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Supply Chain Production Inventory Model: Innovative Study for Shortages Allowed With Partial Backlogging

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					                           International Journal of Modern Engineering Research (IJMER)
              www.ijmer.com        Vol. 2, Issue. 5, Sep.-Oct. 2012 pp-3641-3649       ISSN: 2249-6645

           Supply Chain Production Inventory Model: Innovative Study
                 for Shortages Allowed With Partial Backlogging
                                     Jasvinder Kaur1, Rajendra Sharma2
                          *(Department Of Mathematics, Graphic Era University Dehradun,India)
                         ** (Department Of Mathematics Graphic Era University Dehradun, india)

Abstract: In this paper, we have strived to combine all the    Period before they settle the account with the supplier. This
above mentioned factors into a single problem. We shall        provides an advantage to the customers, due to the fact that
undertake to explore a two echelon supply chain,               they do not have to pay the supplier immediately after
comprising of a vendor and a buyer. The whole environment      receiving the product, but instead, can defer their payment
of business dealings has been assumed to be progressive        until the end of the allowed period. The customer pays no
credit period, which conforms to the practical market          interest during the fixed period they are supposed to settle
situation. The whole combination is very unique and very       the account; but if the payment is delayed beyond that
much practical. The variable holding cost and variable         period, interest will be charged. The customer can start to
setup has been explored numerically as well; an optimal        accumulate revenues on the sale or use of the product and
solution has been reached. The final outcome shows that the    earn interest on that revenue. So it is to the advantage of the
model is not only economically feasible, but stable also.      customer to offer the payment to the supplier until the end
                                                               of the period.
Keywords:    Inventory model, partial backlogging,                       The two famous formulae of EOQ and EPQ are
Progressive permissible delay, Supply chain, Shortages,        treated separately for a buyer and a vendor respectively.
EOQ (Economics Order Quantity).                                From the traditional point of view, the vendor and the buyer
                                                               are two individual entities with different objectives and
I. INTRODUCTION                                                self-interest. Due to rising costs, the globalization trend,
Inventory represents one of the most significant possessions   shrinking resources, shortened product life cycle and
that most businesses possess. It is in direct touch with the   quicker response time, increasing attention has been placed
user department in its day today activities. Inventory         on the collaboration of the whole supply chain system. An
management is playing a key role in setting up efficient       effective supply chain network requires a cooperative
closed loop supply chains. A supply chain is a network of      relationship between the vendor and the buyer. It assumes
facilities and distribution options that performs the          that the buyer must pay off as soon as the items are
functions of procurement of materials, transformation of       received. Suppliers often offer trade credit as a marketing
these materials into intermediate and finished products, and   strategy to increase sales and reduce on-hand stock is
the distribution of these finished products to customers. It   reduced, and that leads to a reduction in the buyer’s holding
consists of a network of companies which are dependent on      cost of finance. In addition, during the time of the credit
each other while making independent decisions. The supply      period, buyers may earn interest on the money. In fact,
chain not only includes the manufacturer and suppliers, but    buyers, especially small businesses which tend to have a
also transporters, warehouses, retailers, and customers        limited number of financing opportunities rely on trade
themselves. Therefore, supply chain analysis tools and         credit as a source of short-term funds. The classical
methodologies have become more and more important. It          inventory models have considered demand rates which
can be a source of great efficiency and cost-savings gains.    were either constant or depended upon a single factor only,
Supply chain speed and flexibility have become key levers      like, stock, time etc. But changing market conditions have
for competitive differentiation and increased profitability.   rendered such a consideration quite unfruitful, since in real
The faster the supply chain, the better a company can          life situation, a demand cannot depend exclusively on a
respond to changing market situation and the less it needs     single parameter. A combination of two or more factors
inventory which resulting in higher return on capital          grants more authenticity to the formulation of the model.
employed. Supply chain management offers a large               Many delivery policies have been proposed in literature for
potential                                                      this problem. Clark and Scarf (1960) presented the
or organizations to reduce costs and improve customer          concept of serial multi-echelon structures to determine the
service performance. In the existing literature, most of the   optimal policy. Goyal (1985) considered a mathematical
inventory models studies only aimed at the determination of    models with a permissible delay in payments to determine
the optimum solutions that minimized cost or maximized         the optimal order quantities. Ha and Kim (1997) used a
profit from the vendor’s and vendor’s side. However, in the    graphical method to analyze the integrated vendor-buyer
modern global competitive market, the buyer and vendor         inventory status to derive an optimal solution. Hwang and
should be treated as strategic partners in the supply chain    Shinn (1997) studied effects of permissible delay in
with a long term cooperative relationship. Recently, many      payments on retailer's pricing and lot sizing policy for
researchers have considered the buyer and vendor as a unit     exponentially deteriorating products. Yang and Wee
to find the optimal EOQ in achieving the minimum total         (2000) developed an integrated economic ordering policy of
cost. In today's business transactions, it is more and more    deteriorating items for a vendor and a buyer. Wang et al.
common to see that the customers are allowed some grace        (2000) analyzed supply chain models for perishable
                                                               products under inflation and permissible delay in payment.

