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The article deals with an economic order quantity (EOQ) inventory model for deteriorating items in which the supplier provides the purchaser a permissible delay in payment. This is so when deterioration of units in the inventory is subject to constant deterioration rate, demand rate is quadratic function of time and salvage value is associated with the deteriorated units. Shortages in the system are not allowed to occur. A mathematical formulation is developed when the supplier offers a permissible delay period to the customers under two circumstances: 1) when delay period is less than the cycle of time; and 2) when delay period is greater than the cycle of time. The method is suitable for the items like state-of-the-art aircrafts, super computers, laptops, android mobiles, seasonal items and machines and their spare parts. A solution procedure algorithm is given for finding the optimal order quantity which minimizes the total cost of an inventory system. The article includes numerical examples to support the effectiveness of the developed model. Finally, sensitivity analysis on some parameters on optimal solution is provided.

It is commonly observed that most of the physical goods in which appreciable deterioration can take place when the item in stock undergoes changes or becomes out of fashion and consequently the loss must be taken into account when analyzing the model. Deterioration is a natural process which is defined as change, decay, evaporation, loss of utility or marginal value of the commodity. Thus, to control and maintain the inventory of deteriorating items to satisfy customer’s demand or retailer’s order is very important in nowadays. Of late, many models have been developed for the control and maintenance of the inventory. Generally, for items like hardware, glassware, steel and toys, the rate of deterioration is too low; and there is little need for considering deterioration of the economic lot-size. But items like seasonal food, vegetables, fruit, blood, fish, meat, radioactive substances, alcohol, chemicals, gasoline, drugs, medicine etc. deteriorate remarkably overtime. In the history, the inventory models for deteriorating items have been continuously modified to become more practicable and realistic. Whitin [

Recently, Khanra et al. [

The remainder of the paper is arranged as follows: In Section 2, the review of literature is presented. In Sections 3 and 4, the notations and fundamental assumptions are used throughout this paper, respectively. Mathematical model with the necessary and sufficient conditions and algorithms of the solution of the model in order to minimize the total relevant inventory costs are given in Sections 5 and 6, respectively. In Sections 7 and 8, numerical examples and the sensitivity analysis of the various parameters are presented to illustrate the model, respectively. Finally, conclusions are drawn and the future research is pointed out in Section 9.

Demand is considered as the driving force of the inventory system. Therefore, its role is important for the development of inventory system of deteriorating items. While developing an inventory model, most of the inventory researchers usually consider the time-dependent demand either linear or exponential for the whole cycle. But in real life situations, the time-varying linear or exponential demand pattern seldom occurs because the linear demand pattern represents the uniform change in demand whereas the other indicates the rapid change in demand. Dash et al. [

In the conventional EOQ inventory model, the costs of the items are assumed to be paid at the time of delivery by the supplier. However, this assumption is not always suitable for business practices, as the supplier allows credit facilities to attract more customers for business competition situations. Such an advantage is likely to motivate customer to order more quantities because paying later indirectly reduces the purchase cost. In business competitions, the practical scenario for the supplier to survive in the market is to offer customers some grace period enabling them to pay later. The customer does not have to pay any interest during this fixed period, but if the payment gets delayed, the supplier will charge interest for the period. Generally, the credit period in which the suppliers offer to the retailers with interest is known as the trade credit period or permissible delay period or delay period. During this period, he may sell the goods, accumulate revenues on the sales and earn interest on that revenue. In other words, trade credit period is a powerful promotional tool by which suppliers encourage and attract the retailers. Therefore, trade credit plays an important role in inventory control for both the supplier and the customers. In business market, the unit selling price should be greater than the unit purchasing price. Generally, suppliers offer delay period on purchase of items to the retailer. During this period, the retailer is encouraged to buy more items and accumulate revenues by selling items and earning interest. Initially, Goyal [

