An Inventory Model For Deteriorating Items With Three Parameter Weibull Deterioration And Price Dependent Demand

DOI : 10.17577/IJERTV2IS50507

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An Inventory Model For Deteriorating Items With Three Parameter Weibull Deterioration And Price Dependent Demand

1R. Amutha

Research Scholar

Department of Mathematics, Presidency College, Chennai-05

2Dr. E. Chandrasekaran

Associate Professor

Department of Mathematics, Presidency College, Chennai-05

Abstract

The paper presents an inventory model for deteriorating items with price dependent demand. Deterioration rate follows a three parameter Weibull distribution. Shortages are allowed and are completely backlogged. The results are illustrated with the help of numerical example. Sensitivity analyses are carried out to analyze the effect of changes in the optimal solution with respect to change in one parameter at a time.

Keywords: Deterioration items, holding cost, Inventory, Price dependent demand time, Shortages.

  1. Introduction

    Inventory models are classified in to three categories (1) Deterioration (2) Obsolescence

    (3) no deterioration/no obsolescence. Deterioration makes the product value dull.

    Ajanta Roy [1] presented an inventory model for time proportional deterioration rate and demand is function of selling price. The Author discussed the model without shortage and also with shortages in which the shortages are completely backlogged. Mukesh Kumar, Anand Chauhan, Rajat Kumar [8] extended Ajanta Roy models with trade credit. Tripathy C.K and L.M. Pradhan [13] gave a model in which the demand of the product decreases with the increase of time and sale price and deterioration rate follows a three parameter Weibull distribution. Now Tripathy C.K and L.M. Pradhan [14] included salvage value and developed an EOQ model for three parameter Weilbull distribution deterioration rate under permissible delay in payments. Padmanabhan.G, Prem Vrat [9] formulated an EOQ model for perishable items under stock dependent selling rate. Sahoo.N.K., Sahoo .C.K. & Sahoo.S.K described an inventory model for price dependent demand and time varying holding cost. Vikas Sharma and Rekha Rani Chaudhary [16] explained and inventory model for two parameter Weibull deterioration rate.

    They found profit for their model. Sanjay JAIN and Mukesh KUMAR [12] explained an inventory model with ramp type demand and three parameter Weibull deterioration rate. The Authors also analyzed and summarized economic order quantity models done by few researchers. There are some products which start deteriorate only after some interval of time. This was explained by taking three parameter Weibull distribution deterioration rate. Anil Kumar Sharma, Manoj Kumar Sharma and Nisha Ramani [2] described an inventory model for two parameter Weibull distribution deterioration rate and demand rate is power pattern. Manoj Kumar Meher, Gobinda Chandra Panda, Sudhir Kumar Sahu [7] adopted a two parameter Weibull distribution deterioration to develop an inventory model under permissible delay in payments. Kun Shan Wu [6] made an attempt in his paper to obtain the optimal ordering quantity of deteriorating items for two parameter Weibull distribution deterioration under shortages and permissible delay in payments. P.K.Tripathy and S.Pradham [15] also define an inventory model with two parameter Weibull distribution as demand rate and deterioration rate increases with time.

    In this present paper, we have developed an inventory model for three-parameter Weibull deterioration rate and price dependent demand. Shortages are allowed and are completely backlogged. Holding cost is assumed to be constant. Our aim is to increase the profit.

  2. Assumptions and notations

    1. The demand rate is a function of selling price.

    2. Shortages are allowed and are completely backlogged

    3. Lead time is zero.

    4. Replenishment is instantaneous

    5. A is the set up cost

    6. C is the unit cost of an item

    7. p is the selling price

    8. Demand D (t) = f (p) = a-p, where a > p.

    9. C2 is the shortage cost per unit time

    (x) (t) = (t-)-1, 0 < 1 , > 0 and

    T 2

    p) [ 1 +

    2

    ( ) 2

    ( 1)( 2)

    2 ( )2 2

    +

    +

    ]

    ]

    4(2 1)( 1)

    – < < is the deterioration rate. At time T1 the Inventory becomes Zero and shortages start occurring.

    (6)

    Stock loss due Deterioration

    1. h is the constant holding cost.

    2. T is the length of the cycle.

    T1

    D = (a-p) e (t ) 0

    T1

    dt (a-p) dt

    0

  3. Mathematical formulation and solution

    (T

    ) 1

    2 (T

    )2 1

    = (a-p) [ 1 + 1 –

    Let I (t) be the inventory at time T ( 0 t T ) the differential equation for the instantaneous state over

    ( ) 1

    1

    2 ( )2 1

    2(2 1)

    (0,T) are given by

    dI (t)

    + (t )

    1

    I (t) = – (a-p), 0 t T1

    ]

    ]

    1 2(2 1)

    (7)

    dt

    (1)

    Order Quality

    dI (t)

    dt

    (2)

    = – (a-p), T1 t T

    T

    Q = [D + (a p)dt ]

