Geometric Programming Geo Raju Mnit Jaipur

7/30/2019 Geometric Programming Geo Raju Mnit Jaipur

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Presented by :MANOJ MEENA (2012PMM5088)GEO RAJU (2012PMM5114)



Let 1, … , denote real positive variables, and ( 1, … , ) a vector withcomponents i.

A real valued function ( ) , with the form … , where >0 and ∈ , is called a monomial function.e.g : 2.3 − . 5

A sum of one or more monomials, i.e., a function of the form …= ,

where > 0 , is called a posynomial function. The term„posynomial‟ is meant tosuggest a combinationof „positive‟ and „polynomial‟. e.g : 0.23 + / , 2 1 + , 2 + 3 + 2

MONOMIAL AND POSYNOMIAL FUNCTIONS

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A geometric program(GP) is an optimization problem of the form ( )

( ) ≤ 1, 1, … , , ( ) 1, 1, … , ,

o where are posynomial functions , are monomials, and are theoptimization variables.Standard form GP is often used in network resource allocation problems. As an example, consider the problem

− − − + 2.3 + 4

( ) − − + − ≤ 1+ 2 + 3 ≤ 1,

( ) 1 o with variables , and . This is a GP in standard form, with n=3 variables, m=2

inequality constraints, and p=1 equality constraints.

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GP in standard form is not a convex optimization problem , because posynomialsare not convex functions.The main trick to solving a GP efficiently is to convert it to a nonlinear but conveoptimization problem, i.e., a problem with convex objective and inequality

constraint functions, and linear equality constraints.The conversion of a GP to a convex problem is based on a logarithmic change of variables(log ; ), and a logarithmic transformation of the objectiveand constraint functions.

log ( ) log ( ) ≤ 0 , 1, … , ,

log ( ) 0, 1, … , ,with variables , … , . Convex form GP is used in problems based on stochastic models such asinformation theoretic problems.In convex form, GP with only monomials reduces to LP.

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Extension 1 : Maximize a monomial subject to posynomial upper bound inequalityconstraints:

( ) ( ) ≤ 1, 1, … , ,

( ) 1, 1, … , ,

Maximizing a monomial is equivalent to minimizing its reciprocal, which is anothermonomial:

( )

( ) ≤ 1, 1, … , , ( ) 1, 1, … , ,

where are posynomial functions , are monomials, and are the optimizationvariables.

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Extension 2 : The right hand side of posynomial inequality and monomialequality constraints can be monomials instead of 1: ( )

( ) ≤ ( ), 1, … , , ( ) ( ), 1, … , ,

By dividing the right hand side monomial on both sides of the constraints,the above problem can be transformed into:

( )

( ) ≤ 1, 1, … , , ( )( )

1, 1, … , ,

where are posynomial functions , & are monomials, and are theoptimization variables.

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Example :Optimize the shape of a box-shaped structure with height , width , and depth .There is limit on the total wall area 2( + ) , and the floor area as well aslower and upper bounds on the aspect ratios / and / . Subject to theseconstraints, maximize the volume of the structure, .

SOLUTION :Here , , and are the optimization variables, and the problem parametersare (the limit on wall area), (the limit on floor area), and , , , (the lower andupper limits on the wall and floor aspect ratios).This leads to the problem :

2 + ≤ , ≤ ,

≤ / ≤ , ≤ / ≤ . It can be transformed to thestandard form GP .

− − − 2 + 2 ≤ 1, 1 ≤ 1,

− ≤ 1, 1 − ≤ 1, −

≤ 1, 1−

≤ 1. 8



Consider the unconstrained minimization problem:

that minimizes the objective function

where c j > 0, x i > 0 & aij are real constants (like +ve, zero, -ve).Two approaches

i. based on differential calculusii. based on the concept of geometric inequality

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SOLUTION OF UNCONSTRAINED GP USING

DIFFERENTIAL CALCULUS

• According to the differential calculus methods the necessary conditions for the minimumof are given by

• By multiplying Eq. (4) by , we can rewrite it as

(5)10



• Let ∗

∗

∗

.

.∗

represents the minimizing vector, then Eq.(5) can be written as

• After dividing by the minimum value of the objective function ∗ , Eq. (6) becomes

where

and denote the relative contribution of j th term to the optimal objective function.• From eq. (8)

• Equations (7) is called theorthogonality condition and Eq. (9) is called thenormality condition.11



• To obtain the minimum value of the objective function ∗, the following procedure canbe adopted. Consider

• Since

• From eq. (11) & (10)

• Substituting the following relation in above eq.