                                                    www.ijmer.com                                                3641 | Page
                            International Journal of Modern Engineering Research (IJMER)
               www.ijmer.com        Vol. 2, Issue. 5, Sep.-Oct. 2012 pp-3641-3649       ISSN: 2249-6645

Teng (2002) modified Goyal (1985) model by considering          2.13 (Cvs  2 t ) = the setup cost for each production
the selling price, instead of purchasing cost, as the base to
calculate the interest. Abad and Jaggi (2003) studied a             cycle for vendor.
seller-buyer model with a permissible delay in payments by      2.14 (Cbs  1t ) = the setup cost per order for buyer.
game theory to determine the optimal unit price and the         2.15 (Chv   2 t ) = holding cost per unit time for vendor.
credit period, considering that the demand rate is a function
of retail price. Huang, Y.F. et al. (2005) considered the       2.16 (Cbh  1t ) = holding cost per unit time for buyer.
optimal inventory policies under permissible delay in           2.17 Cv = the unit cost for vendor.
payments depending on the ordering quantity. Song and           2.18 Cb = the unit purchase cost for buyer.
Cai (2006) has been taken on optimal payment time for a         2.19 Sb = shortage cost per unit time for buyer.
retailer under permitted delay of payment by the                2.20 Lb= lost sale cost per unit time for buyer.
wholesaler. Liao (2007) assumed on an EPQ model for             2.21 VC = the cost of vendor per unit time.
deteriorating items under permissible delay in payments.        2.22 BC = the cost of buyer per unit time.
          In the present study, we have strived to combine      2.23 TC(T) = total cost of an inventory system / time unit.
all the above mentioned factors into a single problem. We       2.24 B= Backlogging rate.
shall undertake to explore a two echelon supply chain,          2.25 The deterioration function
comprising of a vendor and a buyer. The whole
environment of business dealings has been assumed to be
                                                                 (t ,  )  0 ( )t ,      0< 0 ( ) <<1,     t>0
progressive credit period, which conforms to the practical      This is a special form of the two parameter weibull function
market situation. The whole combination is very unique and      considered by Covert and Philip. The function is some
very much practical. The variable holding cost and variable     functions of the random variable  which range over a
setup has been explored numerically as well; an optimal         space and in which a p.d.f. p(  ) is defined such that

                                                                
solution has been reached. The final outcome shows that the
model is not only economically feasible, but stable also.
                                                                     p( )d   1
                                                                 


 II. PROPOSED ASSUMTIONS & NOTATIONS                            III. INDENTATIONS AND EQUATIONS
                                                                MATHEMATICAL FORMULATION
1. ASSUMPTIONS                                                  The actual vendor’s average inventory level in the
The following assumptions are used to develop aforesaid         integrated two-echelon inventory model is difference
model:                                                          between the vendor’s total average inventory level and the
1.1 The demand rate, D(t), is deterministic, the demand         buyer’s average inventory level. Since the inventory level is
    function D(t) is given by D(t) = 0 e , a and b are
                                            t                  depleted due to a constant deterioration rate of the on-hand
                                                                stock, the buyer’s inventory level is represented by the
    positive constants.
                                                                following differential equation:
1.2 Shortages are allowed with partial backlogging.
1.3 If the retailer pays by M, then the supplier does not       Ib (t )  0 ( )tIb (t )  0 e t ,0  t  t1
                                                                 '
                                                                                                                             (1)
    charge to the retailer. If the retailer pays after M and
    before N (N > M), he can keep the difference in the         Ib (t )   B0 et ,
                                                                 '
                                                                                                         t1  t  T          (2)
    unit sale price and unit purchase price in an interest
    bearing account at the rate of Ie/unit/year. During [M,
                                                                The vendor’s total inventory system consisting of
    N], the supplier charges the retailer an interest rate of
                                                                production period and non-production period can be
    Ic1/unit/year on unpaid balance. If the retailer pays
                                                                described as follows:
    after N, then supplier charges the retailer an interest
    rate of Ic2/unit/year (Ic1> Ic2) on unpaid balance.
                                                                I v' 1 (t )  0 ( )tI v1 (t )  ( K  1)0 et , 0  t  T1 (3)
2. NOTATIONS:
2.1  P = the selling price / unit.                              I v' 2 (t )  0 ( )tI v 2 (t )  0 et ,    0  t  T2   (4)
2.2  KD = the production rate per year, where K>1
2.3 C = the unit purchase cost, with C < P.                     The boundary conditions are
2.4 M = the first offered credit period in settling the
    account without any charges.                                I v1 (t )  0, t  0                                          (5)
2.5 N = the second permissible credit period in settling the
    account with interest charge Ic2 on unpaid balance and
    N > M.                                                      I v 2 (t )  0, t  T2                                       (6)
2.6 Ic1 = the interest charged per $ in stock per year by the
    supplier when retailer pays during [M, N].                  Ib (t )  I 0 , t  0                                        (7)
2.7 Ic2 = the interest charged per $ in stock per year by the
    supplier when retailer pays during [N, T]. (Ic1 > Ic2)
2.8 Ie = the interest earned / $ / year.                        Ib (t )  0, t  t1                                          (8)
2.9 T = the replenishment cycle.
2.10 r = the discount rate (r > α)                              I v1 (T1 )  I v 2 (0)                                       (9)
2.11 IE = the interest earned / time unit.
2.12 IC = the interest charged / time unit.
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                                                  International Journal of Modern Engineering Research (IJMER)
                                     www.ijmer.com        Vol. 2, Issue. 5, Sep.-Oct. 2012 pp-3641-3649       ISSN: 2249-6645

And,                                                                                                                                rt12                  t3  t2                 t3             t 4  r 2 t15 0 ( )t14
                                                                                                                                         (r 2  0 ( )) 1 ]  0 [ 1  (  2r ) 1  r (r   ) 1          
                                                                                                                                     2                     6    T 2                6              8    20         12
         T2                                                                                                                                                  0 ( )t15                                        (17)
T                                                                                                                  (10)            (3  2r )
                                                                                                                                                                 30
                                                                                                                                                                           ]
         n
                                                                                                                                Respectively.
The solutions of the above differential equations obtained
                                                                                                                                The annual total holding cost for the buyer and the vendor
are
                                                                                                                                are
                                                                                                                                          1 1                             
                                                                                                                                               t
                       0 ( ) t 2                                     ( ) t 2
                                                 t 2 0 ( )t 3  0 2                                             (11)         HCb          (Cbh  1t )e I b1 (t )dt 
                                                                                                                                                               rt
Ib (t )  I 0 e            2
                                      0 [t                   ]e     ,                      0  t  t1                                T 0
                                                  2        6                                                                                                              
                                                                                                                                                                           
