# View of Optimization of Fuzzy Inventory Model with Deterioration Using Non Linear Programming Methods

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### Abstract

This paper presents an integrated inventory model with deterioration rate. The deterioration rate is playing a vital role in inventory systems. Here, we determine the minimize the total cost and maximize the optimum order time interval using non linear programming method, first we optimized the inventory model using fuzzy geometric programming method; secondly we apply the Lagrangian method for optimization. In both the methods, we use heptagonal fuzzy number for fuzzification and Pascal’s Triangular Graded Mean for defuzzification, Finally, numerical examples, comparative study, sensitivity analysis, and graph are illustrated

### Keywords:

Geometric programming, Lagrangian method, Pascal’s Triangular Graded Mean Method.

### 1. Introduction

An inventory management modeling exercise, demand can be represented as either deterministic or stochastic. In earlier period the uncertainties of inventory models are treated as randomness and are handled by using probability theory. Ya Yanga,c, Huihui Chib, Wei Zhoub,d, Tijun Fana, Selwyn Piramuthue[1],developed Deterioration control decision support for perishable inventory management systems .

L.A. Zadeh,[2] ,introduced fuzzy sets ,operations and applications, Chen and Wang[11] used trapezoidal fuzzy number to fuzzify the order cost, A.Mohammed Shapique[3],derived , Arithmetic Operations on heptagonal fuzzy numbers ,

Many Researches solved Integrated inventory models by Non linear programming methods.

Peterson E. L, Duffin R. J,Zener C. M[6] (1967) is discovered and developed the Geometric Programming Theory and Application, Klain and Jung [7] derived single item inventory problems using geometric problems.K.Kalaiarasi,M.Sumathi,M.Sabina Begum [8],Optimized the a inventory model Using Fuzzy Geometric Programming method .

Chen’s [8] function principle is proposed for arithmetic operation of fuzzy number and Lagrangian method is used for optimization. Graded mean integration is used for defuzzifying the annual integrated total cost for EPQ.

In this paper, the Mathematical model is formulated in crisp and fuzzy environment .A numerical example is encapsulating the solution in crisp and fuzzy environment. we analyse the results. Finally, we conclude this result.

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Ah

x a

fora x a a a

fora x a x a

fora x a a a

a x

x fora x a

a a

fora x a a x

fora x a a a

otherwise

   

 

  



   

 

 

   

  



   

 





### 2.2. Graphical Representation of Heptagonal Fuzzy Number

Fig 2.1 Graphical representation of Heptagonal fuzzy number

h

1

2

3

4

5

6

7

### be a heptaconal fuzzy numbers, Pascal’s triangles formula is

1 6 2 15 3 20 4 15 5 6 6 7

( ) 64

a a a a a a a

d A       

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C

M

C

### 3.2. Mathematical Model

Ya Yanga,c, Huihui Chib, [1] develope the following inventory model .let us consider the proposed inventory model of the minimum total cost is

( ) 2

C M C

O H D T TC TT

,T>0---(3.2.1)

The optimum T and 𝑇𝐶(T) can be obtained by TC 0

T

 

Optimal order quantity

C

M C

###  H D

---(3.2.2) Minimum total cost

C M C ---(3.2.3)

( ) 2

C M C

O H D T TC TT

,T>0---(4.1)

### Min

( ) 2

C M C

O H D T TC TT

---(4.2) T>0

Dual problem is Max

1 2

1 2

( ) 2

t t

C M C

O H D

f t t t

   

    

    ---(4.3)

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Here DD = 2-1-1=0

subject to the normality and orthogonality conditions

1 2

1 2

1 2

###   

---(4.4) Solve the equations (4.4) we get,

1 2

1 t  t 2

---(4..5)

Substitute the value of

1

### ,

2 in equation (4.3), then the dual function is given by,

1 1

2 2

*

*

C M C

C M C

   

   

   

---(4.6)