Most of the inventory models developed assumed that the deterioration of a unit is a complete loss and that these deteriorated units have no sale value. They are considered as lost. But, in real life situations, the supplier can offer a fixed reduced unit cost to his retailer for the deteriorated stock in order to reduce the total inventory cost. In other words, inclusion of salvage value into the deteriorated stock benefits both the supplier and retailer. The proposed strategy can be implemented in inventory control model of selling seasonal items, fashion items, automobiles, smart phones and machines and their spare parts. In the several articles, the models assumed that deteriorated units have salvage values. So they are considered as lost in business. To overcome such loss, supplier can offer to his retailer reduced unit cost for the deteriorated stocks. In this respect, Jaggi and Aggarwal [

The following assumptions are needed for developing the mathematical model:

1) The deterioration rate is constant for the period, which is practically very small.

2) A single type of item is considered over a prescribed period.

3) The delivery lead time (i.e., the length of time between making a decision to replenish an item and its actual addition to stock) is zero.

4) Replacement rate occurs instantaneously.

5) The demand rate is known and is a quadratic increasing function of time.

6) The planning horizon of the inventory system is infinite.

7) No shortages in inventory are allowed.

8) The supplier offers the purchaser a delay period in paying for purchasing cost and the purchaser can accumulate revenues by selling items and by earning interest.

The paper is developed considering the replenishment problem of a single deteriorating item. The inventory system starts at time t = 0 when a lot size of a certain number of units enters the system and ends with zero inventory at time t = T . The depletion of inventory occurs due to combined effect of the time- dependent quadratic demand rate and constant deterioration rate in time period 0 ≤ t ≤ T . Thus, the governing differential equation of the instantaneous state of inventory level I ( t ) at any time t is given by

d I ( t ) d t + θ I ( t ) = − R ( t ) , 0 ≤ t ≤ T , (1)

with the boundary conditions I ( 0 ) = I o (1a)

and

I ( T ) = 0 , (1b)

where R ( t ) = a + b t + c t 2 .

The solution of Equation (1) is given by

I ( t ) = ( a + b T + c T 2 θ − b + 2 c T θ 2 + 2 c θ 3 ) e θ ( T − t ) − a + b t + c t 2 θ + b + 2 c t θ 2 − 2 c θ 3 , 0 ≤ t ≤ T (2)

and the order quantity is

I o = I ( 0 ) = ( a + b T + c T 2 θ − b + 2 c T θ 2 + 2 c θ 3 ) e θ T − a θ + b θ 2 − 2 c θ 3 (3)

Now, the model is analyzed under three possibilities depending upon the relationship between delay period and cycle time.

Case A: t p < T . (Delay period is less than the cycle time).

The total variable cost comprises the sum of the ordering cost, holding cost, deterioration cost and interest chargeable minus the sum of the salvage value of the deteriorated items and interest earned. They are grouped together after evaluating the above costs individually.

1) The ordering cost ( O C ) is

O C = C 3 (4)

2) The deterioration cost ( D C ):

The total demand during the time period [ 0 , T ] is

∫ 0 T R ( t ) d t = a T + b T 2 2 + c T 3 3 (5)

The total number of deteriorated units is given by

I 0 − ∫ 0 T R ( t ) d t = ( a + b T + c T 2 θ − b + 2 c T θ 2 + 2 c θ 3 ) e θ T − a θ + b θ 2 − 2 c θ 3 − a T − b T 2 2 − c T 3 3 (6)

Thus, the deterioration cost ( D C ) for the period [ 0 , T ] is

D C = C 2 [ ( a + b T + c T 2 θ − b + 2 c T θ 2 + 2 c θ 3 ) e θ T − a θ + b θ 2 − 2 c θ 3 − a T − b T 2 2 − c T 3 3 ] (7)

3) The salvage value ( C S V ) for deteriorated items for the period [ 0 , T ] is

C S V = χ [ ( a + b T + c T 2 θ − b + 2 c T θ 2 + 2 c θ 3 ) e θ T − a θ + b θ 2 − 2 c θ 3 − a T − b T 2 2 − c T 3 3 ] (8)

