In: Operations Management

M1_P5. Cattle are sent to a feedlot to be grain-fed before being processed into beef. The owners of a feedlot seek to determine the amounts of cattle feed to buy so that the minimum nutritional standards are satisfied to ensure proper weigh gain, while total feed costs are minimized. The feed mix used is made up of three grains that contain the following nutrients per pound of feed. Nutrient (ounces per pound of feed) FEED A B C D Cost of Mix (per lb.) Feed Mix X 4 2 3 6 $5.00 Feed Mix Y 2 3 0 8 $4.50 Feed Mix Z 5 2 2 5 $3.50 The minimum requirement per cattle per day is 6 lbs. of Nutrient A, 5 lbs. of Nutrient B, 2 lbs. of Nutrient C, and 10 lbs. of Nutrient D. Another restriction is that only 500 lbs. of Feed Mix Z is available per day. Because there are usually 100 cattle to feed at any given time, this means that no more than 5 lbs. of Feed Mix Z can be counted on for use in the feed mix each day. Formulate and solve this problem in Excel to determine the number of lbs. of each type of Feed Mix to include in the total mix that meet the requirements and minimize the cost (Hint: Make sure your LHS and RHS of your constraints are the same units: 1 lb. = 16 ounces)

LPP formulation:

**Decision
Variable**:

For the given situation, decision is regarding how much gram of feeds X, Y, and Z should be blended such that requirement are satisfied and total cost of mix is minimized.

Let,

X = number of lbs of feed X to be mixed

Y = number of lbs of feed Y to be mixed

Z = number of lbs of feed Z to be mixed

**Objective
function**

The objective is to provide feed to cattles by blending three feeds such that total cost is minimized. The cost function is given as follows:

Cost = cost per lbs x lbs of feed i

The objective function is given as follows:

Minimize Z = ($5)X + ($4.5)Y + ($3.5)Z

Subject To:

1 |
Minimum requirement of nutrient A |
Total lbs of A of blending >= 6 |
1/16(4X + 2Y + 5Z >= 6 |

2 |
Minimum requirement of B |
Total lbs of B of blending >= 5 lbs |
1/16(2X + 3Y + 2Z) >= 5 |

3 |
Minimum requirement of C |
Total lbs of C of blending >= 2 lbs |
1/16(3X + 0Y + 2Z >= 2 |

4 |
Minimum requirement of D |
Total lbs of D of blending >= 10 |
1/16(6X + 8Y + 5Z >= 10 |

5 |
Availability of feed Z |
Availability <= 5 |
Z <= 5 |

6 |
Non-negative constraint |
X, Y, Z>= 0 |

Excel Model:

Optimal Solution:

**X = 9.13
lbs**

**Y = 17.25
lbs**

**Z = 5.00
lbs**

**Total cost =
$140.75 per day**

If cattle are processed at 950 pounds. and each head of cattle
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