Stochastic Programming with Cauchy Distribution

(1)

Stochastic Programming with Cauchy Distribution

Manas Kumar Pal

Institute of Management & Information Science Bhubaneswar, Odisha, India

S. Bagh

Department of Statistics, Sambalpur University Orissa, India

Abstract

The aim of this paper is to derive a method for solving a stochastic linear programming problem with Cauchy distribution. Assuming that the coefficients are distributed as Cauchy random variables, the stochastic linear programming is converted to a deterministic non-linear programming problem by a suitable transformation. Then an algorithm can be used to solve the resulting deterministic problem .A numerical example can be considered to illustrate the above methodology.

Keywords and phrases: Stochastic programming, Cauchy random variables, Probability density functions, Levy inversion theorem.

Introduction

Charnes and Cooper [6] first introduced the stochastic programming by taking different objective functions and constraints. Various models have been suggested by several researchers and most of the probabilistic model assumes normal distribution for model coefficients.

Here we consider a stochastic programming problem with Cauchy distribution. A stochastic linear programming problem can be presented as follows:

Max





 n

j j jx c Z

1

, (1)

Subject to the constraints,

m i

b x a

ob _i

n j

i j

ij 1 , 1,2,..., Pr

1

 

    

 







 (2)

, ., ,... 2 , 1 ,

0 j n

x_j  

(2)

First we obtain the probability density function of the linear combination of ‘n’ independent Cauchy random variables, then using the probability density function, the probabilistic constraints are converted to the deterministic constraints, and then the resulting non-linear deterministic problem can be solved.

Probability distribution of





 n

i j ij

i a x

U 1

and the deterministic form of the probabilistic

constraint

Here we will find out the probability density function of random variable





 n

j j ij

i a x

U

1

,

where aij, j=1, 2……… n are independent Cauchy random variables and xj, j=1, 2,……,n are variables .Then using it we find a deterministic form of the probabilistic constraints in a stochastic linear programming problem. The density function of Ui can be obtained by studying the special cases with n=2 and n=3. Let us consider a model involving only two random variables, the ith probabilistic constraints can be stated by



a x a x



b i m

ob i i i 1 i, 1,2,...,

Pr 1 1 2 2    

Where ai1 and ai2 are two independent Cauchy random variables Let Ui = ai1x1 + ai2x2

Then the density function of Ui can be obtained as,

 

U e t dt

f u

itu ) ( 2

1





 



 

 _{(Using Levy inversion theorem)}



e itue x x tdt 

 

  

 1 2

2 1



    _

  

 







   

 

 

dt e

e dt e

e itu x x t itu x x t 0

0

2 1 2

1

2 1



    _

  

 







  



   

dt e

e dt e

eitu x x t itu x x t 0

0

2 1 2

1

2 1



   





















 



   

dt

e

dt

e

itu x x t itu x x t 0

0

2 1 2

1

2

1



    _

  

 



 



  



   

dt e

dt

e tx x iu t x x iu 0

0

2 1 2

1

2 1



 





 





 

  

 

   

 

       

0 2

1 2

1

2 1 2

1

2 1

iu x x e iu

x x

e t x x iu t x x iu





 



_

  

 

     

iu x x iu x x

1 1

2 1

(3)

Stochastic Programming with Cauchy Distribution















  

 

 



 ₂ ₂ ₂

2 1

2 1 2 2

1

u i x x

x x

















  

 

 

 

2 2 2 1

2 1

u x x

x x



(4)

Hence ith probabilistic constraints can be made deterministic constraints by integrating the probability density function of ui as stated below,

 

u du i m

f _i _i

b i i

..., ,... 2 , 1 1

0

 







This can be simplified as follows,

 













i

b

i i

b

i du

u x x

x x du

u f

i i



 _  _

0

2 2 2 1

2 1

0

1

 (5)

If u_i a_i₁x₁a_i₂x₂ Then

1 2 2 1

i i i

a x a u

x   and

2 1 1 2

i i i

a x a u

x  

Now

2 1 1 1

2 2 2

1

i i i i

i i

a x a u a

x a u x

x     









2 1

2 2 2 1 2 1 2 1

i i

i i i

a a

x a x a a a

u   



Let

2 1

2 2 2 1 2 1 2

1 2 1

i i

a a

x a x a B and a

a a a

A   

Then, x1x2  Aui B p(Say) Now putting the above values in (5)