4) The holding cost ( H C ) for the period [ 0 , T ] is

H C = C 1 ( e θ T − 1 ) θ 2 ( a + b T + c T 2 − b + 2 c T θ + 2 c θ 2 ) − C 1 T θ [ a + b T 2 + c T 2 3 − b + c T θ + 2 c θ 2 ] (9)

5) The interest payable ( P I A ) for the period [ 0 , T ] is

P I A = C 2 I c θ 2 ( e θ ( T − t p ) − 1 ) ( a + b T + c T 2 − b + 2 c T θ + 2 c θ 2 ) − C 2 I c ( T − t p ) θ [ a + b ( T + t p ) 2 + c ( T 2 + T t p + t p 2 ) 3 − b + c ( T + t p ) θ + 2 c θ 2 ] (10)

where C 1 = h 1 C 2 .

6) The interest earned ( E I A ) for the period [ 0 , T ] is

E I A = p I e ∫ 0 T t R ( t ) d t = p I e T 2 ( a 2 + b T 3 + c T 2 4 ) . (11)

The total cost function for the period [ 0 , T ] is defined as

T C A ( T ) = O C + H C + ( D C − C S V ) + P I A − E I A . (12)

The total variable cost per unit time ( T V C A ( T ) ) for the period [ 0 , T ] is

T V C A ( T ) = 1 T [ O C + H C + ( D C − C S V ) + P I A − E I A ] = C 3 T + C 1 θ T [ ( a + b T + c T 2 − b + 2 c T θ + 2 c θ 2 ) ( e θ T − 1 θ ) − a T − b T 2 2 − c T 3 3 + b T + c T 2 θ − 2 c T θ 2 ] + ( C 2 − χ ) T [ ( a + b T + c T 2 θ − b + 2 c T θ 2 + 2 c θ 3 ) e θ T − a θ + b θ 2 − 2 c θ 3 − a T − b T 2 2 − c T 3 3 ]

+ C 2 I c θ T ( a + b T + c T 2 − b + 2 c T θ + 2 c θ 2 ) ( e θ ( T − t p ) − 1 θ ) − C 2 I e T ( a 2 + b T 3 + c T 2 4 ) − C 2 I c ( T − t p ) θ T [ a + b ( T + t p ) 2 + c ( T 2 + T t p + t p 2 ) 3 − b + c ( T + t p ) θ + 2 c θ 2 ] (13)

The objective of the problem is to determine the optimal value of T so that T V C ( T ) is minimized. The necessary condition to minimize T V C A ( T ) for a given value of t p is

d [ T V C A ( T ) ] d T = 0 (14)

provided it satisfies the condition d 2 [ T V C A ( T ) ] d T 2 > 0 .

From Equation (14), the respective non-linear equation is

d [ T V C A ( T ) ] d T = ( a + b T + c T 2 ) [ ( C 1 θ + C 2 − χ ) ( e θ T − 1 ) + C 2 I c θ ( e θ ( T − t p ) − 1 ) − C 2 I e T ] − [ T V C A ( T ) ] = 0 (15)

The second order of T V C A ( T ) with respect to T is as follows:

d 2 [ T V C A ( T ) ] d T 2 = ( a + b T + c T 2 ) T [ ( C 1 θ + C 2 − χ ) θ e θ T + C 2 I c e θ ( T − t p ) − C 2 I e ] − ( 2 a + b T ) T 2 [ ( C 1 θ + C 2 − χ ) ( e θ T − 1 ) + C 2 I c θ ( e θ ( T − t p ) − 1 ) − C 2 I e T ] + 2 T 2 [ T V C A ( T ) ] (16)

Case B: t p > T . (Delay period is greater than the cycle time)

Here, total variable cost comprises the sum of the ordering cost, holding cost and deterioration cost interest chargeable minus the sum of the salvage value of the deteriorated items and interest earned as interest chargeable is zero. The ordering cost, holding cost, deterioration cost and the salvage value of the deteriorated items are same as Case A.

