is 12

)'2(122 ≤+ IE xx



Opposite case: decrease of abundant resourcesAssume (3) and (4) associated with (abundant) resourcesHow much can we reduce their avalability without affecting the current optimal solution?

Constraint (3)

(4)

(2)

(1)

(3)

1

2

B

C

xE

xI

1 2 3 4

A

F

E D

X

A

F

E D

X’

1≤+− IE xx2−≤+− IE xx



Opposite case: decrease of abundant resourcesAssume (3) and (4) associated with (abundant) resourcesHow much can we reduce their avalability without affecting the current optimal solution?

Constraint (4)

1

2

A

F

B

C

xE

xI

1 2 3 4

X

(2)

(4)

(1)

DE DE

X’

2≤Ix

34≤Ix



Which among resource A and B is worth increasing first? (the company may have a limited budget to invest)

• The Unit Value of a resource yi (also called Shadow Price):

• yi = the increase of objective for a unitary increase of resource iavailability

variation resourcemaxvariationmax

iZyi =



Examples• Resource A:

• Resource B:

• Resource B is the most convenient to increase

(K€/ton)31

33839

673

3813=−=

−

−=Ay

34

43

3854

8123

3818=

−

=−

−=By (K€/ton)



LP - Graphic solution and post-optimalityConsider a possible variation in the product selling pricesHow the optimal solution change? • Changing coefficient cE and cI the slope of objective function changes

F1

2

A

E D

B

C

xE

xI

1 2 3 4

X

cI increasecE decrease(1)

slope23−=−

I

Ecc


slope23−=−

I

Ecc

F1

2

A

E D

B

C

xE

xI

1 2 3 4

X

cI decreasecE increase

(2)

Changing cE e cI point C remainsthe optimal solution until theslope of the objective functionequals the one of either (1) or (2)



• Compute the ranges for cE and cIFirst case: change cE with fixed cI=2

– If cE = 1 C and D are optimal. If cE becomes < 1 only D is optimal

– If cE = 4 C and B are optimal. If cE becomes > 4 only B is optimal

21

2−=− Ec

= slope of (1) 1= Ec

= slope of (2)22

−=− Ec 4= Ec

41 ≤≤ Ec



• Compute the ranges for cE and cISecond case: change cI with fixed cE=3

– If cI = 3/2 C and B are optimal. If cI becomes < 3/2 only B is optimal

– If cI = 6 C and D are optimal. If cI becomes > 6 only D is optimal

= slope of (1)

= slope of (2)


213 −=−

Ic

23 −=−Ic

1= Ic

23= Ic

623 ≤≤ Ic

operations research –massimo paolucci –dibris universityof ... didattico 2017... · operations...

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