# Introduction

he Economic Lot Scheduling Problem (ELSP) is assumed that the production facility in the incontrol state producing items of high quality. It finds the problem of production sequence, production times and idle time of several products. It will minimize the inventory and setup cost also. In this model, the items are produced and consumed simultaneously for a portion of the cycle time. The rate of consumption of items is varying throughout the month. The cost of production per unit is same irrespective of production lot size. Here stockout is permitted. It is assumed that the stockout will be satisfied from the units produced at a later date with a penalty. The items are not produced between the period, while the inventory consumes and the next cycle begins. Then another item might be produced. There must be a setup time between the two items. The total cycle length is T.

A particular product is produced at a rate of P, the demand of that product is D. Then, inventory will built up at a rate of P-D. Because the product consumes while production. The built up inventory will consume at a certain number of period, then cycles begins again. The operation of this model is shown in Fig. 1.

In the economic lot scheduling problem, it is not assumed that changeover times is sequence dependent. So, when the changeover times are sequence independent, then the economic lot scheduling problem essentially tries to minimize the total cost which is the sum of the ordering cost and carrying cost (Srinivasan, G., Quantitative Models in Operations  


# Development of Disaggregation Method with Stockout

Three products are produced such as sacking, Hessian, CBC (Carpet Baking Clothe). The daily demand and the inventory of these products remain constant. The demand and inventory of the products are summarized form the last three years data as shown in Table 1. The capacity in each of the 3 days are 35, 37, 33 tons/day, respectively. Allocation time have to find for making the products.

The value of r represents the demand that can be met with the existing inventory. The production of jproduct has to be started before r j hours. The products are sorted according to increasing value of r. The order is found as Sacking-Hessian-CBC. The products will be produced in the said order. The process flow also depends upon the value of r. The maximum value of r is 0.666, from which the cycle time is counted considering the demand constant.

It is assumed that, the reasonable upper limit of the cycle time is (?? +    -------------------------------------------------(   The maximum value of r is 6.65, from which the cycle time is counted, that is 1 day. Again, it is assumed that, the upper limit of the cycle time is (  --------------------------------------------------(  ---------------(  ----------------(5) ?? ?????? ? 0.857[ production time limit for CBC] --------------( 6) ?? ???? ? 6.5[ production time limit for Hessian] -------------( 7)

?? ???? , ?? ?????? , ?? ???? , ?? ? 0 TORA software is used for solving the problem. The optimal solution of our problem is given by, 
?? 1 = ?? ???? = 0 ? 0, ??2

# Result and Discussion

This paper translates the ELS of a Jute industry for time varying demand with Stock out. At the beginning of every cycle, the existing inventories are worked out. The expected inventory at the end of each cycle has been calculated. These values are used to compute r j for finding the sequence of production. The order of production has changed in the two cycles because of the values of r j . Again the shortage of inventory in each cycle has also changed the order of production. Therefore, it is observed that, the order of production does not depend upon the values of r j , but depends on the values of stockout of inventories. The change of order satisfies all the constraints and factors. Finally the starting time of each product is calculated by LP. This is acceptable and the model provides flexibility in this regard.   