Now, the total interest earned ( E I B ) during the cycle time is given by the sum of the interest earned during the period [ 0 , T ] and the interest earned during the delay period [ T , t p ] . Thus,

E I B = C 2 I e ∫ 0 T t R ( t ) d t + C 2 I e ( t p − T ) ∫ 0 T R ( t ) d t = C 2 I e T [ ( a + b T 2 + c T 2 3 ) t p − a T 2 − b T 2 6 − c T 3 12 ] (17)

Here, the total cost function in this case is defined as

T C B ( T ) = O C + H C + ( D C − C S V ) − E I B (18)

The total variable cost per unit time ( T V C B ( T ) ) in this case is

T V C B ( T ) = 1 T [ O C + H C + ( D C − C S V ) − E I B ] = C 3 T + C 1 θ T [ ( a + b T + c T 2 − b + 2 c T θ + 2 c θ 2 ) ( e θ T − 1 θ ) − a T − b T 2 2 − c T 3 3 + b T + c T 2 θ − 2 c T θ 2 ] + ( C 2 − χ ) T [ ( a + b T + c T 2 θ − b + 2 c T θ 2 + 2 c θ 3 ) e θ T − a θ + b θ 2 − 2 c θ 3 − a T − b T 2 2 − c T 3 3 ] − C 2 I e [ ( a + b T 2 + c T 2 3 ) t p − a T 2 − b T 2 6 − c T 3 12 ] (19)

The necessary condition to minimize T V C B ( T ) for a given value of t p is

d [ T V C B ( T ) ] d T = 0 (20)

provided it satisfies the condition d 2 [ T V C B ( T ) ] d T 2 > 0 .

From (20), the respective non-linear equation is

d [ T V C B ( T ) ] d T = ( a + b T + c T 2 ) ( e θ T − 1 ) ( C 1 θ + C 2 − χ ) − C 2 I e [ t p ( a + b T + c T 2 ) − a T − b T 2 2 − c T 3 3 ] − [ T V C B ( T ) ] = 0 (21)

The second order derivative of T V C B ( T ) with respect to T is as follows:

d 2 [ T V C B ( T ) ] d T 2 = ( a + b T + c T 2 ) T [ ( C 1 θ + C 2 − χ ) θ e θ T + C 2 I c e θ ( T − t p ) − C 2 I e ] − ( 2 a + b T ) T 2 [ ( C 1 θ + C 2 − χ ) ( e θ T − 1 ) + C 2 I c θ ( e θ ( T − t p ) − 1 ) − C 2 I e ] + 2 T 2 [ T V C B ( T ) ] (22)

Case C: t p = T . (Delay period is equal to the cycle time).

The respective cost function is obtained from either Equation (13) or Equation (19) by substituting T = t p because of both of ( T V C A ( T ) ) and ( T V C B ( T ) ) are identical.

Based on the results above, a procedure is derived to locate the optimal cycle time for the two cases.

The following Solution procedure is recommended for the calculation of EOQ and optimal solution.

Step I: Perform (1)-(9).

1) Assign values to the parameters.

2) Solve equation (15) for T 1 * .

3) Test the respective sufficient condition.

4) Compare T 1 * with t p .

5) If T 1 * > t p , then substitute the value of T 1 * in Equation (13) to get T V C A ( T 1 * ) .

6) Solve Equation (21) for T 2 * .

7) Test the respective sufficient condition.

8) Compare T 2 * with t p .

9) If T 2 * < t p , then substitute the value of T 2 * in Equation (19) to get T V C B ( T 2 * ) .

Then the following decision will be held.

Decision I. If T 1 * > t p > T 2 * is satisfied, then the optimal cost T V C ( T * ) , i.e., the minimum cost is obtained by comparing both T V C A ( T 1 * ) and T V C B ( T 2 * ) and evaluate the corresponding optimal order quantity I o * from Equation (3).

