Lect Class 4

7/24/2019 Lect Class 4

1/28

ME F344/MF F344

ENGINEERING OPTIMIZATION

Rajesh P Mishra,

1228A

[email protected]


2/28

Mute ur call

Plz add Engineering

Optimization group of

facebook (Rajesh Mishra)


3/28

Reddy Mikks Problem

The Reddy Mikks company owns a small paint factory that produces bothinterior and exterior house paints for wholesale distribution. Two basicraw materials, A and B, are used to manufacture the paints.

The maximum availability of A is 6 tons a day; that of B is 8 tons a day. Thedaily requirements of the raw materials per ton of interior and exteriorpaints are summarized in the following table.

A market survey has established that the daily demand for the interiorpaint cannot exceed that of exterior paint by more than 1 ton. The survey

also showed that the maximum demand for the interior paint is limited to2 tons daily.

The wholesale price per ton is $3000 for exterior paint and $2000 perinterior paint. How much interior and exterior paint should the companyproduce daily to maximize gross income?

Tons of Raw Material per Ton of Paint

Exterior Interior Maximum Availability (tons)

Raw Material A 1 2 6

Raw Material B 2 1 8


4/28

Reddy Mikks Problem Formulation

Define:

XE= Tons of exterior paint to be produced

XI= Tons of interior paint to be produced

Thus, the LP formulation of the Reddy-Mikks Company is as follows:

Maximize z = 3XE+ 2XI

Subject to:

XE+ 2XI 6 (1) (availability of raw material A)

2XE+XI 8 (2) (availability of raw material B)

-XE+XI 1 (3) (Restriction in production)

XI 2 (4) (Demand Restriction)

XE,XI 0


5/28

Graphical Solution of an LP Problem

Used to solve LP problems with two (and sometimes

three) decision variables

Consists of two phases

Finding the values of the decision variables for which allthe constraints are met (feasible region of the solution

space)

Determining the optimal solution from all the points in the

feasible region (from our knowledge of the nature of theoptimal solution)


6/28

Finding the Feasible Region (2D)

Steps

Use the axis in a 2-dimensional graph to represent thevalues that the decision variables can take

For each constraint, replace the inequalities with

equations and graph the resulting straight line on the 2-dimensional graph

For the inequality constraints, find the side (half-space) ofthe graph meeting the original conditions (evaluate

whether the inequality is satisfied at the origin) Find the intersection of all feasible regions defined by all

the constraints. The resulting region is the (overall)feasible region.


7/28

Graphical Solution of the Ready

Mikks Problem

A solutionis any

specification of values for

the decision variables.

A feasible Solutionis a

solution for which all the

constraints are satisfied.

The feasible regionis the

set of all feasible

solutions.

Notice that the feasible

region is convex

I

1 2 3E

4 5 6 7

0+ 2(0)

6

Constraint 4: XI2

Constraint 3: -XE + XI1

Constraint 2: 2XE + XI8

Feasible

Region

Constraint 1: XE + 2XI6


8/28

Finding the Optimal Solution

Determine the slope of the objective function (aninfinite set of straight lines-isoclines) Select a convenient point in the feasible region

Draw the corresponding straight line (a single isocline)

Determine the direction of increase of the objectivefunction (we are maximizing) Select a second point in the feasible region and simply

evaluate the objective function at that point

Follow the direction of increase until reaching the(corner) point beyond which any increase of theobjective function would take you outside of thefeasible region

a line on a diagram or map connecting points of equal gradient or inclination


9/28

Graphical Solution of the Ready

Mikks Problem

An Optimal Solutionis a

feasible solution that

has the most favorable

value of the objectivefunction.

A Corner-point feasible

(CPF) solution is a

solution that lies at a

corner of the feasible

region.

The optimal solution is a

corner point feasible

solution (why?)

Max z = 3XE + 2XI

Point 1: XE

=2, XI =

0: Z = 6

Point 2:

XE =4/3,

XI =1

Z = 6Point: XE =3.33, XI =1.33(How can we get this point?)

Z = 12Z = 9

Z = 12.66

I

1 2 3E

4 5 6 7


10/28

Exercise (five minutes)

For the Ready Mikks problem, find all the corner-point feasible solutions

Suppose that another constraint is added to the

problem : XE + XI1, and the problem is changedfrom maximization to minimization. For this new

problem, find the new optimal solution

Discuss and answer the following question: Is it

possible to get a non-convex feasible region from the

addition of a linear constraint?