 





i b

i i

b

i du

u B Au

B Au du

u f

i i



 _  _

0

2 2 0

1



dp

A B p p A

p

B Ab_i



 

    

  

   

    

B

2 2

1







    

  

   

    

B Ab_i

A B A p p A

pdp B

2 2

1







    

  

                           

B Ab_i

A B A p A

B A

p p A

pdp B

2 1

2 2

2

(4)





   

 

           

   

B Ab_i

A B p A

B p

A A

pdp B

2 2 2

2 2

2 1

1 1







   

 

           



 



B Abi

A B p A

B p

A A A

pdp B

2 2 2

2 1

1







   

 

           



 



B Ab_i

A B p A

B p

A A

pdp B

2 2

2

2 1

1



Let R=a+bt+ct2, where

A B a and A

B b A A c

2 2

2 ,

1 _ _



 then,



 

    

R dt c b R c R tdt and B

b ac

2 ln 2

1 0

4

4 2 2 (c.f [42] , Page 68)

    

 

    

 

   

 

           



 

   



 

          

 

           



 

      

      

   



 







B Abi

A B p A

B p

A A

dp

A A

A B

A B p A

B p

A A

A

A B

2 2

2

2 1

1 2

2 2

1 ln

1 2

1 1



B Ab

B

i

B

p A A A

B

arctg B

A B A

B p A

B p

A A



    

 

    

 

   



 



     

 

           



 

      

      

   



 



2 2

2

4 1 2

2

4 2 1 2

1 ln

1 2

1 1



Because 2 2 , 0

  

 

  





arctg b ct if

R dt

(c.f [42], Page 68)

B Ab

B

i

p AB A A arctg A

A B p A

B p

A A



    

 

    

 

   

 

   



 

 

    

 

           



 

      

      

   



 

 1 1

1 1 2

1 ln

1 2

1

1 2

2 2

2















    

 

    

 

   

 

     

 

 

    

 

              

 

      

      

    

 

 Ab B

AB A A arctg A

A B B Ab A

B B

Ab A A

A

A i i i

1 1

1 1 2

1 ln

1 2

1

1 2

2 2 2

2

(5)

Stochastic Programming with Cauchy Distribution

 

                                                                            

 AB B

A A arctg A A B B A B B A A A A 1 1 1 1 2 1 ln 1 2 1 1 2 2 2 2 2 2 

















                                                                                 AB B AB B A A arctg A A B A B A B A B A A A AB B Ab B A b A A arctg A A B A B A B Ab A B Ab B b A A A A i i i i i 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 1 1 1 2 ln 1 2 1 1 1 1 2 2 2 1 ln 1 2 1  





                                                 A A B b B b A A arctg A A B B b A B Ab B b A B b A B A b A A A i i i i i i i 1 1 1 1 2 2 2 ln 1 2 1 2 2 2 2 2 2 3 2 2 2 4 2 



 







               

 arctg A

A AB A A 1 1 ln 1 2 1 2 2 2 





 

                                                              

 A arctg A

B b B b A arctg A AB A B B b B b A B Ab B b A AB b A A A i i i i i i i 2 2 2 2 2 2 2 2 2 3 2 1 1 ln 2 2 2 ln 1 2 1 









 

                                                     

 A arctg A

B b B b A arctg A AB AB B b A Ab A A A i i i i 2 2 2 2 2 2 2 2 1 1 2 1 ln 1 2 1 













 

                                          

 B A arctg A

b B b A arctg A AB Ab A AB Bb A AB Ab A A

A i i i i i

(6)









 

  

 

  

 

   

 

 

   



 _ _

     

 

 

   



 



 B A A arctg A

b arctg A

B Ab b

B A A

A i i

i 1

1 1 1

2 1

ln 1 2

1 ₂ 2 ₂ 2

2 2



On substituting the values of A and B,

 

i

b

i du u f i



0

=

             

 

             

 

