Reclaiming the Solid Waste

Case 2: Reclaiming the solid waste

INTRODUCTION

        This case talks about the Save-It company who operates the reclamation center which collects the solids waste of the four types which is mixed and forms the salable product. From this material three different types of the product are formed. Although this product formed have different proportion of the material with the different quality standards.

Green Earth organization who owes the Save-It co, which is devoted in dealing with the environmental issues. So the profits of the Save-It co helps to support the organization. So the organization have raised $30000 per week as the contribution and grants to cover the entire treatment cost for the solid material. The board of directors of Green Earth has instructed the management of Save-It to divide this money among the materials in such a way that at least half of the amount available of each material is actually collected and treated.

Within this restriction the management wants to determine the amount of each product grade to produce and the exact mix of materials to be used for each grade. The objective is to maximize the net weekly profit (total sales income minus total amalgamation cost), exclusive of the fixed treatment cost of $30,000 per week that is being covered by gifts and grants.

FACTS AND FINDINGS

        The following are the tables which shows the cost of the amalgamation and selling price for each of the product is the given. And the other table shows the quantities available for collection and treatment each week, as well as the cost of treatment, for each type of material.

PRODUCT DATA FOR SAVE-IT COMPANY

SOLID WASTE MATERIAL DATA FOR THE SAVE-IT CO

Analysis

The problem could be solved through linear programing. Through which we would find our objective of maximizing the net weekly profit by selling the salable product.

OBJECTIVE:  To maximize the profit (by selling salable product)

DECISION VARIABLE :

• Let X1 be material 1 of A
• Let X2 be material 2 of A
• Let X3 be material 3 of A
• Let X4 be material 4 of A
• Let X5 be material 1 of B
• Let X6 be material 2 of B
• Let X7 be material 3 of B
• Let X8 be material 4 of B
• Let X9 be material 1 of C
• Let X10 be material 2 of C
• Let X11 be material 3 of C
• Let X12 be material 4 of C

OBJECTIVE FUNCTION:  To maximize profit,

        P= {8.50 * (X1+X2+X3+X4) + 7 * (X5+X6+X7+X8) + 5.50 * (X9+X10+X11+X12)} – {3 * (X1+X2+X3+X4) + 2.50 * (X5+X6+X7+X8) + 2 * (X9+X10+X11+X12)}

Subject to constraints,

MIXTURE SPECIFICATION:  

GRADE A

• 0.70X1-0.30X2-0.30X3-0.30X4 <= 0
• -0.40X1+0.60X2-0.40X3-0.40X4>=0
• -0.50X1-0.50X2+0.50X3-0.50X4<=0
• -0.20X1-0.20X2-0.20X3+0.80X4=0

GRADE B

• 0.50X5-0.50X6-0.50X7-0.50X8<=0
• -0.10X5+0.90X6-0.10X7-0.10X8>=0
• -0.10X5-0.10X6-0.10X7+0.90X8=0

GRADE C

• 0.30X9-0.70X10-0.70X11-0.70X12<=0

AVAILABILTY

• MATERIAL 1 : X1+X5+X9<=3000
• MATERIAL 2: X2+X6+X10<=2000
• MATERIAL 3: X3+X7+X11<=4000
• MATERIAL 4: X4+X8+X12<=1000

ADDITIONAL RESTRICTION

• MATERIAL 1 : X1+X5+X9>=1500
• MATERIAL 2: X2+X6+X10>=1000
• MATERIAL 3: X3+X7+X11>=2000
• MATERIAL 4: X4+X8+X12>=500

TREATMENT COST OF MATERIAL

• 3X1+6X2+4X3+5X4+3X5+6X6+4X7+5X8+3X9+6X10+4X11+5X12=3000

Non-negativity function: Xi >= 0

        Where, i= 1,2,3,4,,5,6,7,8,9,10,11,12

Now this would be solved through MS Excel;