    0

    With boundary condition I (T1) = 0

    (T

    ) 1

    2 (T

    )2 1

    = (a-p) [ 1 + 1 + T –

    Solving equation (1)

    ( ) 1

    1

    ]

    ]

    2 ( )2 1

    2(2 1)

    1

    1

    I (t) = – (a-p) [(t-T1) +

    2

    1

    ((t- ) 1 – (T – ) 1 ) +

    1 1

    1 1

    (8)

    1

    2(2 1)

    2(2 1)

    (3)

    ((t- )2 1 – (T – )2 1 ) ], 0 t T

    CQ

    Purchase cost =

    T

    Solving equation (2)

    C (T

    ) 1

    2 (T )2 1

    = – (a-p) (t-T ), T t T

    = (a-p)

    [ 1 + 1 + T –

    1 1

    (4)

    T 1

    2(2 1)

    Shortage Cost

    ( ) 1

    2 ( )2 1

    ] (9)

    C T 1

    2(2 1)

    SC = 2 (a p) (t-T1) dt

    T T1

    Total Profit Per unit time is

    1

    C2

    = (a-p) (T-T1)2

    = p (a-p) –

    T

    [Ordering cost + purchase cost +

    2T

    (5)

    shortages cost + holding cost]

    Holding Cost

    1

    K(p,T, T1) = p (a-p) –

    T

    { A +

    C2

    (a-p) (T-T1)2 + C (a-p)

    2

    h T1

    (T

    ) 1

    ( ) 1

    HC =

    T

    I (t) dt

    0

    ( 1 –

    1

    1

    • h

    (T ) 2

    2 ( )2 1

    2 (T ) 1

    = (a-p) [

    T

    1

    ( 1 – T1 (T1- ) ) +

    1

    1

    1

    1

    +

    +

    2

    2(2 1)

    2(2 1)

    + T ) h (a-p)

    2 (T

    )2 2 [

    ( 1 – T (T – )2 1 )] + h (a-

    2(2 1) 2( 1) 1 1

    1

    (T ) 2

    2

    ( ) 2

    2 ( )2 2

    ( 1 -T (T – )

    1 ) + – ]

    2 1 1

    1 )

    1 )

    (T )2 2

    2(2 1)

    T 2

    ( 1)( 2)T 2

    4T 2 (2 1)( 1)

    (10)

    2 1 1

    ( – T1 (T1- )] + h (a-p) [ +

    K ( p,T )

    1 C (T Tv)2

    2( 1)

    ( ) 2

    2

    + ]}

    + ]}

    2 ( )2 2

    = (a-2p) –

    p

    { 2 – C

    T 2

    ( 2)( 1) 4(2 1)( 1)

    <>Let T1= vT, 0 < v < 1

    (Tv ) 1

    (

    (

    1

    ( ) 1

    1

    1

    K (p,T) = p (a-p) –

    T

    { A +

    C2

    (a-p) (T-vT)2 + C (a-p)

    2

    2 ( )2 1

    2(2 1)

    (Tv ) 2

    2 (Tv )2 1

    +

    +

    2(2 1)

    + T) + h [ (

    1

    2

    (

    (

    (vT ) 1 ( ) 1 – Tv (Tv- ) 1 ) + (

    1

    2 ( )2 1

    1

    +

    +

    2 (vT )

    2 1

    2

    (Tv )2 2

    • Tv (Tv – )

      2 1

      2(2 1)

      (Tv)2

      )] h [ +

      2(2 1)

      2(2 1)

      + T ) h (a-p) [

      2( 1) 2

      (

      (

      1

      (Tv ) 2

      2

    • vT (vT – )

      1 ) +

      ( ) 2

      +

      +

      ( 2)( 1)

      2 ( )2 2

      4(2 1)( 1)

      ]} = 0. (11)

      2 For the maximization of profit we set,

      2(2 1)

      (

      (

      (vT )2 2

      + ].

      + ].

      2( 1)

    • vT (vT – )

    2 1

    )] + h(a-p) [

    K ( p,T )

    T = 0 &

    Provided

    K ( p,T )

    p = 0

    (vT)2

    ( ) 2

    2 ( )2 2

    2 K ( p,T )

    ( ) (

    ( ) (

    < 0,

    2 K ( p,T ) )

    2 ( 2)( 1) 4(2 1)( 1)

    T 2

    p2

    < 0 and

    K ( p,T )

    A C2 2

    ( 2 K ( p,T ) )( 2 K ( p,T ) ) ( 2 K ( p,T ) )

    > 0

    > 0

    = –

    T T 2 2

    (a-p) (1- v)

    – C(a-p) [

    T 2

    p2

    Tp

    +

    +

    T ( 1)v(vT ) (vT ) 1

    T 2 ( 1)

  4. Numerical Example

    + +

    + +

    ( ) 1

    2 ( )2 1

    Let us consider C = 5, v = 0.2, C2

    = 5, = 0.4,

    T 2 ( 1) 2T 2 (2 1)