       

       

   

 

  

     

 

     

 

   

 

     

 



    

  

    

 

 

   

 

 

      

      

    

 

    

 

 

   

 

     

 



    

  

    

 

    

 



2 1 2

1 1

2 2 2 1 1

2

2 1 2

2 1

1 2 2 2 1 1

2 1 2

2

1 2 2 2 1 1

2

2 1 2

2 1

1 1 1

1 1 1 1

1 1 1

1

1 1

1 2 1

1 1 ln 1 1 1 2

1 1

1

i i i

i i

i i i i

i i

i i i

i

i i i i

i i i

i i

a a arctg a

a

a a x a a x

a a b

arctg

a a

a a x a a x

b a a b

a a x a a x

a a



(6)

Proceeding as in above for the cases of three variables, the ith probabilistic constraints can be presented as,



a x a x a x



b i m

ob i i i i 1 i, 1,2,...,

Pr 1 1 2 2 3 3    

Where ai1, ai2 and ai3 are three independent Cauchy random variables

3 3 2 2 1

1x a x a x

a U

Let i  i  i  i

Then the density function of Ui can be obtained as,

 

U e t dt

f itu _u( )

2 1





 



 

 _{(Using Levy inversion theorem )}



e

itu

e

x x x t

dt



 

   



1 2 3

2

1



    _

  

 







    

 

  

dt e

e dt e

e itu x x x t itu x x x t 0

0

3 2 1 3

2 1



    _

  

 



 



  



    

dt e

e dt e

eitu x x x t itu x x x t 0

0

3 2 1 3

2 1



    _

  

 







    



   

dt e

e dt e

eitu x x x t itu x x x t 0

0

3 2 1 3

2 1



    _

  

 



 



   



    

dt e

dt

e t x1 x2 x3 iu t x1 x2 x3 iu

2 1

(7)

Stochastic Programming with Cauchy Distribution

 





 





 

   

  

   

 

         

0 3

2 1 3

2 1

3 2 1 3

2 1

iu x x x

e iu

x x x

e t x x x iu t x x x iu





 



_

  

 

       

iu x x x iu x x

x₁ ₂ ₃ ₁ ₂ ₃

1 1

2 1

















  

 

  

 

 ₂ ₂ ₂

3 2 1

3 2 1 2 2

1

u i x x x

x x x







_

  

 

  

  

2 2 3 2 1

3 2 1

u x x x

x x x

 (7)

Hence ith probabilistic constraints can be made deterministic constraints by integrating the probability density function of ui as stated below,

 

u du i m

f _i _i

b i i

..., ,... 2 , 1 1

0

 







This can be simplified as follows,

 













i

b

i i

b

i du

u x x x

x x x du

u f

i i



 _ _  _

0

2 2 3 2 1

3 2 1

0

1



(8)

If u_i a_i₁x₁a_i₂x₂a_i₃x₃ Then

1 3 3 2 2 1

i i i

i

a

x a x a u

x    ,

2 3 3 1 1 2

i i i i

a

x a x a u

x   

and

3 2 2 1 1 3

i i i i

a

x a x a u

x   

Now

3

2 2 1 1 2

3 3 1 1 1

3 3 2 2 3

2 1

i i i

i i

i

a

x a x a u a

x a x a u x x

x           

   

 

   

 

 

   

 

 

   

 

 

   

 

  

2 3 1 3 1 1 2 3 2 2 3 1 2 1 1 3

2 1

1 1 1

i i i i i

i i i i

i i i

a a a a x a a a a x a a a a x a

a a

u

Let

   

 

   

 

 

   

 

 

   

 

 

  

2 3

1 3 1 1 2

3 2 2 3 1

2 1 1

3 2 1

i i i i i

i i i i

i i

i i i i

i i i

a a a a x a

a a a x a

a a a x B and

a a a

a a a A

Then, x₁  x₂ x₃  Au_i B p(Say)

Now putting the above values in (8)

 

b



i



_i i

b

i du

u B Au

B Au du

u f

i i



 1 _ ₂ _ ₂

(8)