Step II: Decision II. If T 1 * > t p and T 2 * > t p are satisfied, then the optimal cost T V C ( T * ) is T V C A ( T 1 * ) and evaluate the corresponding optimal order quantity I o * from Equation (3).

Decision III. If T 1 * < t p and T 2 * < t p are satisfied, then the optimal cost T V C ( T * ) is T V C B ( T 2 * ) and evaluate the corresponding optimal order quantity I o * from Equation (3).

The following numerical examples are presented in order to demonstrate the above solution procedure which can be applied to determine the optimal solution.

Example 1. (Case A and Case B): Minimum average cost is T V C A ( T 1 * ) .

Let a = 1000 , b = 150 , c = 15 , θ = 0.20 , C 1 = 0.12 , C 2 = 20 , C 3 = 200 , χ = 0.02 , I c = 0.15 , I e = 0.13 , and t p = 0.25 in appropriate units. Solving Equation (13), we get T 1 * = 0.351257 year and putting T 1 * = 0.351257 year in Equation (13), the corresponding average cost is T V C A ( T 1 * ) = $ 939.98 which

satisfies the sufficient condition (16), i.e., d 2 [ T V C A ( T 1 * ) ] d T 2 = 15048.8 > 0 .

Similarly, solving Equation (21), we get T 2 * = 0.238718 year and putting T 2 * = 0.238718 year in Equation (19), the corresponding average cost is

T V C B ( T 2 * ) = $ 1001.42 which satisfies the sufficient condition (22), i.e.,

d 2 [ T V C B ( T 2 * ) ] d T 2 = 30273.846 > 0 .

In this case, by Step I, the optimal average cost is T V C A ( T 1 * ) = $ 939.98 and the corresponding cycle length is T 1 * = 0.351257 year and the corresponding EOQ is I o * ( T 1 * ) = 373.846 .

Example 2. (Case A and B): Minimum average cost is T V C B ( T 2 * ) .

Let a = 1000 , b = 150 , c = 15 , θ = 0.20 , C 1 = 0.12 , C 2 = 20 , C 3 = 200 , χ = 0.02 , I c = 0.15 , I e = 0.13 and t p = 0.35 in appropriate units. Solving Equation (15), we get T 1 * = 0.401514 year and putting T 1 * = 0.401514 year in Equation (13), the corresponding average cost is T V C A ( T 1 * ) = $ 884.336 which

satisfies the sufficient condition (16), i.e., d 2 [ T V C A ( T 1 * ) ] d T 2 = 13280.9 > 0 .

Similarly, solving Equation (21), we get T 2 * = 0.239385 year and putting T 2 * = 0.239385 year in Equation (19), the corresponding average cost is T V C B ( T 2 * ) = $ 736.681 which satisfies the sufficient condition (22), i.e.,

d 2 [ T V C B ( T 2 * ) ] d T 2 = 30026.2 > 0 .

In this case, by Step I, the optimal average cost is T V C B ( T 2 * ) = $ 736.681 and the corresponding cycle length is T 2 * = 0.239385 year and the corresponding EOQ is I o * ( T 2 * ) = 249.717 .

The effect of changing of the several parameters on the optimal cycle time and the optimal total cost is studied. The sensitivity analysis of the parameters present in this model is also performed. The optimal values of the total average cost T V C ( T * ) change significantly with changes (−50%, −25%, −10%, +10%, +25%, +50%) of different parameters value in