* * 1

* * 2

2

C

M C

C M C

---(4.7) Min

C M C ---(4.8)

### 4.1. Mathematical Model in fuzzy sence and the Geometric programming solution

Suppose O and DCC

are taken as a fuzzy heptagonal fuzzy numbers The primal problem is

Min

C M C

### 

---(4.1.1) T>0

Dual problem is Max

1 2

1 2

t t

C M C

###  

 

---(4.1.2)

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We use Pascal’s Triangular Graded Mean for Defuzzification ,

1 2 3 4 5 6 7

1 2 3 4 5 6 7

( 6 15 20 15 6 )

64

( 6 15 20 15 6 )

64

C C C C C C C

C

C C C C C C C

C

O O O O O O O

O

D D D D D D D

D

     

     

---(4.1.3) We apply GP method we get optimum order and total cost is

2 1 2 3 4 5 6 7

1 2 3 4 5 6 7

2( 6 15 20 15 6 )

( 6 15 20 15 6 )

C C C C C C C

M C C C C C C C

O O O O O O O

T H D D D D D D D

     

       ---(4.1.4)

## 

2 ( 1 6 2 15 3 20 4 15 5 6 6 7)( 1 6 2 15 3 20 4 15 5 6 6 7)

## 

### ( ) 1 64

HM OC OC OC OC OC OC OC DC DC DC DC DC DC DC

--- --(4.1.5)

### 5. Mathematical Model in Crisp sence and the Lagrange solution

let us consider the proposed inventory model of the minimum total cost is

( ) 2

C M C

O H D T TC TT

,T>0---(5.1)

To apply Pascal’s Triangular Graded Mean method to defuzzify the fuzzy total cost, and then obtain the optimal order quantity Tby using new arithmetic operation

1 1 2 2

3 3 4 4

5 5 6 6 7 7

### 2 2 2

C M C C M C

C M C C M C

C M C C M C C M C

###  

---(5.2)

Computation of T at which TC(T) is minimum, when

( ) 0 dTC T

dT

and where 2

2

### dT 

2 1 2 3 4 5 6 7

1 2 3 4 5 6 7

2( 6 15 20 15 6 )

( 6 15 20 15 6 )

C C C C C C C

M C C C C C C C

O O O O O O O

T H D D D D D D D

     

       ---(4.1.4)

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## 

2 ( 1 6 2 15 3 20 4 15 5 6 6 7)( 1 6 2 15 3 20 4 15 5 6 6 7)

## 

### ( ) 1 64

HM OC OC OC OC OC OC OC DC DC DC DC DC DC DC

### 5.1. Mathematical Model in Fuzzy sence and the Lagrange solution

let us consider the proposed inventory model of the minimum total cost is

( ) 2

C M C

O H D T TC TT

,T>0---(5.1)

To apply Pascal’s Triangular Graded Mean method to defuzzify the fuzzy total cost, and then obtain the optimal order quantity Tby using new arithmetic operation

1 1 1 2 2 2

7 6

3 3 3 4 4 4

5 4

5 5 5 6 6 6 7 7 7

3 2 1

2 6 2

( ) 1 15 20

64 2 2

15 6

2 2 2

C M C C M C

C M C C M C

C M C C M C C M C

O H D T O H D T

T T

O H D T O H D T

TC T T T

O H D T O H D T O H D T

T T T

    

  

    

   

 

     

 

        

     

         

 

   

 

 ---(5.2)

with

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5

6

### T

7 ,

If we replace inequality conditions

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4

5

6

### T

7 into the following inequality

1 2 3 4 5 6 7

2 1 3 2 4 3 5 4 6 5 7 6 1

0 .

0, 0, 0, 0, 0, 0, 0

T T T T T T T

T T T T T T T T T T T T T

      

            

In the following stages, extension of the Lagrangian method is used to find the solutions of

1 2 3 4 5 6 7

to minimize

 

##   T)

.