11/28

Solution

I

1 2 3E

4 5 6 7

Feasible

Region

New

Constraint:

XE + XI1

Max: XE =3.33, XI =1.33

Z = 2: XE =0, XI =1


12/28

GE produces two types of electric motors,

each on a separate assembly line. The

respective daily capacities of the two linesare 150 and 200 motors. Type I motor uses

2 units of a certain electronic component,

and type II motor uses only 1 unit. Thesupplier of the component can provide 400

pieces a day. The profits per motor of types

I and II are $8 and $5 respectively.Formulate the problem as a LPP and find

the optimal daily production.


13/28

Let the company producex1type I motors

andx

2 type II motors per day.

1 28 5z x x

Subject to the constraints1

2

1 2

1 2

150

2002 400

, 0

x

x

x x

x x

The objective is to findx1andx2so as to

Maximize the profit


14/28

Graphical solution

z=1800

z=1700

z=1200

z=400

z=1000

(100,200)

(150,100)

(150,0)

0,200)

x1

x2

Maximize z=8x1

+5x2Subject to the constraints

2x1+x2400

x1 150

x2200

x1,x20

Optimum

=1800 at


15/28

Diet Problem

Minimize z = 2x1+ 3x2

Subject to x1+ 3x215

2x1+ 2x220

3x1+ 2x224

x1,x20

Optimum = 22.5 at (7.5, 2.5)


16/28(7.5,2.5)

(4,6)

(0,12)

(15,0)Minimum at

z = 42

z = 39

z= 36

z = 30z = 26

z = 22.5

Graphical Solution of Feed Mix Problem


17/28

Maximize z = 2x1+x2

Subject to x1+x2 40

4x1+x2 100

x1,x20

Optimum = 60 at (20, 20)


18/28

(20,20)

(0,40)

(25,0)

z maximum at


19/28

Maximizez= 2x1+x2

Subject to 2

1 2

1 2

1 2

1 2

10

2 5 60

18

3 44

, 0

x

x x

x x

x x

x x


20/28

[1]

[2]

(5,10)

[3]

(10,8)

[4]

(13,5)

(14.6,0)

(0,10)

z is maximum at (13, 5)

Maxz= 31

z = 4

z = 10

z = 20

z = 28

z = 31


21/28

Exceptional Cases

1. Usually a LPP will have a unique optimal

solution.2. But there are problems where there may be no

solution,

3. may have alternative optimum solutions and

unboundedsolutions.

4. We graphically explain these cases in the

following slides.

5. We note that the (unique) optimum solutionoccurs at one of the corners of the set of all

feasible points.


22/28

Alternative optimal solutions

1 210 5z x x

Subject to the constraints 1

2

1 2

1 2

150

200

2 400

, 0

x

x

x x

x x

Consider the LPP

Maximize


23/28

Graphical solution

z=2000

z=1500

z=400

z=1000

(100,200)

(150,100)

(150,0)

0,200)

x1

x2

Maximize z=10x1

+5x2Subject to the constraints

2x1+x2400

x1 150

x2200

x1,x20

z maximum

=2000 atz=600


24/28

1. Thus we see that the objective function z is

maximum at the corner (150,200) and also

has an alternative optimum solution at thecorner (100,200).

2. It may also be noted that z is maximum at

each point of the line segment joiningthem.

3. Thus the problem has an infinite number of

(finite) optimum solutions.4. This happens when the objective function

is parallel to one of the constraint

equations.


25/28

Maximize z = 5x1+ 7x2

Subject to2x1- x2 -1

-x1+ 2x2 -1x1,x20

No feasible solution


26/28

Maximize z =x1+x2

Subject to

-x1+ 3x2 30

- 3x1+ x2 30

x1,x20

unbounded solution z=20

z=30

z=50

z=70


27/28

Multiple period production Inventory model

A Production Planning Problem Acme ManufacturingCompany has received a contract to deliver home windowsover the next 6 months. The successive demands for the sixperiods are 100, 250, 190, 140, 220, and 110, respectively.Production cost for window varies from month to monthdepending on labor, material, and utility costs. Acmeestimates the production cost per window over the next 6

months to be $50, $45, $55, $48, $52, and $50,respectively. To take advantage of the fluctuations inmanufacturing cost, Acme may elect to produce more thanis needed in a given month and hold the excess units fordelivery in later months. This, however, will incur storage

costs at the rate of $8 per window per month assessed onend-of-month inventory. Develop a LP to determine anoptimum production schedule for Acme.


28/28

Questions/Queries?

Documents

Lect Class 4