On the basis of sensitivity analysis of the parameters, the following features

Changing parameters | Change in parameters | Cycle length ( T 1 * ) | Total cost ( T V C A ( T 1 * ) ) | Cycle length ( T 2 * ) | Total cost ( T V C B ( T 2 * ) ) | Optimal cycle length ( T * ) | Optimal total cost ( T V C ( T * ) ) |
---|---|---|---|---|---|---|---|

a | +50 | 0.311248 | 1079.73 | 0.196504 | 1041.62 | 0.196504 | 1041.62 |

+25 | 0.328076 | 1012.97 | 0.214591 | 1030.43 | 0.214591 | 1030.43 | |

+10 | 0.341003 | 970.087 | 0.228140 | 1015.52 | 0.228140 | 1015.52 | |

−10 | 0.363206 | 908.398 | 0.250887 | 983.430 | 0.363206 | 908.398 | |

−25 | 0.385391 | 857.595 | 0.273124 | 947.763 | 0.385391 | 857.595 | |

−50 | 0.441043 | 759.892 | 0.327654 | 856.993 | 0.441043 | 759.892 | |

b | +50 | 0.347954 | 945.380 | 0.237367 | 1003.39 | 0.347954 | 945.380 |

+25 | 0.349579 | 942.696 | 0.238036 | 1002.41 | 0.349579 | 942.696 | |

+10 | 0.350579 | 941.070 | 0.238444 | 1001.82 | 0.350579 | 941.070 | |

−10 | 0.351943 | 938.885 | 0.238994 | 1001.01 | 0.351943 | 938.885 | |

−25 | 0.352989 | 937.232 | 0.239411 | 1000.41 | 0.352989 | 937.232 | |

−50 | 0.354780 | 934.451 | 0.240117 | 999.384 | 0.354780 | 934.451 | |

c | +50 | 0.354369 | 957.299 | 0.238689 | 1001.45 | 0.354369 | 957.299 |

+25 | 0.352804 | 948.581 | 0.238704 | 1001.43 | 0.352804 | 948.581 | |

+10 | 0.351873 | 943.406 | 0.238712 | 1001.42 | 0.351873 | 943.406 | |

−10 | 0.350643 | 936.572 | 0.238723 | 1001.41 | 0.350643 | 936.572 | |

−25 | 0.349727 | 931.494 | 0.238732 | 1001.40 | 0.349727 | 931.494 | |

−50 | 0.348215 | 923.120 | 0.238746 | 1001.38 | 0.348215 | 923.120 | |

θ | +50 | 0.288355 | 1261.25 | 0.208512 | 1237.54 | 0.288355 | 1261.25 |

+25 | 0.314936 | 1104.92 | 0.222126 | 1123.26 | 0.314936 | 1104.92 | |

+10 | 0.335154 | 1006.81 | 0.231656 | 1051.16 | 0.335154 | 1006.81 | |

−10 | 0.370299 | 872.666 | 0.246449 | 950.183 | 0.370299 | 872.666 | |

−25 | 0.407349 | 774.140 | 0.259560 | 870.214 | 0.407349 | 774.140 | |

−50 | 0.528423 | 674.344 | 0.286847 | 727.005 | 0.528423 | 674.344 | |

C 1 | +50 | 0.351931 | 966.618 | 0.237652 | 1008.85 | 0.351931 | 966.618 |

+25 | 0.351593 | 953.285 | 0.238183 | 1005.14 | 0.351593 | 953.285 | |

+10 | 0.351391 | 945.299 | 0.238503 | 1002.91 | 0.351391 | 945.299 | |

−10 | 0.351122 | 934.666 | 0.238933 | 999.925 | 0.351122 | 934.666 | |

−25 | 0.350902 | 926.703 | 0.239256 | 997.685 | 0.350902 | 926.703 | |

−50 | 0.350585 | 913.455 | 0.239798 | 993.945 | 0.350585 | 913.455 | |

C 2 | +50 | 0.310932 | 1047.57 | 0.196518 | 1036.24 | 0.196518 | 1036.24 |