Stage 1 : To find the min

 

##   T)

,we have to find the derivative of

 

##   T)

with respect to

1 2 3 4 5 6 7

### , , , , , , T TT T T T T

. equal to zero

Let 1 2 3 4 5 6 7

0, 0, 0, 0, 0, 0, 0,

P P P P P P P

T T T T T T T

Then

7 6 5

1 2 3

1 2 3

4 3 2 1

4 5 6 7

4 5 6 7

C C C

M C M C M C

C C C C

M C M C M C M C

###    

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Because the above show that

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2

3

4

5

6

### T

7 it does not satisfy the constraint

1 2 3 4 5 6 7

### 0  T  T  T  T .    TTT

,so K = 1 and go to Stage 2.

Stage 2 : Convert the inequality constraint

2

1

### 0

into equality constraint T2 − T1 = 0. We have Lagrangian function as

1

2

3 4

5 6

7

 

##   )   T

2

1

### .

we have to find the derivative of L T T T T T

### 

1, 2, 3, 4, 5,T ,6 T7,

with respect to

1 2

3 4 5 6 7,

. equal to zero.

7 6

1 2

1 2

5 4 3 2 1

3 4 5 6 7

3 4 5 6 7

### , , , ,

C C

M C C

C C C C C

M C M C M C M C M C

###     

Because the above show that

3

4

5

6

### T

7 it does not satisfy the constraint

1 2 3 4 5 6 7

### 0  T  T  T  T .    TTT

,so K = 2 and go to Stage 3.

Stage 3 : Convert the inequality constraints

2

1

,

3

2

### 0

, into equality constraints T2−T1

= 0 and T3−T2 = 0. Then the Lagrangian method is

1

2

3 4

5 6 7

1 2

1

2 1

2

3 2

 

###   T  T   T  T

, we have to find the derivative of L T T T T T

### 

1, 2, 3, 4, 5,T ,6 T7,

1, 2

with respect to

1 2

3 4 5 6 7,

1, and

### 

2,and equal to zero,

7 6 5

1 2 3

1 2 3

4 3 2 1

4 5 6 7

4 5 6 7

C C C

M C C C

C C C C

M C M C M C M C

###    

Because the above show that

4

5

6

### T

7 it does not satisfy the constraint

1 2 3 4 5 6 7

### 0  T  T  T  T .    TTT

,so K = 3 and go to Stage 4.

Stage 4 : Convert the inequality constraints

2

1

,

3

2

,

4

3

### 0

,into equality constraints T2 − T1 = 0, T3 − T2 = 0 and T4 − T3 = 0. We optimize The Lagrangean function is given by

1

2

3 4

5 6 7

1 2

3

1

2 1

2

3 2

3

4 3

 

###   T  T   T  T   T  T

we have to find the derivative of L T T T T T

### 

1, 2, 3, 4, 5,T ,6 T7,

1, 2, 3

with respect to

1 2 3 4 5 6 7

### , , , , , , T TT T T T T

, ,and equal to zero,

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7 6 5 4

1 2 3 4

1 2 3 4

3 2 1

5 6 7

5 6 7

C C C C

M C C C C

C C C

M C M C M C

###   

Because the above show that

5

6

### T

7 it does not satisfy the constraint

1 2 3 4 5 6 7

### 0  T  T  T  T .    TTT

,so K = 4 and go to Stage 5.

Stage 5 : Convert the inequality constraints

2

1

,

3

2

,

4

3

,and

5

4

### 0

into equality constraints T2 − T1 = 0, T3 − T2 = 0 ,T4 − T3 = 0 and T5-T4=0. The Lagrangean function is given by

###  

1 2 3 4 5 6 7 1 2 3 4 1 2 1 2 3 2 3

4 5 4

4 3

###  

 

we have to find the derivative of L T T T T T

### 

1, 2, 3, 4, 5,T ,6 T7,

1, 2, 3, 4

with respect to

1 2 3 4 5 6 7

### , , , , , , T TT T T T T

, ,and equal to zero

7 6 5 4 3

1 2 3 4 5

1 2 3 4 5

2 1

6 7

6 7

C C C C C

M C C C C C

C C

M C M C

###  

Because the above show that

6

###  T

7 it does not satisfy the constraint

1 2 3 4 5 6 7

### 0  T  T  T  T .    TTT

,so K = 5 and go to Stage 6.