                      B0 t1 t
Ib (t )                e  e  ,                                                 t1  t  T                                         0                                        t 5 r 2  ( )t 4  ( )t 5 (3  2r )
                                                                                                                 (12)                       t2 t3             t4
                                                                                                                                        Cbh [ 1  1 (  2r )  1 r (r   )  1  0 1  0 1                           ]
                                                                                                                                       T        2 6               8               20        12               60
                                                                                             0 ( ) t 2
                                                     t2       0 ( )t 3               
I v1 (t )  ( K  1)0 [t                                                        ]e            2
                                                                                                           , 0  t  T1              10 t13 t14       t5              ( )t15 Cbh I 0      rt 2
                                                      2               6                                                                 [  (  2r )  1 r (r   )  0       ]       [t1  1
                                                                                                                    (13)              T 3 8             10                15       T           2

                                           ( )
I v 2 (t )  0 [(T2  t )  (T22  t 2 )  0 (T23  t 3 )],                                                   0  t  T2       
                                                                                                                                      (r 2  0 ( ))t13    I t 2 rt 3 (r 2  0 ( ))t14
                                                                                                                                                         ] 1 0 [ 1  1                   ]
                            2                6                                                                                                6             T     2   3         8
                                                                                                                    (14)
                                                                                                                                                                                                                             (18)
Using the condition that one can get,
                                                                                                                                And
                                        t12         0 ( )t13
I 0  0 [t1                                                   ]                                                 (15)                   1 1
                                                                                                                                                T                                          T2
                                                                                                                                                                                                                              
                                           2               6                                                                    HCv            (Chv   2 t )e rt I v1 (t )dt  e rT1  (Chv   2t )e  rt I v 2 (t )dt   I b
                                                                                                                                           T2  0
                                                                                                                                                                                           0                                 
                                                                                                                                                                                                                              
If the product of the deterioration rate and the
replenishment interval is much smaller than one, the                                                                             1                T2 T3             T4             T 5 r 2  ( )T 4  ( )T 5 (3  2r )
buyer’s and the vendor’s actual average inventory                                                                                [( K  1)0Chv { 1  1 (  2r )  1 r (r   )  1  0 1  0 1                          }]
                                                                                                                                 T2                2 6               8              20         12              60
level, I b and I v , are
                                                                                                                                     2 ( K  1)0 T13 T14       T5              ( )T 5 C  e rT1 T 2 T 3 (  2r )
                  t1                                                                                                                             [  (  2r )  1 r (r   )  0 1 ]  bh 0 [ 2  2
           1  rt                                                                                                                          T2       3 8          10                15       T         2        6

           T
Ib           e I b (t )dt
            0
                                                                                                                                     r 2T24 0 ( )T24 0 ( )T25 (4  3r ) 0 e rT1 2 T23 T24 (3  2r ) I 0
                                                                                                                                                                         ]           [                 ]  [t1
     I0       rt                  2
                                  t    t              t          3                            2                            3
                                                                                                                                       24       24              60               T2        6        24         T
       [t1      (r 2  0 ( )) ]  0 [  (  2r )
                                 1                               1                            1                            1

     T         2                   6  T 2               6
                                                                                                                                    rt12                  t3  t2                 t3             t 4  r 2 t15 0 ( )t14
                                                                                                                                         (r 2  0 ( )) 1 ]  0 [ 1  (  2r ) 1  r (r   ) 1          
                                                                                                                                     2                     6 T 2                   6              8    20         12
                              t14  r 2 t15 0 ( )t14              ( )t15                                                                             0 ( )t1  5
                                                                                                                                    (3  2r )                                                                 (19)
 r (r   )                                          (3  2r ) 0        ]                                                                               30
                                                                                                                                                                      ]
                               8    20         12                     30
                                                                                                                                 respectively.
                                                                                                                    (16)        The annual deterioration cost for the buyer and the vendor
and                                                                                                                             are
                                                                                                                                               Cb      t1                       
                                                                                                                                                         0 ( )te I b1 (t )dt  =
                                                                                                                                                                     rt
                     rtT1
                                                                T2                                                             DCb 
          1
                     e I v1 (t )dt  e 1  e I v 2 (t )dt   I b                                                                                   0                         
                                         rT    rt
Iv                                                                                                                                            T                                
          T2       0
                                            0              
                                                            
                                                                                                                                    I 00 ( ) t12 rt13 r 4 t14 00 ( ) t13 t14 (  2r ) t15
    1               T T      2
                                  T    3
                                                T r 0 ( )T 0 ( )T (3  2r )
                                                       4                 5 2                      4             5               [             {              }        {                 r (r   )}]
      [( K  1)0 (  (  2r )  r (r   ) 
                            1         1
                                                     1
                                                                       1
                                                                                  )              1             1                        T       2 6       8        T       3        8       10
    T2               2 6           8             20      12            60                                                                                                                                                    (20)

               T22 T23         r 2T 4  rT 4  ( )T 4  ( )T 5 (4  3r ) I 0                                                 and
e rT1 0 (       (  2r )  2  2  0 2  0 2                          )]  [t1
                2 6              24      8      24             60              T



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                                        International Journal of Modern Engineering Research (IJMER)
                           www.ijmer.com        Vol. 2, Issue. 5, Sep.-Oct. 2012 pp-3641-3649       ISSN: 2249-6645

           Cv  1                                                                             Regarding interest charged and interest earned based on the
                T                                     T2

DCv            0 ( )te rt I v1 (t )dt  e  rT1  0 ( )te  rt I v 2 (t )dt           length of the cycle time t1, three cases arise:
           T2  0
                                                      0                            
                                                                                    
                                                                                               IV. FIGURES AND TABLES
= C [ ( K  1)00 ( ) {T1  T1 (  2r )  T1 r (r   )}  e 00 ( ) {T2
                           3    4              5                rT1          3
                                                                                               Regarding interest charged and interest earned based on the
   v
             T2           3        8              10                 T2     6                  length of the cycle time t1, three cases arise:

      T2 4 (3  2r )                                                                          Case I: M ≥ t
                                                                                                           1
                    }]                                                         (21)
             24
                                                                                               Inventory level
respectively.