+25 | 0.327754 | 1010.45 | 0.214539 | 1027.76 | 0.327754 | 1010.45 | |

+10 | 0.340807 | 969.102 | 0.228094 | 1014.46 | 0.340807 | 969.102 | |

−10 | 0.363554 | 909.325 | 0.250998 | 984.432 | 0.363554 | 909.325 | |

−25 | 0.386787 | 859.717 | 0.273639 | 950.069 | 0.386787 | 859.717 | |

−50 | 0.448166 | 762.779 | 0.330790 | 859.988 | 0.448166 | 762.779 |

C 3 | +50 | 0.402373 | 1214.36 | 0.290496 | 1379.22 | 0.402373 | 1214.36 |
---|---|---|---|---|---|---|---|

+25 | 0.377880 | 1081.84 | 0.265993 | 1199.53 | 0.377880 | 1081.84 | |

+10 | 0.362194 | 997.984 | 0.250019 | 1083.26 | 0.362194 | 997.984 | |

−10 | 0.339889 | 880.094 | 0.226802 | 915.493 | 0.339889 | 880.094 | |

−25 | 0.321920 | 786.265 | 0.207535 | 777.356 | 0.207535 | 777.356 | |

−50 | 0.288966 | 616.802 | 0.170235 | 512.707 | 0.170235 | 512.707 | |

χ | +50 | 0.351334 | 939.621 | 0.238753 | 1001.17 | 0.351334 | 939.621 |

+25 | 0.351215 | 939.801 | 0.238736 | 1001.29 | 0.351215 | 939.801 | |

+10 | 0.351272 | 939.908 | 0.238725 | 1001.37 | 0.351272 | 939.908 | |

−10 | 0.351241 | 940.052 | 0.238711 | 1001.47 | 0.351241 | 940.052 | |

−25 | 0.351218 | 940.160 | 0.238700 | 1001.54 | 0.351218 | 940.160 | |

−50 | 0.351179 | 940.339 | 0.238682 | 1001.66 | 0.351179 | 940.339 | |

I c | +50 | 0.329930 | 955.000 | 0.238718 | 1001.42 | 0.329930 | 955.000 |

+25 | 0.339316 | 948.231 | 0.238718 | 1001.42 | 0.339316 | 948.231 | |

+10 | 0.346109 | 943.489 | 0.238718 | 1001.42 | 0.346109 | 943.489 | |

−10 | 0.357000 | 936.148 | 0.238718 | 1001.42 | 0.357000 | 936.148 | |

−25 | 0.366988 | 929.687 | 0.238718 | 1001.42 | 0.366988 | 929.687 | |

−50 | 0.388717 | 916.463 | 0.238718 | 1001.42 | 0.388717 | 916.463 | |

I e | +50 | 0.411444 | 694.678 | 0.219984 | 821.307 | 0.411444 | 694.678 |

+25 | 0.377918 | 821.947 | 0.228784 | 921.857 | 0.377918 | 821.947 | |

+10 | 0.361248 | 893.754 | 0.234593 | 966.376 | 0.361248 | 893.754 | |

−10 | 0.342033 | 985.006 | 0.243064 | 1035.91 | 0.342033 | 985.006 | |

−25 | 0.329441 | 1050.48 | 0.250046 | 1086.58 | 0.329441 | 1050.48 | |

−50 | 0.311183 | 1154.69 | 0.263125 | 1167.84 | 0.311183 | 1154.69 | |

t p | +50 | 0.415372 | 877.827 | 0.239553 | 670.495 | 0.239553 | 670.495 |

+25 | 0.381666 | 899.200 | 0.239134 | 835.958 | 0.239134 | 835.958 | |

+10 | 0.362924 | 921.031 | 0.238884 | 935.234 | 0.362924 | 921.031 | |

−10 | 0.340300 | 962.853 | 0.238552 | 1067.60 | 0.340300 | 962.853 | |

−25 | 0.325361 | 1005.35 | 0.238303 | 1166.87 | 0.325361 | 1005.35 | |

−50 | 0.305092 | 1101.34 | 0.237891 | 1332.32 | 0.305092 | 1101.34 |

are observed.

