Jhun Dhon Mari G. Abihay December 12, 2013

Quantitative Methods in Management Linear Programming

The Whole Feed Nutrition Center uses three bulk grains to blend a natural cereal that it sells by the pound. The store advertises that each 2-ounce serving of the cereal when taken with 0.5 cup of whole milk, meets an average adult’s minimum daily requirement for protein, riboflavin and phosphorus.

The minimum adult daily requirement for protein is 3 units; for riboflavin, 2 units; and for phosphorus, 1 unit. Whole Food wants to select the blend of grains that will meet the adult daily requirement at a minimum cost.

Whole Food’s Natural Cereal RequirementsGRAIN COST PER

POUND(CENTS)

PROTEIN(UNITS/LB)

RIBOFLAVIN(UNITS/LB)

PHOSPHORUS(UNITS/LB)

A 33 22 16 8B 47 28 14 7C 38 21 25 9

Let:

X A=poundsof grain A∈one2−ounce servingof cereal

X B=poundsof grainB∈one2−ounce servingof cereal

XC=pounds of grainC∈one2−ounce servingof cereal

C=total cost of mixing a2−ounceserving of cereal

Objective function:

C=0.33 X A+0.47 XB+0.38 XC

Constraints:

22 X A+28 XB+21 XC ≥3

16 X A+14 X B+25 XC≥2

8 X A+7 XB+9 XC≥1

X A , XB , XC≥0

Matrix Form:

D=|22 28 2116 14 258 7 9 |

Solving for D, making sure that Cramer’s Rule is applicable:

D=22|14 257 9 |−28|16 25

8 9 |+21|16 148 7 |

D=22 [ (14 ) (9 )−(25 ) (7 ) ]−28 [(16 ) ( 9 )−(25 ) (8 ) ]+21 [ (16 ) (7 )− (14 ) (8 ) ]

D=490

Since,

D≠0

Then, Cramer’s Rule will be used which is:

X A=DX A

D, X B=

DX B

D, XC=

DXC

D

Solving forDX A,DX B andDX C:

DX A=|3 28 21

2 14 251 7 9 |

DX A=3|14 257 9 |−28|2 25

1 9 |+21|2 141 7 |

DX A=3 [ (14 ) ( 9 )−(25 ) (7 ) ]−28 [ (2 ) (9 )−(25 ) (1 ) ]+21 [ (2 ) (7 )−(14 ) (1 ) ]

DX A=49

DX B=|22 3 21

16 2 258 1 9 |

DX B=22|2 251 9 |−3|16 25

8 9 |+21|16 28 1|

DX B=22 [(2 ) (9 )−(25 ) (1 ) ]−3 [ (16 ) (9 )− (25 ) (8 ) ]+21 [ (16 ) (1 )− (2 ) (8 ) ]

DX B=14

DX C=|22 28 316 14 28 7 1|

DX C=22|14 27 1|−28|16 2

8 1|+3|16 148 7 |

DX C=22 [ (14 ) (1 )−(2 ) (7 ) ]−28 [ (16 ) (1 )−(2 ) (8 ) ]+3 [(16 ) (7 )−(14 ) (8 ) ]

DX C=0

Solving for XA, XB and XC:

X A=DX A

D

X A=49

490

X A=0.1 poundsof grain A

X B=DX B

D

X B=14490

X B=0.0286 poundsof grain B

XC=DXC

D

XC=0

490

XC=0 pounds of grainC

Proving that the solved amounts satisfies the given constraints:

22 X A+28 XB+21 XC ≥3

22 (0.1 )+28 (0.0286 )+21 (0 )≥3

3≥3(satisfied !)

16 X A+14 X B+25 XC≥2

16 (0.1 )+14 (0.0286 )+25 (0 )≥2

2≥2(satisfied !)

8 X A+7 XB+9 XC≥1

8(0.1)+7 (0.0286)+9(0)≥1

1≥1(satisfied !)

X A , XB , XC≥0

0.1 ,0.0286 ,0≥0 (satisfied !)

Proving that the solved values can give the minimum cost required by comparing to other values.

For X A=0.1 ; X B=0.0286 ; X C=0 :

C=0.33XA+0.47XB+0.38XC

C=0.33 (0.1 )+0.47 (0.0286 )+0.38 (0 )

C=$0.04644

For X A=0.04287 ; X B=0.04287 ; XC=0.04287 :

C=0.33XA+0.47XB+0.38XC

C=0.33 (0.04287 )+0.47 (0.04287 )+0.38 (0.04287 )

C=$0.05059

For X A=0.05 ; X B=0.0286; XC=0.05 :

C=0.33XA+0.47XB+0.38XC

C=0.33 (0.05 )+0.47 (0.0286 )+0.38 (0.05 )

C=$0.048942

Therefore, the solved values of XA, XB and XC are the most economical in terms of cost and nutrition.

Reference:

Render, B., and R. M. Stair Jr. “Linear Programming Applications,” Quantitative Analysis for Management, 1997.