The annual set-up cost for the buyer and the vendor are
                   t                             T
            1 1
            T 
OCb         [ (Cbs  1t )dt   (Cbs  1t )dt ]
              0                 t1



       =C              1T                                                     (22)
           bs   
                        2

and
                       T                             T2
            1 1
            T2                                       (C
OCv          [ (Cvs  2 t )dt                          vs    2 t )dt ]
               0                                     0




      = [Cvs   2   2 (T1  T2 ) ]
                               2    2

               1
                                                                               (23)
                                               2

respectively.
                                                                                                                                                   t1              M Time
The annual shortage cost for the buyer is
                                                                                                   0

                                                                                                                                    Fig 1: t1 ≥ M
           Sb e rt1        T
                                 0
SCb 
              T             
                            t1
                                   et1  et  e rt dt
                                             
                                                                                               In the first case, retailer does not pay any interest to the
                                                                                               supplier. Here, retailer sells I0 units during (0, t1) time
    Sb 0 e rt1 e(  r )t1   e( t1  rT )   e(  r )T                      (24)
                                                                                               interval and paying for CI0 units in full to the supplier at
               [                                        ]                                   time M ≥ t1, so interest charges are zero, i.e.
       T         r (  r )         r          (  r )
                                                                                               IC1 = 0                                                                    (28)
The annual lost sale cost for the buyer is
                                                                                               Retailers deposits the revenue in an interest bearing account
            Lb e  rt1      T
LCb 
               T             (1  B)
                            t1
                                          0   et e  rt dt                                    at the rate of Ie / $ / year. Therefore, interest earned IE1, per
                                                                                               year is

= Lb e 0 (1  B) [e(  r )T  e(  r ) t ]
       rt  1                                                                                                     t                                 t
                                                                                (25)                       PI e 1  rt                     1

                                                                                                               [  e D(t )tdt  ( M  t1 )  e  rt D(t )dt ]
                                                               1

        T (  r )                                                                             IE1 
                                                                                                           T2 0                            0

The different costs associated with the system are set-up
                                                                                                       = PI e 0 [(M  t )t  {1  (  r )(M  t )} t1  t1 (  r ) ] (29)
                                                                                                                                                       2    3
costs, holding costs, deterioration cost and shortage cost.
                                                                                                                        1 1                      1
Our aim is to minimize the total cost.                                                                    T2                                          2        3

From (9), one can derive the following condition:                                              Total cost per unit time of an inventory system is

                    T12 0 ( )T13  (2)T            T 2  ( )T23  (2)T                TCb (t1 ,  ) = OCb +HCb + DCb + SCb + IC1 – IE1
                                                 2                                         2
                                          0     1                                0     2

( K  1)0 [T1                   ]e         0 [T2  2  0         ]e
                     2       6                           2      6
                                                                                               = [Cbs  1T  0 Cbh [ t1  t1 (  2r )  t1 r (r   )   t1 r  0 ( )t1
                                                                                                                         2    3              4                 5 2           4
                                                                               (26)
                                                                                                         2 T            2 6                 8                20        12
By Taylor’s series expansion, (4.26) is derived as

          1                                                                                     0 ( )t15 (3  2r ) 10 t13 t14       t5              ( )t15
 T1          T2            1  2 T2 
                                                                                (27)                                   ]   [  (  2r )  1 r (r   )  0        ]
         K 1                                                                                              60            T 3 8            10                15

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                                                                                                                At15          ( K  1)0 A T13 T14 (  2r ) T15 r (r   )
                                                                                                    (3  2r )      ]  Cv [             {                               }
    Cbh I 0       rt   (r  0 ( ))t
                       2             2
                                        I t    3
                                                 rt   (r  0 ( ))t
                                                            2       3      2                4
                                                                                                                 30                T2       3        8             10
           [t1     1
                                      ] 1 0 [ 
                                               1
                                                          1       1
                                                                     ]                     1

     T             2         6           T 2      3         8                                         e rT1 0 A T23 T2 4 (3  2r )                               (34)
                                                                                                                 {                          }]
                                                                                                          T2         6                24
     I 00 ( ) t12 rt13 r 4 t14 00 ( ) t13 t14 (  2r ) t15
[             {              }        {                 r (r   )}]
         T       2 6       8        T       3        8       10                                  To minimize the total cost per unit time, the optimum value
                                                                                                 of t1, T2 is the solution of following equation.
    Sb 0 e rt1 e(r )t1 e(t1 rT ) e(r )T Lb e rt1 0 (1  B) (r )T ( r )t1
               [                               ]                 [e      e         ]        Case II: M < t1< N
       T  r (  r )           r        (  r )    T (  r )
                                                                                                Inventory level
     PI e 0                                         t 2 t 3 (  r )
            [( M  t1 )t1  {1  (  r )( M  t1 )} 1  1           ]]
      T2                                              2        3

                                                                                   (30)

Hence the mean cost                                                                             Q

< TCb >=
                 TC           b   (t1 ,  ) p( )d                               (31)


< TCb
>= [Cbs  1T  0 Cbh [ t1  t1 (  2r )  t1 r (r   )   t1 r  At1
                           2    3              4                 5 2     4


           2 T            2 6                 8                20     12
  At15 (3  2r ) 10 t13 t14                t5          At 5
                   ]     [  (  2r )  1 r (r   )  1 ]
          60           T 3 8                  10          15
  Cbh I 0       rt12 (r 2  A)t13 1 I 0 t12 rt13 (r 2  A)t14
         [t1                  ]     [                    ]
   T             2         6        T 2         3       8
  I A t 2 rt 3 r 4 t 4  A t 3 t 4 (  2r ) t15
[ 0 { 1  1  1 }  0 { 1  1                     r (r   )}]
   T 2 6               8       T 3          8       10
                                                                                                                                     M                 t1     N Time
    Sb 0 e rt1 e(r )t1 e(t1 rT ) e(r )T Lb e rt1 0 (1  B) (r )T (r )t1
               [                               ]                 [e      e        ]                                    Fig: 2 M < t1 < N
       T  r (  r )           r        (  r )    T (  r )
     PI e 0                                         t 2 t 3 (  r )                            In the second case, supplier charges interest at the rate Ic1
            [( M  t1 )t1  {1  (  r )( M  t1 )} 1  1           ]]
      T2                                              2        3                                 on unpaid balance.