1) If the initial rate of demand (a) increases, then the optimal cycle length ( T * ) decreases and the optimal average cost ( T V C ( T * ) ) increases. Here T * and T V C ( T * ) are moderately sensitive to change in a.

2) If the rate of increasing demand (b) increases, then the optimal cycle length ( T * ) decreases and the optimal average cost ( T V C ( T * ) ) increases. Here T * and T V C ( T * ) are lowly sensitive to change in b.

3) If the rate of changing demand (c) increases, then both the optimal cycle length ( T * ) and the optimal average cost ( T V C ( T * ) ) increase. Here T * and T V C ( T * ) are lowly sensitive to change in c.

4) If the rate of deterioration ( θ ) increases, then the optimal cycle length ( T * ) decreases and the optimal average cost ( T V C ( T * ) ) increases. Here T * and T V C ( T * ) are highly sensitive to change in θ .

5) If the holding cost per unit time ( C 1 ) increases, then both the optimal cycle length ( T * ) and the optimal average cost ( T V C ( T * ) ) increase. Here T * and T V C ( T * ) are lowly sensitive to change in C 1 .

6) If the unit purchase cost ( C 2 ) increases, then the optimal cycle length ( T * ) decreases and the optimal average cost ( T V C ( T * ) ) increases. Here T * and T V C ( T * ) are highly sensitive to change in C 2 .

7) If the ordering cost ( C 3 ) increases, then both the optimal cycle length ( T * ) and the optimal average cost ( T V C ( T * ) ) increase. Here T * and T V C ( T * ) are highly sensitive to change in C 3 .

8) If the salvage value ( χ ) increases, then the optimal cycle length ( T * ) increases and the optimal average cost ( T V C ( T * ) ) decreases. Here T * and T V C ( T * ) are lowly sensitive to change in χ .

9) If the interest charged per unit ( I c ) increases, then the optimal cycle length ( T * ) decreases and the optimal average cost ( T V C ( T * ) ) increases. Here T * and T V C ( T * ) are lowly sensitive to change in I c .

10) If the interest earned per unit ( I e ) increases, then the optimal cycle length ( T * ) increases and the optimal average cost ( T V C ( T * ) ) decreases. Here T * and

11) If the permissible delay period (

The proposed model assumes an EOQ model for deteriorating items with the time-dependent quadratic increasing demand under permissible delay in payment. The reason for considering quadratic demand rate is that it depicts different phases of market demand including accelerated rise or fall in demand. Shortages are not permitted in this inventory system. Salvage value is included in the deteriorated units. The model is suitable for the demand of items such as state-of-the-art aircrafts, super computers, laptops, android mobiles, seasonal items and machines and their spare parts and also newly launched fashion goods, seasonal items, cosmetics etc. for which the demand rate accelerates as they are launched into the market and declines when the season ends. The objective of the model is to optimize the cycle time, ordering cost and the total system costs. Further, several numerical examples and sensitivity analysis with respect to various parameters are presented to validate the theoretical results.

In the future study, it is hoped to extend and incorporate the proposed model into several situations, such as, varying deterioration rate like Weibull distribution and Gamma distribution. Also, we could generalize the model to incorporate quantity discounts, inflation rates and allow for shortages and partial backlogging and other things. The present idea can be extended to consider the parameter as fuzzy or stochastic fuzzy. In addition, we could extend the deterministic demand to stochastic fluctuating demand patterns.

The authors would like to express deep felt gratitude to the Editor-in-Chief and the referees for their invaluable suggestions and guidance.

The authors declare no conflicts of interest regarding the publication of this paper.

Singh, T., Muduly, M.M., Mallick, C., Gupta, R.K. and Pattanayak, H. (2019) An Ordering Policy for Deteriorating Items with Time-Dependent Quadratic Demand and Salvage Value under Permissible Delay in Payment. American Journal of Operations Research, 9, 201-218. https://doi.org/10.4236/ajor.2019.95013

The following symbols are needed for developing the mathematical model:

T: Length of the cycle time (decision variable).