Stage 6 : Convert the inequality constraints

2

1

,

3

2

,

4

3

,

5

4

and

6 5

### T   T

into equality constraints T2 − T1 = 0, T3 − T2 = 0 ,T4 − T3 = 0, T5-T4=0 and T6-T5=0.

The Lagrangean function is given by

###    

1 2 3 4 5 6 7 1 2 3 4 5 1

4 5 4 5 6

2 1 2 3 2 3 4 3

5

 

### 

we have to find the derivative of L T T T T T

### 

1, 2, 3, 4, 5,T ,6 T7,

1, 2, 3, 4, 5

with respect to

1 2 3 4 5 6 7

### , , , , , , T TT T T T T

, ,and equal to zero

7 6 5 4 3 2

1 2 3 4 5 6

1 2 3 4 5 6

1 7

7

C C C C C C

M C C C C C C

C M C

### 

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Because the above show that

6

###  T

7 it does not satisfy the constraint

1 2 3 4 5 6 7

### 0  T  T  T  T .    TTT

,so K = 6 and go to Stage 7.

Stage 7 : Convert the inequality constraints

2

1

,

3

2

,

4

3

,

5

4

6 5

and

7

6

### 0

into equality constraints T2 − T1 = 0, T3 − T2 = 0 ,T4 − T3 = 0, T5-T4=0 ,T6-T5=0 and T7-T6=0. The Lagrangean function is given by

###      

1 2 3 4 5 6 7 1 2 3 4 5 6 1 2 1

4 5 4 5 6

2 3 2 3 4 3

6

5 7 6

###  

 

we have to find the derivative of L T T T T T

### 

1, 2, 3, 4, 5,T ,6 T7,

1, 2, 3, 4, 5, 6

with respect to

1 2 3 4 5 6 7

,

### 

6 ,and equal to zero

7 6 5 4 3 2 1

1 2 3 4 5 6 7

1 2 3 4 5 6 7

C C C C C C C

M C C C C C C C

7 6 5 4 3 2 1

1 2 3 4 5 6 7

C C C C C C C

M C C C C C C C

###       

6. Numerical Example Example:

Crisp Model:

Let us consider the following data:

OC= 50/- Per unit , DC = 84 unit/year, HM = Rs.15/- Per unit,

By Minimizing The Total Cost ,Using Geometric Programming And Lagrange Method,We Obtain The Optimal Value

T =0.2817 and TC=Rs.354.96/- Fuzzy Model:

Let us Take the fuzzy data:

OC

= (35,40,45,50,55,60,65) , DC

= (54,64,74,84,94,104,114), HM = Rs.15/- Per unit,

By Minimizing The Total Cost ,Using Geometric Programming And Lagrange Method,We Obtain The Optimal Value

T =0.2817 and TC=Rs.354.96/-

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M

OC

DC

OC

DC

### T TC(T) T TC(T)

1 15 0.2817 354.97 0.2817 354.97

2 20 0.2440 409.88 0.2440 409.88

3 25 0.2182 458.26 0.2182 458.26

4 30 0.1992 501.99 0.1992 501.99

5 35 0.1844 542.22 0.1844 542.22

From the above table we observed that:

(i) The economic order time interval obtained by pascals Graded Mean Integration is equal to crisp economic order time interval

(ii) Total cost obtained by Graded Mean Integration is equal to crisp total cost T value for various HM values

M

M

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