                                                                                                 Interest earned, IE2 during [0, M] is
                                                                                   (32)
                                                                                                                  M
                                                                                                    IE2  PI e  e rt D(t )tdt
Where A=                 0   ( ) p( )d                                         (33)
                                                                                                                  0


                                                                                                                        2   3            4
 <TCv >= OCv +HCv + DCv - IC1                                                                            = PI e 0 [ M  M (  r )  M (  r ) 2 ]                (35)
                                                                                                                      2   6            8

                                                                                                 Retailer pay for I0 units purchased at time t = 0 at the rate of
                      2 (T12  T22 ) 1                  T2 T3             T4
= [Cvs 1  2                    ]  [( K  1)0Chv { 1  1 (  2r )  1 r (r   )         C / $ / unit to the supplier during [0, M]. The retailer sells D
                             2          T2                2 6               8                    (M).M units at selling price P/ unit. So, he has generated
  T15 r 2 AT14 AT15 (3  2r )  2 ( K  1)0 T13 T14                                           revenue of P D(M).M + IE2. Then two sub cases may arise:
                               }]              [  (  2r )
   20       12         60                  T2        3 8                                         Sub Case: 2.1
 T 5
                AT 5
                       Cbh 0 e rT1 T2 2 T23 (  2r ) r 2T2 4 AT2 4                            Let P D(M).M + IE2 ≥ CI0, i.e. retailer has enough money to
 1 r (r   )  1 ]               [                        
 10             15          T         2         6         24     24                              settle his account for all I0 units procured at time t = 0. Then
                                                                                                 interest charge will be
                                                                                                 IC2.1 = 0                                                 (36)
     AT25 (4  3r ) 0 e rT1  2 T23 T2 4 (3  2r ) I 0     rt 2
                   ]            [                  ]  [t1  1                                and interest earned
           60            T2         6         24         T      2
              t13 0 t12          t3             t 4  r 2 t15 At14
(r 2  A)       ]  [  (  2r ) 1  r (r   ) 1          
               6 T 2               6              8    20       12
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               IE2                                                                    To minimize the total cost per unit time, the optimum value
IE2.1                                                                                of t1, T2 is the solution of following equation.
                T2
                                                                                      Sub Case: 2.2
        = PI e 0 [ M  M (  r )  M (  r ) 2 ] (37)
                       2   3            4
                                                                                      Let P D(M).M + IE2 < CI0. Here, retailer will have to pay
           T2        2   6            8                                               interest on unpaid balance U1 = CI0 – (P D(M).M + IE2) at
                                                                                      the rate of Ic1 at time M to the supplier. Then interest paid
So, total cost TC2.1 per unit time of inventory system is                             per unit time is given by
                                                                                                                   t1
                                                                                                      U12 Ic1
                                                                                                                   e
                                                                                                                          rt
<TCb >= OCb + HCb + DCb + SCb + LCb +IC2.1 – IE2.1                                    IC2.2                                    I (t )dt
                                                                                                       PI 0        M


                                                                                      = U1 Ic1 [ (t1  M )  (  r )(t1  M )  (  r ) (t1  M ) ]
                                                                                          2         2   2               3   3            2   4    4


S= [Cbs  1T  0 Cbh [ t  t (  2r )  t r (r   )   t r  At
                                        2        3    4                5 2        4
                                       1        1    1                1          1       PI 0         2               2                    8
               2          T            2        6    8                20       12                                                                (40)

                                                                                      Where,
 At 5 (3  2r ) 10 t13 t14       t5             At 5
 1             ]   [  (  2r )  1 r (r   )  1 ]                               U1 = CI0 – (P D(M).M + IE2)
        60         T 3 8            10             15
                                                                                             =CI0–
                                                                                                            Ie          2 I (  r ) 3  3 I e (   r ) 2
    Cbh I 0      rt 2 (r 2  A)t13 1 I 0 t12 rt13 (r 2  A)t14                       P0 [ M  (            )M 2  (  e          )M  (                )M 4 ]
           [t1  1              ]     [                    ]                                           2          2       2           6       8
     T            2         6        T 2       3         8                                                                                                                      (41)

                                                                                      And interest earned
     I 0 A t12 rt13 r 4 t14    A t 3 t 4 (  2r ) t15
[        {               } 0 { 1  1            r (r   )}]
      T 2       6     8        T 3          8       10                                                IE2
                                                                                      IE2.2 
                                                                                                       T2


                                                                                               = PI e 0 [ M  M (  r )  M (  r )2 ]
                                                                                                              2   3            4

 S  e rt1 e(r )t1 e(t1 rT ) e(r )T Lb e rt1 0 (1  B) (r )T (r )t1                                                                                              (42)
 b 0 [                                    ]                 [e      e        ]               T2        2   6            8
   T  r (  r )          r        (  r )    T (  r )
                                                                                      So, total cost TC2.2 per unit time of inventory system is
        PI e 0 M             2
                                      M     3
                                                    M     4
                                                                               (38)   <TCb >= OCb + HCb + DCb + SCb + LCb + IC2.2 – IE2.2
              [                        (  r )     (  r ) 2 ]
         T2      2                     6             8


