EML 4550 - Spring’08 EML4552 - Engineering Design Systems II (Senior Design Project) Optimization Theory and Optimum Design Linear Programming Hyman: Chapter

EML 4550 - Spring’08

EML4552 - Engineering Design Systems II(Senior Design Project)

Optimization Theory andOptimum DesignLinear Programming

Hyman: Chapter 10


Optimization

Minimize (Maximize) an Objective Function of certain Variables subject to Constraints

0),...,(

...

0),...,(

0),...,(

),...,(min

21

212

211

21

nm

n

n

n

xxxg

xxxg

xxxg

xxxf


Linear Programming

Maximize a Linear Objective Function of i positive variables subject to j Linear Constraints

0

,...,2,1

),...,(max

1

121

i

n

ijiij

n

iiin

x

mj

rxa

xkxxxU


Linear Programming

Almost all linear optimization problems can be reduced to the canonical (standard) form Minimization, change sign on objective

function Variables that can be negative, define:

Constraint reversal, ‘slack’ and ‘surplus’ variables

0,

ii

iii

xx

xxx


L.P.: Geometric Interpretation

0

0

)4(....................5

)3..(..........4

)2(..........42

)1(..........93

2),(max

2

1

2

21

12

21

2121

x

x

x

xx

xx

xx

xxxxU


L.P.: Geometric Interpretation

x1

x2(1)

(2)

(3)

(4)

U= 0 2 4 6.5


L.P.: The “Simplex” Algorithm

Standard method to solve linear programming optimization problems

Geometric interpretation of linear programming problems (simple example)

Based on principle that optimum will be ON a constraint (typically a “corner”)


Simplex Algorithm: Example

0,,

4204

46023

4302

523max

321

21

31

321

321

xxx

xx

xx

xxx

xxxU



VariablesSlackxxx

xxxxxx

xxx

xxx

xxxx

xxxU

_"",,

0,,,,,

4204

46023

4302

523max

654

654321

621

531

4321

321

Add x4 to make the Inequality an equality



The Simplex “Algorithm” consists of a number of steps carried out in ‘cycles’ until an optimum is reached

“Geometrically”, the Simplex method explores the objective function by ‘walking’ along the constraints in the ‘steepest’ direction until it reaches the optimum

Based on: Unique solution Optimum on a constraint



Cycle 1 Select a starting point (a ‘corner’) In this example a corner is set by 3 variables (the

other 3 can be solved from the constraints)

Step 1: Select the 3 ‘basis’ variables such that they appear in only one constraint and have coefficient of unity (x4, x5, x6 in this case). Assign a value of 0 to the ‘non-basis’ variables.

Step 2:Determine coordinates of initial ‘corner’ (x1=0, x2=0, x3=0, x4=430, x5=460, x6=420)



Step 3: Calculate objective function at the corner point (x1=0, x2=0, x3=0, x4=430, x5=460, x6=420), U=0

Step 4: Move from corner to an ‘adjacent’ corner along the ‘edges’ represented by the constraints. Vary the non-basis variables to move along the edges. Move along the edge which will increase “U” the most. To do this, express the objective function in terms of the non-basis (U=3x1+2x2+5x3, already, if not, solve the constraints to do so)



Step 5: From the present ‘corner’ (x1=0, x2=0, x3=0) move along the variable that will produce the largest increase in U (highest coefficient), in this case x3. We re-formulate the basis by introducing x3 and deleting from the basis one of the variables.

Step 6: We want to increase x3 by as much as possible without violating any constraint (i.e., move to the next ‘corner’). To do this, reformulate the constraints for the ‘incoming’ variable to the basis



Step 6 (cont.): In this case for x1=x2=0 and the incoming x3, the constraints are:

Step 7: Determine how much x3 can increase before driving any variable negative. Limits are: (a) 430, (b) 230, (c) therefore, x3 can be increased up to 230 and x5 leaves the basis

).......(..........420

)........(4602

)(..........430

6

53

43

cx

bxx

axx



Step 8: x3, x4, and x6 are the new basis, set the non-basis variables to zero, x1=0, x2=0, x5=0

Prepare to start the next cycle (repeat procedure)



Cycle 2Step 1: The new basis was chosen as the last

step of the previous cycle (x3, x4, x6).Step 2:Determine coordinates of new ‘corner’

by re-writing constraints:

(x1=0, x2=0, x3=230, x4=200, x5=0, x6=420)

).......(..........420

)......(..........4602

)(..........430

6

3

43

cx

bx

axx



Step 3: Calculate objective function at the corner point (x1=0, x2=0, x3=230, x4=200, x5=0, x6=420), U=3(0)+2(0)+5(230)=1150

Step 4: Prepare to choose new basis. Re-write U in terms of non-basis variables (using the constraints to eliminate the basis variables): 521 )2/5(2)2/9(1150 xxxU



Step 5: Move away from present corner along the edge that will increase U the most. From the objective function, only an increase in x2 will yield improvement (all other coefficients are negative). Introduce x2 to the basis.

Step 6: Re-write constraints in terms of new basis variable:

).(..........4204

).........(..........4602

)...(4302

62

3

432

cxx

bx

axxx



Step 7: Determine how much x2 can increase before driving any variable negative. Limits are: (a) 100 (for x3=230),

(b) , (c) 105 therefore, x2 can be increased up to 100 and x4 leaves the basis

Step 8: x2, x3, and x6 are the new basis, set the non-basis variables to zero, x1=0, x4=0, x5=0

Prepare to start the next cycle (repeat procedure)



Cycle 3Repeat the same steps now with x2, x3, x6

as the basis. After steps 2 and 3 the solution is x1=0, x2=100, x3=230, x4=0, x5=0, x6=20, U=1350 (further improvement). At step 4, however, all coefficients on the objective function, U, are negative, so no further improvements are possible, we are done! We have found the optimum (x1=0, x2=100, x3=230, U=1350).



Simplex Algorithm

Found the optimum with exploration of just 3 ‘corners’

Efficiency increases tremendously with the complexity of the problem (number of variables and constraints)

“New” L.P. algorithms may be even more efficient in extremely large problems

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EML 4550 - Spring’08 EML4552 - Engineering Design Systems II (Senior Design Project) Optimization Theory and Optimum Design Linear Programming Hyman: Chapter