                                                                                      _ _                                          _        _        _                _ _                                                                                              _        _        _        _ _                           3 2 1 3 2 1 2 3 1 3 1 1 2 3 2 2 3 1 2 1 1 2 3 2 1 2 3 2 1 2 3 1 3 1 1 2 3 2 2 3 1 2 1 1 3 2 1 2 2 2 3 1 3 1 1 2 3 2 2 3 1 2 1 1 2 3 2 1 2 3 2 1 3 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 2 1 1 1 1 ln 1 1 1 1 2 1 1 1 1 i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i i a a a arctg a a a a a a a x a a a a x a a a a x a a a b arctg a a a a a a a x a a a a x a a a a x b a a a b a a a a x a a a a x a a a a x a a a a a a a a a  (9)

Finally we generalize the results to ‘n’ independent random variables and the result may be stated as follows



a x a x a x



b i m

ob i i ... in n i 1 i, 1,2,...,

Pr 1 1 2 2     

Where ai1 ,ai2 ,………,ain are ‘n’ independent Cauchy random variables Let Ui = ai1x1 + ai2x2 +………+ ainxn

 





i

b i i i i b i du u B Au B Au du u f Then i i



 _  _ 0 2 2 0 1 ,                                                                                                                                                                                                                              















                 n j ij n j ij n j k k n j ik ij j n j ij i n j ij n j k k n j ik ij j i n j ij i n j k k n j ik ij j n j ij n j ij n j ij a arctg a a a x a b arctg a a a x b a b a a x a a a 1 1 1 1 2 1 2 1 1 1 1 2 2 1 1 2 1 2 1 1 1 1 1 1 1 1 1 1 1 2 1 1 ln 1 1 2 1 1

(9)

Stochastic Programming with Cauchy Distribution

Deterministic model of stochastic linear programming

Assuming aij to be independent Cauchy random variables, the stochastic linear programming model as stated in (1), (2) and (3) can be converted to a deterministic linear programming problem as

Maximize

 



 



 n

j

j j x c E Z

E

1

(11)

i

n

j ij n

j k k

n

j ik ij j n

j ij i

n

j ij

n

j k k

n

j ik ij j

i n

j ij i

n

j k k

n

j ik ij j n

j ij

n

j ij n

j ij

a arctg

a

a a x a b

arctg

a

a a x

b a b

a a x a

a a



  

                 

 

                 

 

   

   

 

   

   

 

         

      

 

      

 

        

  

 

  

 

    

  

        

         

   

   

 

   

   

 



  

 

  

 

       



  

 

  

 

        

   

   

        

       

















 



 





 





 



 

1

1 1

1

1 1

2 1

1 ln

1 1 2

1

1 1

1 1 2

1 2

1

1 1 1 2

2

1 1 2

1 2

1 1

(12)

x_j 0 ,j1,2,...,n (13)

Then we present a numerical example to illustrate the methodology.

Let, a11=5, a12=4, a13=8, a21=10, a22=2, a23=20 b1=10, b2=20, 1 0.05and 2 0.1

Maximize, Z=5x₁+6x₂+3x3



 





 







05 . 0 575 . 0 575

. 0 6

. 3 3 . 1 875 . 1

0625 . 133 330625

. 1

1

1 6 . 3 3 . 1 875 . 1

5 . 11 6

. 3 3 . 1 875 . 1

0625 . 133 ln

66125 . 2

575 . 0 1

3 2

1

3 2

1 2

3 2

1 _

    

 

    

 

   

 

 

 

 

  

  

    

  

 

 

 

arctg x

x x arctg

x x













 







01 . 0 65

. 0 65

. 0 12

3 . 0 5 . 5

45 . 28 4225

. 1

1

1 12 3

. 0 5 . 5

26 12

3 . 0 5 . 5

9 . 56 ln

845 . 2

65 . 0 1

3 2

1

3 2

1 2

3 2

1 _

    

 

    

 

   

 

 

 

 

  

  

   

 

 

 

 

arctg x

x x

arctg

x x



(10)

By using the Genetic Programming we found the optimum solution for the above program is

 

23.886 727

. 1 ,

748 . 1 ,

741 .

1 ₂ ₃

1  x  x  and EZ 

x

References

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