                                                                                      = [Cbs  1T  0 Cbh [ t1  t1 (  2r )  t1 r (r   )   t1 r  At1
<TCv >= OCv + HCv + DCv - IC2.1                                                                                 2    3              4                 5 2     4


                                                                                                2 T            2 6                 8                20     12
= [C      2 (T1  T2 ) ]  1 [( K  1) C {T1  T1 (  2r )  T1 r (r   )
                      2    2                        2    3              4

    vs 1   2                                  0 hv
                       2         T2                2 6                 8                   At15 (3  2r ) 10 t13 t14       t5             At 5
                                                                                                         ]   [  (  2r )  1 r (r   )  1 ]
     T r AT AT (3  2r )  2 ( K  1)0 T T
         5 2          4            5                             3      4
                                                                                                  60         T 3 8            10             15
       1
                   1
                          }]     1
                                         [  (  2r )          1      1

      20   12    60               T2       3 8
                                                                                            Cbh I 0      rt 2 (r 2  A)t13 1 I 0 t12 rt13 (r 2  A)t14
 T5             AT 5 C  e rT1 T2 2 T23 (  2r ) r 2T2 4 AT2 4                                   [t1  1              ]     [                    ]
 1 r (r   )  1 ]  bh 0    [                                                          T            2         6        T     2   3         8
 10             15        T      2         6         24     24

        AT25 (4  3r ) 0 e rT1  2 T23 T2 4 (3  2r ) I 0     rt 2                     I 0 A t12 rt13 r 4 t14    A t 3 t 4 (  2r ) t15
                      ]            [                  ]  [t1  1                  [        {               } 0 { 1  1            r (r   )}]
              60            T2         6         24         T      2
                                                                                            T 2       6     8        T 3          8       10
               t13 0 t12          t3             t 4  r 2 t15 At14
(r 2  A)        ]  [  (  2r ) 1  r (r   ) 1                                     Sb 0 e rt1 e( r )t1 e( t1 rT ) e( r )T   L e rt1 0 (1  B) ( r )T
                6 T 2               6              8    20       12                                   [                                 ] b                 [e         e( r )t1 ]
                                                                                              T         r (  r )      r        (  r )      T (  r )
                   At15          ( K  1)0 A T13 T14 (  2r ) T15 r (r   )
 (3  2r )            ]  Cv [             {                               }           PI e 0 M 2   M3            M4                                                        (43)
                   30                 T2       3        8             10                         [        (  r )     (  r ) 2 ]
                                                                                            T2      2    6             8

     e rT1 0 A T23 T2 4 (3  2r ) -
               {                  }]                                                <TCv >= OCv + HCv + DCv - IC2.2
         T2       6         24
PI e 0 M 2 M 3            M4
               (  r )     (  r ) 2 ]                                   (39)
                                                                                      =S [C      2 (T1  T2 ) ]  1 [( K  1) C {T1  T1 (  2r )  T1 r (r   )
       [                                                                                                     2    2                        2    3              4
 T2      2   6             8                                                               vs 1   2                                  0 hv
                                                                                                              2         T2                2 6                 8

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     T15 r 2       AT14 AT15 (3  2r )      ( K  1)0 T13 T14                in account at M and total money
                                      }]  2          [       (  2r )
       20           12         60                 T2       3   8
                                                                                 PD(N).N=P 0 e
                                                                                                                 N
                                                                                                                          N at N, there are three sub cases may
  T   5
                  AT    C e 5         rT1
                                               T2
                                                   T (  2r ) r T2
                                                     3         2    4
                                                                     AT     4
                                                                                 arise:
    r (r   ) 
     1
                     ]  bh 0
                            1
                                              [ 2  2               2
  10              15        T                   2      6        24    24
                                                                                 Sub Case 3.1 Let P D(M).M + IE2 ≥ CI0

    AT25 (4  3r ) 0 e rT1  2 T23 T2 4 (3  2r ) I 0     rt 2               This case is same as sub case 2.1, here 3.1 designate
                  ]            [                  ]  [t1  1                 decision variables and objective function.
          60            T2         6         24         T      2
                                                                                 Sub Case 3.2 Let P D(M).M + IE2 < CI0                                                           and
        t3  t2                 t3             t 4  r 2 t15 At14
(r  A) 1 ]  0 [ 1  (  2r ) 1  r (r   ) 1          
      2                                                                                                                        N

         6 T 2                   6              8    20       12                 PD( N  M ).( N  M )  PI e  D(t )dt  CI 0  ( PD(M ).M  IE2 )
                                                                                                                               M

                    At 5
                                ( K  1)0 A T T (  2r ) T r (r   )
                                                 3    4             5
(3  2r )           1
                       ]  Cv [             {  1    1
                                                                  1
                                                                        }                                                        
                    30               T2       3    8           10                P0 e ( N M ) ( N  M )  PI e 0 [( N  M )  ( N 2  M 2 )]  CI 0  ( P0 eM M  IE2 )
                                                                                                                                 2
                                                                                 This case similar to sub case 2.2.

     e rT1 0 A T23 T2 4 (3  2r )     U 2 Ic (t 2  M 2 )                     Sub Case 3.3 Let P D(M).M + IE2 < CI0                                                           and
               {                   }]  1 1 [ 1                                                                                    
         T2       6         24            PI 0        2                          P0 e   ( N M )
                                                                                                     ( N  M )  PI e 0 [( N  M )  ( N 2  M 2 )]  CI 0  ( P0 eM M  IE2 )
                                                                                                                                     2
     (  r )(t13  M 3 ) (  r ) 2 (t14  M 4 )                                Here, retailer does not have enough money to pay off total
                                                ]                      (44)
              2                      8                                           purchase cost at N. He will not pay money of P D(M).M +
                                                                                 IE2 at M and PD( N  M ).( N  M )  PI [( N  M )   ( N 2  M 2 )] at
                                                                                                                                               e
To minimize the total cost per unit time, the optimum value                                                                                                        2
of t1, T2 is the solution of following equation.                                 N. That’s why he has to pay interest on unpaid balance U1 =
                                                                                 CI0 – (P D(M).M + IE2) with Ic1 interest rate during (M, N)
Case III: t1 ≥ N                                                                                                                                                        N

                                                                                                                       e 
                                                                                 and U  U  PD( N  M ).( N  M )  PI D(t )dt
                                                                                      2   1
                                                                                                                                                                        M
                                                                                 with interest rate Ic2 during (N, t1).