    ]

    ]

    2T (2 1)v(vT )2 2 (vT )2 1

    2T 2 (2 1)

    + h (a-p)

    [

    [

    p

    T

    TP

    Q

    0.5

    32.4215

    5.9714

    141.5371

    108.3591

    0.6

    32.4534

    5.9869

    140.8515

    109.2179

    0.7

    32.4866

    6.0031

    140.1385

    110.1126

    0.8

    32.5210

    6.0198

    139.3986

    111.0397

    p

    T

    TP

    Q

    0.5

    32.4215

    5.9714

    141.5371

    108.3591

    0.6

    32.4534

    5.9869

    140.8515

    109.2179

    0.7

    32.4866

    6.0031

    140.1385

    110.1126

    0.8

    32.5210

    6.0198

    139.3986

    111.0397

    T ( 2)v(vT ) 1 (vT ) 2

    = 2, = 1, h = 1, a = 50, A = 1000 in proper units we get p = 32.3909, T = 5.9564, TP = 142.1949, Q = 107.5330

  5. Sensitivity Analysis

    ( 1)( 2)T 2

    v2 (vT ) +

    v 2T (2 2)(vT )2 1 2 (vT )2 2

    4(2 1)( 1)T 2

    2 v2 (vT )2

    2

    v 2

    ]- h (a-p) [ –

    2

    a

    p

    T

    TP

    Q

    60

    36.7900

    5.2035

    346.5236

    124.2395

    70

    41.3870

    4.6959

    605.7523

    138.6358

    80

    46.0894

    4.3198

    918.4367

    151.5396

    H

    p

    T

    TP

    Q

    2

    32.43.03

    5.9277

    139.9951

    106.7867

    3

    32.4698

    5.8998

    137.8099

    106.0562

    4

    32.5093

    5.8125

    135.6390

    105.3385

    5

    32.5488

    5.8458

    133.4822

    104.6336

    H

    p

    T

    TP

    Q

    2

    32.43.03

    5.9277

    139.9951

    106.7867

    3

    32.4698

    5.8998

    137.8099

    106.0562

    4

    32.5093

    5.8125

    135.6390

    105.3385

    5

    32.5488

    5.8458

    133.4822

    104.6336

    From this table, it can be observed that

    C2

    p

    T

    TP

    Q

    4

    31.8084

    6.5334

    177.8724

    121.6404

    3

    31.1694

    7.3674

    218.8579

    141.8183

    2

    30.4342

    8.6899

    267.7428

    174.0637

    1

    29.4950

    11.0218

    329.7243

    234.6714

    C2

    p

    T

    TP

    Q

    4

    31.8084

    6.5334

    177.8724

    121.6404

    3

    31.1694

    7.3674

    218.8579

    141.8183

    2

    30.4342

    8.6899

    267.7428

    174.0637

    1

    29.4950

    11.0218

    329.7243

    234.6714

    1. p(t, p) is slightly sensitive to changes in , h and it is highly sensitive to changes in a, C2, C2 and v.

    2. p is slightly sensitive to changes in , h

      and moderately sensitivity to changes in a, C2, C2 and v.

    3. Q is slightly sensitive to changes in the values of , h and C and it is highly

      sensitive to changes in a, C2 and v.

    4. T is slightly sensitive to changes in the values of , a, h and C and moderately sensitive to changes in C2 and v

  6. Conclusion

    C

    p

    T

    TP

    Q

    6

    32.9898

    6.0645

    124.4512

    105.7209

    7

    33.5930

    6.1782

    107.3297

    103.8447

    8

    34.2008

    6.2979

    90.8342

    101.8983

    9

    34.8138

    6.4246

    74.9689

    99.8804

    C

    p

    T

    TP

    Q

    6

    32.9898

    6.0645

    124.4512

    105.7209

    7

    33.5930

    6.1782

    107.3297

    103.8447

    8

    34.2008

    6.2979

    90.8342

    101.8983

    9

    34.8138

    6.4246

    74.9689

    99.8804

    A deterministic inventory model for deteriorating inventory model with three parameter Weibull distribution deterioration rate has been developed. Demand rate is function of selling price and holding cost is constant occurring shortages and completely backlogged. A numerical example is also given in support of the theory. A future research it may be consider to extend the model under permissible delay in payments.

  7. References

v

p

T

TP

Q

0.3

31.6714

6.3800

179.1983

121.7266

0.4

30.8940

6.1710

202.9896

130.9089

0.5

30.2681

5.4801

206.8682

129.9765

0.6

29.9177

4.7713

193.7094

123.3320

v

p

T

TP

Q

0.3

31.6714

6.3800

179.1983

121.7266

0.4

30.8940

6.1710

202.9896

130.9089

0.5

30.2681

5.4801

206.8682

129.9765

0.6

29.9177

4.7713

193.7094

123.3320

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