Feed Mixture Arithmetic: Check a Single Nutrient Target

Feed mixture arithmetic: check a single nutrient target

Two concentration labels can support a short mixture calculation only when they describe the same nutrient on the same basis. Before drawing a Pearson square, identify the denominator, test whether the target is possible and decide what the requested mixture mass actually represents.

The components below are unnamed fictional materials with invented crude-protein percentages. Every target is a mathematical exercise, not an animal requirement. Meeting one concentration does not produce a complete ration, certify feed safety or authorise real mixing or feeding.

State the basis before using a percentage

A 12% crude-protein value on a dry-matter basis means 12 kg of reported crude-protein equivalent per 100 kg of dry matter. A 12% as-fed value instead uses 100 kg of the material including its water. Those denominators differ. Equal printed percentages need not mean equal composition on a common basis.

In the first two cases, every percentage is dry-matter based and every mixture mass is kilograms of dry matter. The model assumes concentrations remain constant, component masses add exactly and no nutrient or dry matter is lost. It provides no information about actual product composition, palatability or safe inclusion limits.

Colorado State University’s Pearson-square explanation, hosted by Cornell Cooperative Extension, identifies the same-basis and within-range requirements. The calculations here are independently derived mass balances using original teaching numbers.

Worked case 1: derive a feasible mixture

Component A contains 12% crude protein and B contains 30%, both on a dry-matter basis. Find their contributions to a 60 kg dry-matter mixture at 18%. The target lies between 12 and 30, so a non-negative mixture is possible.

Let x be kilograms of A dry matter. Then B contributes 60 − x kilograms of dry matter. The target crude-protein amount is 0.18 × 60 = 10.8 kg. Balancing that amount gives:

0.12x + 0.30(60 − x) = 10.8
0.12x + 18 − 0.30x = 10.8
18 − 0.18x = 10.8
0.18x = 7.2, so x = 40 kg.

Therefore A contributes 40 kg dry matter and B contributes 20 kg dry matter. Check the total and the nutrient independently rather than accepting the algebra’s final line on its own.

A simple mean of 12% and 30% is 21%, but that would describe equal dry-matter contributions. The required contributions are unequal. Weighted concentration = total nutrient equivalent ÷ total dry matter × 100, which gives 18% here.

Cross-check using the Pearson square

Text alternative: place A’s 12% and B’s 30% on the left and the 18% target in the centre. The opposite difference 30 − 18 = 12 gives parts of A; 18 − 12 = 6 gives parts of B. Thus A:B = 12:6 = 2:1, or two-thirds A and one-third B on the stated dry-matter basis.

There are eighteen parts altogether, so one part represents 60 ÷ 18 = 3⅓ kg dry matter. A receives 12 × 3⅓ = 40 kg; B receives 6 × 3⅓ = 20 kg. “Parts” express a ratio, not kilograms until the total mass sets the scale.

The larger share belongs to the lower-concentration component because the 18% target is closer to 12% than to 30%. Reversing the contributions gives (20 × 0.12 + 40 × 0.30) ÷ 60 × 100 = 24%, so the nutrient audit catches the reversal.

Worked case 2: reject an impossible target first

Keep the same A and B concentrations but ask for 35%. No mixture using non-negative amounts of only these components can exceed B’s 30%. A weighted average must remain between the two concentrations.

The equation confirms the problem: 0.12x + 0.30(60 − x) = 0.35 × 60. It gives x = −16⅔ kg of A and 76⅔ kg of B. Negative material is not a feasible component, so the algebra describes an impossible requirement under the model.

Taking absolute differences blindly does not repair the case. Differences of 5 and 23 would produce the ratio A:B = 5:23, whose weighted concentration is (5 × 12 + 23 × 30) ÷ 28 ≈ 26.79%, not 35%. The range check must come before the square.

If the target equals an endpoint, such as 12%, the feasible result is all A and no B. A strictly intermediate target uses positive amounts of both distinct components. These boundary cases are mathematical facts, not advice that a single real feed is nutritionally complete.

Worked case 3: unlike bases stop the first calculation

A revised card gives A as 12% crude protein on an as-fed basis, B as 30% on a dry-matter basis and the target as 18% dry-matter based. The familiar-looking 12, 30 and 18 must not go directly into the square. A’s denominator differs.

Initially, A’s dry-matter fraction is missing, so the calculation remains pending. A supplementary fictional record then gives A as 80% dry matter. In 100 kg as-fed A, there are 80 kg dry matter and 12 kg crude-protein equivalent. Its dry-matter concentration is 12 ÷ 80 × 100 = 15%.

Now use 15%, 30% and 18% on one basis. Opposite differences give A:B = 12:3 = 4:1. For 60 kg dry matter, A contributes 48 kg and B 12 kg. The check is 48 × 0.15 + 12 × 0.30 = 7.2 + 3.6 = 10.8 kg, exactly 18% of sixty.

A’s 48 kg dry matter corresponds to 48 ÷ 0.80 = 60 kg as-fed A. But B’s dry-matter fraction is not supplied, so its as-fed mass and the total as-fed mixture mass remain unknown. Do not label the 48 and 12 dry-matter masses as as-fed quantities. University of Minnesota’s basis discussion explains why changing water content matters to this distinction.

Three solved practice tasks

Task 1: solve and audit another target

Find an 80 kg dry-matter mixture at 16% from components at 10% and 26%, all on the same basis. Solution: low:high parts = 10:6 = 5:3. Use 50 kg low and 30 kg high. Nutrient equivalent = 5 + 7.8 = 12.8 kg; 12.8 ÷ 80 × 100 = 16%.

Task 2: interpret an endpoint

Using Case 1’s components, find a 60 kg dry-matter mixture at 30%. Solution: use zero A and 60 kg B in the mathematical model. Adding any positive A would lower the concentration below 30%. This endpoint result still says nothing about a complete ration.

Task 3: handle equal concentrations

Both components are 20% on the same basis. What happens with a 20% target, and with a 22% target? Solution: any non-negative split totaling the requested mass gives 20%, so no unique ratio is determined. A 22% target is impossible. Dividing zero square-parts by zero is not a valid solution method.