                                                                                 Therefore, total interest charged on retailer, IC3.3 per unit
                                                                                 time is

                                                                                                     U1 Ic1 ( N  M ) U 2 Ic1
                                                                                                                        2                       t1

                                                                                                                                                e
                                                                                                                                                        rt
                                                                                 IC3.3                                                                      I b (t )dt
                                                                                                             T2        PI 0                     N


                                                                                         = U1 Ic1 ( N  M )  U 22 Ic1 [ t12  N 2  (t13  N 3 )(  2r )  (t14  N 4 ) r (r   )
                                                                                                      T2           PI 0    2                   6                    8



                                                                                     (t15  N 5 ) 2 0 ( )(t14  N 4 ) 0 ( )(t15  N 5 )(3  2r ) U 22 Ic1
                                                                                                r                                                ]        [(t1  N )
                                                                                          20                12                       12                  P
                                                                                         r (t12  N 2 ) (r 2  0 ( ))(t13  N 3 )
                                                                                                                                  ]                                           (45)
                                                                                                2                    6

                                                                                  Interest earned per unit time is

                                                                                                     IE2
                                                                                 IE3.3 
                                                                                                      T2

                                                                                           = PI e 0 [ M 2  M 3 (  r )  M 4 (  r ) 2 ]                                   (46)
                                  Fig 3: t1 ≥ N                                                      T2        2           6                       8

In the final case, retailer pays interest at the rate of Ic2 to the              So, total cost TC3.3 per unit time of inventory system is
supplier. Based on the total purchased cost, CI0, total
money             P            D(M).M              +            IE2              <TCb >= OCb + HCb + DCb + SCb + LCb + IC3.3 – IE3.3
  PI e 0 M 2 M 3                M4
=        [           (  r )      (  r )2 ]
    T2     2       6              8

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                       www.ijmer.com        Vol. 2, Issue. 5, Sep.-Oct. 2012 pp-3641-3649       ISSN: 2249-6645

                                                                                      NUMERICAL ILLUSTRATION: THE PRECEDING
        T         t2 t3             t4              t 5 r 2 At 4
= [Cbs  1  0 Cbh [ 1  1 (  2r )  1 r (r   )  1  1
                                                                                      THEORY CAN BE ILLUSTRATED BY THE
                                                                                      FOLLOWING NUMERICAL EXAMPLE WHERE
         2 T         2 6               8               20      12
                                                                                      THE PARAMETERS ARE GIVEN AS FOLLOWS:
                                                                                      Demand parameters, a = 500, b = 5, c = 2
 At 5 (3  2r ) 10 t13 t14       t5             At 5
 1             ]   [  (  2r )  1 r (r   )  1 ]                               Selling price, P = 30
        60         T 3 8            10             15
                                                                                      Buyer’s purchased cost, Cb = 35
  C I       rt 2 (r 2  A)t13 1 I 0 t12 rt13 (r 2  A)t14
 bh 0 [t1  1              ]     [                    ]
    T        2         6        T 2       3         8                                 Buyer’s percentage holding cost per year per dollar,
  I A t 2 rt 3 r 4 t 4  A t 3 t 4 (  2r ) t15
[ 0 { 1  1  1 }  0 { 1  1                r (r   )}]                           Cbh = 0.2
   T 2 6          8       T 3           8      10
                                                                                      Buyer’s ordering cost per order, Cbs = 500
  S  ert1 e(r )t1 e(t1 rT ) e(r )T Lb ert1 0 (1  B) (r )T (r )t1
 b 0 [                                      ]              [e      e ]           Buyer’s shortage cost, Sb = 50
    T  r (  r )         r          (  r )     T (  r )
+ U1 Ic1 ( N  M )  U 2 Ic1 [ t1  N 2  (t1  N )(  2r )  (t1  N ) r (r   )
                        2         2              3  3                4   4
                                                                                      Vendor’s unit cost, Cv = 20
          T2           PI 0 2                         6                8
                                                                                      Vendor’s percentage holding cost per year per dollar,
          (t  N ) 2 0 ( )(t  N ) 0 ( )(t  N )(3  2r ) U Ic
            5    5             4    4            5    5                2
          1
                  r         1
                                               1
                                                              ]    [(t1  N )
                                                                       2 1             Cvh = 0.2
             20              12                 12               P
                                                                                      Vendor’s setup cost per order, Cvs = 1000
 r (t 2  N 2 ) (r 2  0 ( ))(t13  N 3 )
 1                                        ]                                         Vendor’s production rate per year, K = 5
        2                    6
 PI  M 2 M 3                   M4                                                                          0 ( )
 e 0[              (  r )       (  r )2 ]                             (47)     Deterioration rate,             = 0.01
   T2      2     6                8
                                                                                      First delay period, M= 0.2
TCv = OCv + HCv + DCv - IC3.3
                                                                                      Second delay period, N= 0.4
= [C      2 (T1  T2 ) ]  1 [( K  1) C {T1  T1 (  2r )  T1 r (r   )
                      2    2                        2    3              4

    vs 1   2                                  0 hv                                    The interest earned, Ie = 0.05
                       2         T2                2 6                 8
                                                                                      The interest charged, Ic1 = 0.10
     T15 r 2    AT 4 AT 5 (3  2r )      ( K  1)0 T13 T14
                1  1               }]  2          [       (  2r )              The interest charged, Ic2 = 0.20 (Ic1 > Ic2)
       20        12         60                 T2       3   8
                                                                                      Backlogging rate, B=0
 T5             AT 5 C  e rT1 T2 2 T23 (  2r ) r 2T2 4 AT2 4
 1 r (r   )  1 ]  bh 0    [                        
 10             15        T      2         6         24     24                                                    Table 1:

    AT25 (4  3r ) 0 e rT1  2 T23 T2 4 (3  2r ) I 0     rt 2
                  ]            [                  ]  [t1  1                      N      T2            t1             VC        BC         TC
          60            T2         6         24         T      2
                                                                                      1   0.827183     0.800625         1757.09   1405.95    3163.03

                t13 0 t12          t3             t 4  r 2 t15 At14
(r 2  A)         ]  [  (  2r ) 1  r (r   ) 1                               2   0.942755     0.456282         2086.02   1517.28    3603.30
                 6    T 2            6              8    20       12
                                                                                      3      1.02889   0.331991         2274.14   1790.02    4064.16
           At 5        ( K  1)0 A T13 T14 (  2r ) T15 r (r   )
(3  2r ) 1 ]  Cv [             {                               }
           30               T2       3        8             10                        4      1.10067   0.266369         2425.17   2083.54    4508.71

                                                                                      5      1.16312   0.225188         2559.84   2374.64    4935.49
  e rT1 0 A T23 T2 4 (3  2r )
            {                   }]          -
      T2       6         24
PI e 0 M 2     M3            M4                                               (48)
         [        (  r )      (  r ) 2 ]
     T2           2       6                    8

To minimize the total cost per unit time, the optimum value
of t1, T2 is the solution of following equation.


                                                                             www.ijmer.com                                              3648 | Page
                             International Journal of Modern Engineering Research (IJMER)
                www.ijmer.com        Vol. 2, Issue. 5, Sep.-Oct. 2012 pp-3641-3649       ISSN: 2249-6645



                               Table 2:                           AKNOWLEGEMENT
                                                                  We are thankful to Dr.A.P.Singh of SGRR (PG) College
          T2              t1         VC        BC        TC       Dehradun for his constant support and review of our
N                                                                 research paper. We also thank Prof. Yudhveer Singh
                                                                  Moudgil of Uttranchal Institute of Technology for
 1    0.792393     0.745431        1966.30   1774.98   3741.28    reviewing our research paper.
 2    0.921355     0.433375        2435.64   1877.95   4313.59
                                                                  OBSERVATION
 3     1.01214     0.317385        2708.53   1969.85   4678.39
                                                                  The data obtained clearly shows that individual optimal
 4     1.08612     0.255438        2927.92   2215.75   5143.68    solutions are very different from each other. However, there
                                                                  exists a solution which ultimately provides the minimum
 5     1.14978     0.216326        3121.89   2474.02   5595.9     operating cost to the whole supply chain. All the
                                                                  observations can be summed up as follows:
                                                                  1. An increase in the interest charged, increases the buyer
                               Table 3:                              cost BC and decrease the vendor cost VC of the
                                                                     commodity.
N       T2            t1            VC         BC        TC
                                                                  2. Optimal solution for the buyer is n=1 in table first while
1    0.792393     0.745431        1780.22    1823.14   3603.36       for the vendor, it is n=5 in table 4. The overall optimal
                                                                     solution which ultimately minimizes the cost across the
2    0.921355     0.433375        1934.79    1957.21   3892.00       whole supply chain is n=5 in table 4

3     1.01214     0.317385        2265.29    2049.26   4314.55    REFERENCES
                                                                  1)    Hwang H. S. (1997). A Study on an inventory model
4     1.08612     0.255438        2315.26    2320.36   4635.62          for items with Weibull ameliorating. Com. & Ind.
                                                                        Eng. 33, 701-704.
5     1.14978     0.216326        2497.69    2546.68   5044.37    2)    Hwang H. S. (1999). Inventory Model for Items for
                                                                        both Deteriorating and Ameliorating Items. Com. &
                                                                        Ind. Eng. 37, 257-260.
                               Table 4:                           3)    Hwang, H. S. (2004): A Stochastic Set-covering
                                                                        Location Model for Both Ameliorating and
N       T2           t1             VC         BC        TC             Deteriorating Items. Com. & Ind. Eng. 46, 313-319.
                                                                  4)    Law S. T. and Wee H. M. (2006). An integrated
1    1.43526      0.40970         1524.28    6918.29   8442.57          production-inventory model for ameliorating and
                                                                        deteriorating items taking account of time
2    2.22410      0.41285         1328.68    4504.47   5833.15
                                                                        discounting. Math. & Comp. Mod., 43, 673-685.
                                                                  5)    Mondal B., Bhunia A. K. and Maiti M. (2003). An
3    2.69166     0.416398         1270.01    3076.3    4346.31
                                                                        inventory system of ameliorating items for price
4    3.01918     0.420088         1172.2     2283.4    3455.6           dependent demand rate. Com. & Ind. Eng. 45, 443-
                                                                        456.
5    3.27318     0.423883         1032.2     1856.66   2888.86    6)    Moon I., Giri B. C. and Ko B. (2005). Economic order
                                                                        quantity models for ameliorating/ deteriorating items
                                                                        under inflation and time discounting. Euro. Jour.
V. CONCLUSION                                                           Oper. Res. 162, 773-785.
Here we have studied a two echelon supply chain with
some very realistic assumptions. We studied our model in a
progressive credit period. No doubt, this assumption
imparts an economic viability to the whole study. In real
world, it is noted that, as a result of progressive permissible
delay in settling the replenishment account, the economic
replenishment interval and order quantity generally increase
marginally, although the annual cost decreases
considerably. The saving in cost as a result of permissible
delay in settling the replenishment account largely come the
ability to delay payment without paying any interest. As a
result of increasing order quantity under conditions or
permissible delay in payments, we need to order less often.
So this EOQ model is applicable when supplier gives the
trade credit to the retailer.

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