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• Eq. (12) becomes

(∵ ∆ ∗

= ,∀ from eq.(7))

Degree of Difficulty :The quantity N −n − 1 is termed adegree of difficulty in geometric programming.In the case of a constrained GP , N denotes the total number of terms in all the posynomials andn represents the number of design variables.If N − n − 1 = 0 , the problem is said to have azero degree of difficulty

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Finding the Optimal Values of Design Variables :• Since ∗and ∆∗(j =1, 2, . . . ,N) are known, the optimal values of the design variables

can be determined from the relations.

• The simultaneous solution of these equations will yield the desired quantities x ∗ i (i =1, 2, .

. . , n). To simplify the simultaneous solution of Eq. (14), we rewrite them as

• Taking logarithms on both the sides of Eq. (15), we obtain

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• By letting

• Eq. (16) can be written as

• Once are found, the desired solution can be obtained as

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Example :It has been decided to shift grain from a warehouse to a factory in an open rectangular box of length x 1 meters, width x 2 meters, and height x 3 meters. The bottom, sides, and the ends of thebox cost, respectively, $80, $10, and $20/m 2. It costs $1 for each round trip of the box.

Assuming that the box will have no salvage value, find the minimum cost of transporting 80m3 of grain.

SOLUTION:The total cost of transportation is given bytotal cost = cost of box + cost of transportation

= (cost of sides + cost of bottom + cost of ends of the box) +

(number of round trips required for transporting the grain× cost of each round trip)

where 1, 2, and 3 indicate the dimensions of the box,as shown in figure.

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Comparing Eq. (E1) with the general posynomials, we obtainc 1 = 80, c 2 = 40, c 3 = 20, c 4 = 80

The orthogonality and normality conditions are given by

that is

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To find the optimal values of the design variables, using Eq. (14) as

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Example :Determine the optimal pipe diameter for the minimum installed plus operatingcosts by geometric programming for 100 ft. of pipe conveying a given flow

rate of water.The installed cost in dollars is 150 D and the lifetime pumping cost in dollarsis 122,500/D 5. The diameter D is in inches.Solution:

minimize: y = 150D + 122,500/D5

N = 2, n = 1 ; N - (n+1) = 0 degree of difficulty∆1+ ∆2 = 1 normality condition∆1 – 5∆2 = 0 orthogonality conditionSolve simultaneously:∆1 = 5/6 ; ∆2 = 1/6We know that



y* = $719.62

Solve for D:

∆1 = 150D / 719.62 = 5/6D = 4.0 inches



SOLUTION OF UNCONSTRAINED GP USING

• The arithmetic mean –geometric mean inequality is given by :...............(20)

with ..…….......(21)

• Using the inequality of (20), the objective function can be written as (by setting∆ , 1,2, … , )

. ……....……….(22)

• The left-hand side of the inequality (22) is called the primal functionand the right handside of the inequality (22) is called the predual function.

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• By using

the predual function can be expressed as ,

……..(24)

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• If we select the weights∆ so as to satisfy the normalization condition, Eq. (21), and also

the orthogonality relations

.……(25)• Therefore, Eq. (24) reduces to

…… .…...(26)

… …….(27)

• In this inequality, the right side is called the dual function ,(∆ , ∆ , … , ∆ ).Theinequality (27) can be written simply as

≥ … … (28) A basic result is that the maximum of the dual function equals the minimum of the primalfunction.

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PRIMAL – DUAL RELATIONSHIP AND SUFFICIENCY CONDITIONS IN THE UNCONSTRAINEDCASE :-

• If f ∗ indicates the minimum of the primal function andv ∗denotes the maximum of

the dual function, Eq. (28) states that…………(29)

• For convenience of notation, let us denote the objective functionf ( X ) by x 0 and make theexponential transformation.

………….(30)

where the variables are unrestricted in sign.• Define the new variables∆ , also termed weights, as

…………..(31)

which can be seen to be positive and satisfy the relation

......……….(32)

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• By taking logarithms on both sides of Eq. (13), we obtain

……………(33)

Or …………….(34)

• Thus the original problem of minimizing ( ) with no constraints can be replaced by one of minimizingw 0 subject to the equality constraints given by Eqs. (32) and (34).

• The objective function x 0 is given by as following.

………………(35)

• Since the exponential function ( ) is convex with respect tow i , the objective function x 0,which is a positive combination of exponential functions, is also convex. Hence there is only onestationary point for x 0 and it must be the global minimum.

