Question

Coffee Blending Model: A coffee manufacturer blends three component coffee beans into three final blends of coffee. The Table below summarizes the very precise recipes for the final coffee blends, the cost and availability information for the three components, and the wholesale price per pound of the final blends. The percentages in the body of the table indicate the percentage of each component to be used in each blend.

Table: Percentage of each component to be used in each blend

_____________________________________________________________________________                                   Final Blend (percent)                

                                                                                 Cost Per            Maximum Availability

           Component                   1        2         3            Pound           Each Week (Pound)

_____________________________________________________________________________                                                                                                 

               1                 80       0         20       $0.60      40,000

               2                 50       15       35       $0.80      25,000

               3                 0         40       60       $0.55      20,000

_____________________________________________________________________________  

           Wholesale Price/lb.       $1.25   $1.50   $1.40 _____________________________________________________________________________

Weekly capacity for the processor's plant is 100,000 pounds, and the company wishes to operate at capacity. There is no problem in selling the final blends, although there is a requirement that minimum production level of 10,000, 25,000 and 30,000 pounds, respectively, be met for blends 1, 2 and 3. The manufacturer wishes to determine the number of pounds of each component that should be used in each blend so as to maximize total weekly profit.

Formulate the problem as an L.P. model.  

Could you present it as the set up in Excel?

Answer 1

Sol:

xij = number of pounds of component i used in final blend j, i= 1, 2,3,4; j =1,2,3.

            Therefore, we have

Final Blend number of pounds  

Component (pounds)	1	2	3	Cost Per  Pound	Maximum Availability Each Week (Pound)
1	x11	x12	x13	$0.60	40,000
2	x21	x22	x23	$0.80	25,000
3	x31	x32	x33	$0.55	20,000
Wholesale Price/lb.	$1.25	$1.50	$1.40

Objective fucntion is

Subject to :

For Blend Type 1=

x 11= .20(x11+x21+x31) i.e.           -0.8 x11 + 0.2 x21 + 0.2 x31 = 0

0.4 x11 - 0.6x21 + 0.4x31 = 0

0.15 x11 + 0.15 x21 -0.85 x31  = 0

0.25 x11 + 0.25 x21 + 0.25 x31 = 0

Final Blend of Type 2.

                                                -.65 x12 + .35 x22 + .35 x32 = 0

                                                .15 x12 - .85 x22 + .15 x32 = 0

                                                .20 x12 + .20 x22 - .80 x32 = 0

                                                .30 x12 + .30 x22 + .30 x32 = 0

            3.         Final Blend of Type 3.

                                                -.90 x13 + .10 x23 + .10 x33 = 0

                                                .35 x13 - .65 x23 + .35 x33 = 0

                                                .40 x13 + .40 x23 - .60 x33 = 0

                                                .15 x13 + .15 x23 + .15 x33 = 0

            4.         Plant Capacity.

            (x11 + x12 + x13) + (x21 + x22 + x23) + (x31 + x32 + x33) = 100,000

            5.         Component Availabilities.               

                                                x11 + x12 + x13 ≤ 40,000

                                                x21 + x22 + x23 ≤ 25,000

                                                x31 + x32 + x33 ≤ 20,000

  

6. Minimum Production Level Requirements:                                              

x11 + x21 + x31≥ 10,000

                                                x12 + x22 + x32 ≥ 25,000

                                                x13 + x23 + x33 ≥ 30,000

            Nonnegativity.           All   xij ≥ 0 for   1 = 1, 2, 3, j = 1, 2, 3.

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