# IV.
1![Fig.1 : (a) Manufacturing model of inventory,(b) Total cycle time for a particular facility](image-2.png "Fig. 1 :")
1![= 2.166 . Therefore, capacity for 2.166 days,the equivalent daily demands are 18, 6 and 10 tons/day for the three products respectively and equivalent daily capacity is 35 tons/day. Now, using the demand of each product to construct a manufacturing model with shortages (Panneerselvam, R., Production and operations management, 2nd edition, Chapter-9, Page-214), For Sacking, D = 18 tons/day, P = 25 tons/day, C 0 = 0.25, C c = 0.10, C s = 0.50 Economic Batch Quantity (EBQ), Maximum inventory, ?? 1 = ? 2?? 0 ?? ?? × ??(?????) ?? × ?? ?? ?? ?? +?? ?? = ? 10+0.50) = 4.5825 tons/day Maximum stockout, ?? 2 = ? 2?? 0 ?? ?? ?? ?? (?? ?? +?? ?? )](image-3.png "1 ??)")
1![?? 2 + ?? 3 = 0.6546 + 0.2545 + 0.0509 = 0.96 ?? = ?? 1 + ?? 2 + ?? 3 + ?? 4 = 0.6546 + 0.2545 + 0.0509 + 0.1309 = 1.091 Similarly for CBC, D = 6 tons/day, P = 10 tons/day, C 0 = 0.25, C c = 0.](image-4.png "??? = ?? 1 +")
![1) (?? ???? ? ?? ???? )35 + 1.091 × 35 ? T × 18[ production time for Sacking] ---------(2) (?? ?????? ? ?? ???? ) 35 + 1.5811 × 35 ? T × 10[ production time for Hessian] -----(3) (?? ???? + ?? ? ?? ?????? ) 35 + 1.3416 × 35 ? T × 7[ production time for CBC] -------(4) ?? ???? ? 0.555[ production time limit for Sacking] -------------(5) ?? ?????? ? 0.666[ production time limit for CBC]-----------------(6) ?? ???? ? 0.636[ production time limit for Hessian]-------------(7) ?? ???? , ?? ?????? , ?? ???? , ?? ? 0 Where, ?? ???? = Production start time for sacking ?? ???? = Production start time for Hessian ?? ?????? = Production start time for CBC ?? = Total cycle time Here TORA software is used for solving the problem.The optimal solution is given by, ?? 1 = ?? ???? = 0 ? 0.555, ?? 2 = ?? ?????? = 0.0143 ? 0.666, ?? 3 = ?? ???? = 0.636 ? 0.636, ?? 4 = ?? = 3.3581 This becomes the first cycle and this is implemented for ?? = 3.3581 days. The inventories of the three products at the end of 3.3581 days aresummarized in the table 2.](image-5.png "")
23![Fig. 2 : Inventory model with stockout for (a) Sacking, (b) CBC and (c) Hessian](image-6.png "Fig. 2 :Fig. 3 :")
11![= 6.65 . Therefore, capacity for 6.65 days, the equivalent daily demands of the three products are found 7, 3 and 5 tons/day respectively and equivalent daily capacity is 35 tons/day. Again, the model gives the following values considering the stockout. For Sacking, D = 7 tons/day, P = 10 tons/day, C 0 = 0.25, C c = 0.10, C s = 0.50 ?? = 11.83, ?? 1 = 2.95, ?? 2 = 0.59, ?? = 1.69, ?? 1 = 0.9833, ?? 2 = 0.4214, ?? 3 = 0.0845, ?? 4 = 0.1966, ??? = 1.4892 For CBC, D = 3 tons/day, P = 7 tons/day, C 0 = 0.25, C c = 0.10, C s = 0.50 ?? = 5.61, ?? 1 = 2.67, ?? 2 = 0.53, ?? = 1.81, ?? 1 = 0.6675, ?? 2 = 0.89, ?? 3 = 0.1766, ?? 4 = 0.1325, ??? = 1.7341 For Hessian, D = 5 tons/day, P = 8 tons/day, C 0 = 0.25, C c = 0.10, C s = 0.50 ?? = 8.94,?? 1 = 2.79, ?? 2 = 0.559, ?? = 1.788, ?? 1 = 0.93, ?? 2 = 0.558, ?? 3 = 0.1118, ?? 4 = 0.1863, ??? = 1.599 LP formulation is Objective function: maximize T Subject to ?? ?? ? ?? ??](image-7.png "1 + 1 ??)")
1![?? ?????? ? ?? ???? )35 + 1.69 × 35 ? T × 7[ production time for Sacking] -----------(2) (?? ???? ? ?? ?????? ) 35 + 1.87 × 35 ? T × 3[ production time for CBC]](image-8.png "1 )(")
![3) (?? ???? + ?? ? ?? ???? ) 35 + 1.78 × 35 ? T × 5[ production time for Hessian] -------(4) ?? ???? ? 0[ production time limit for Sacking]](image-9.png "")




1SackingCBCHessianInventory (tons/day)1047Demand (tons/day), D 118611P 1 = 35 tons/dayD 220710P 2 = 37 tons/dayD 31698P 3 = 33 tons/dayr = Inventory/Demand0.5550.6660.636
2Year 201615( ) Volume XVI Issue I Version Iof Researches in EngineeringGlobal JournalSackingCBCHessianInventory (tons/day)-50665G © 2016 Global Journals Inc. (US)
			Year 2016 G © 2016 Global Journals Inc. (US)
		
		

			

## Acknowledgements

The author wishes his thanks to Managing Director of Khalishpur Jute Mill for his kind helped in this regard.

			

			

				


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