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• The global minimum point of w 0 can be obtained by constructing the followingLagrangian function and finding its stationary point:

……….(36)

where

……….(37)

• Withλ denoting the vector of Lagrange multipliers. At the stationary point of L, wehave.

…………..(38)

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These equations yield the following relations:

Equations (39), (41), and (42) give the relation

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• Thus the values of the Lagrange multipliers are given by

………....( 45)

• By substituting Eq. (45) in Eq. (36), we obtain

……….…( 46)

• The function given in Eq. (46) can be considered as the Lagrangian functioncorresponding to a new optimization problem whose objective function is given by

…….…… (47)

• and the constraints by…..……….( 48)

……………(49)

• This problem will be the dual for the original problem. The quantities (1 − w 0),w 1 , w 2 , . . . ,w

n can be regarded as the Lagrange multipliers for the constraints given by Eqns. (48)

and (49). 29



• Now it is evident that the vector ∆ which makes the Lagrangian of Eq. (46) stationary willautomatically give a stationary point for that of Eq. (36). It can be proved that the function

is convex since∆ j is positive.• Since the function is given by the negative of a sum of convex functions, it will be a

concave function.• Hence the minimum of the original primal function is same as the maximum of the function given

by Eq. (47) subject to the normality and orthogonality conditions given by Eqs. (48) and (49) with

the variables∆

j constrained to be positive.• By substituting the optimal solution∆∗, the optimal value of the objective function becomes

• By taking the exponentials and using the transformation relation (30), we get

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Example :In a certain reservoir pump installation, the first cost of the pipe is givenby (100D + 50D 2), where D is the diameter of the pipe in centimeters. The cost of thereservoir decreases with an increase in the quantity of fluid handled and is given by

20/Q, where Q is the rate at which the fluid is handled (cubic meters per second).The pumping cost is given by (300Q 2 /D5). Find the optimal size of the pipe and theamount of fluid handled for minimum overall cost.

SOLUTION :

We can see that

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• Since N > (n + 1), these equations do not yield the required ∆ j (j = 1 to 4) directly. But anythree of the ∆ j‟s can be expressed in terms of the remaining one.

• By solving

• The dual problem can now be written as

• The orthogonality and normality conditions are given by

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• Since the maximization of is equivalent to the maximization of , we will maximize for convenience

• Since is expressed as a function of ∆ 4 alone, the value of ∆ 4 that maximizes must beunique (because the primal problem has a unique solution).

• The necessary condition for the maximum of is

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After simplification

i.e.,

from which the value of ∆∗4 can be obtained by using a trial-and-error process as

follows:

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we find that ∆∗4 ≃ 0.147 Eqn. (E2) gives

The optimal value of the objective function is given by

The optimum values of the design variables can be found from

(E4)

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From Eqs. (E1) and (E4), we have

These equations can be solved to find the desired solutionD∗= 0.922 cm, Q∗= 0.281m3/s.

Here D = dia. of pipe in cm

Q = is the rate at which the fluid is handled in m3

/sec.

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MOSEK : MOSEK is a software package for the solution of linear, mixed-integer linear, quadratic, mixed-integer quadratic, quadratically constraint, conic andconvex nonlinear mathematical optimization problems.

TOMLAB : The TOMLAB Optimization Environment is a powerful optimizationplatform and modeling language for solving applied optimization problems inMATLAB.GGPLAB : GGPLAB is a MATLAB-based toolbox for specifying and solving

geometric programs (GPs) and generalized geometric programs (GGPs).YALMIP : A free MATLAB toolbox for rapid prototyping of optimization problemIt have a simple interface that recognizes some GGPs, and automatically formsand solves the resulting GPs.

SOFTWARES

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Early applications of geometric programming include nonlinear network flow problems,optimal control, optimal location problems, and chemical equilibrium problems.Geometric programming has been used in a variety of problems in digital circuit design.Environment quality control.

Resource allocation in communication and network systems.Information theory.Probability and statistics.Structural design .Computer system architecture design.Inventory control.Production system optimization.Computational finance.Systems and control theory.Geometric modeling.

APPLICATIONS

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S. Boyd, S.-J. Kim, L. Vandenberghe, and A. Hassibi,“ATutorial onGeometric Programming,” Optimizationand Engineering, 8(1):67-127,2007.

M. Chiang, “Geometricprogramming for communicationsystems,”Foundations and Trends in Communications and Information Theory,2005.Singiresu S.R, “EngineeringOptimization: Theory andPractice,” Fourth

Edition,2009

REFERENCES

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Geometric Programming Geo Raju Mnit Jaipur

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