“At least 30% of the blend is naphtha” has a decision in its denominator, and it is linear anyway: multiply out and the row becomes (1 − t) on the component named and −t on every other. Clearing is legal because the total is a quantity that cannot be negative, which is a condition to check rather than a step to take — and this lesson shows one ratio where it fails and declines it.
Core ideas
One algebraic move, and the condition that makes it a rearrangement rather than a new statement. The condition is the lesson.
The batch is N + R + B litres, so “at least 30% naphtha” is N ≥ 3/10 (N + R + B). The denominator is not data; it is the sum of the decisions, which is why the requirement looks non-linear and why writing N ≥ 3/10 instead is about a different problem.
Multiply out N ≥ t(N + R + B) and collect: (1 − t)N − tR − tB ≥ 0. Every variable appears, including the ones the requirement never mentioned, and that is the visible sign that a share is a statement about the whole batch.
The total litres cannot be negative, so the move is legal here. N ÷ B ≥ 2 is the same move applied to a denominator that a feasible blend can drive to zero, and N − 2B ≥ 0 is then a different requirement rather than the same one rewritten.
The material
The requirement as stated, the row it becomes, the percentages achieved at the optimum, and the condition that makes the move legal.
A refinery blends naphtha, reformate and butane into a 100-litre batch. At least 30% of the batch must be naphtha and at most 25% of it may be butane; naphtha costs 5 a litre, reformate 3 and butane 2. Minimise the cost of the batch.
For decisions x₁, …, xₙ measured in the same unit, the requirement “component i is at least a fraction t of the total” is xᵢ ≥ t(x₁ + … + xₙ). Collecting terms gives the linear row (1 − t)xᵢ − t(the rest) ≥ 0, and “at most t” gives the same row with the inequality reversed.
The right-hand side of the cleared row is 0, always. A share requirement carries no constant: it compares one decision with a multiple of the sum of all of them, and scaling the whole batch up or down does not change whether it holds.
The cheapest batch is 30 litres of naphtha, 45 of reformate and 25 of butane, costing 335. Both share requirements are met exactly: the achieved shares are 30%, 45% and 25%, printed as the exact fractions 3/10, 9/20 and 1/4 rather than as rounded percentages. The blend takes as much of the cheap butane as the ceiling allows and as little of the expensive naphtha as the floor allows, which is what the two bounds were written to control.
Write the naphtha requirement as N ≥ 3/10 instead and something worse than an error happens: the model stays feasible and solvable, and it answers a different question. The cheapest batch becomes N = 3/10 litres, R = 747/10 litres and B = 25 litres at a cost of 1378/5 — and the naphtha share is 0.3%, not 30%. The variable is measured in litres and the requirement, being a proportion, is measured in nothing; the two sides of that row do not carry the same unit and no arithmetic will complain.
A subtler version takes the share of the wrong total: N ≥ 3/10 (R + B) asks for 30% of the others rather than 30% of the batch. It is a legal linear row and a weaker requirement — it is satisfied at N = 300/13 litres, which is a naphtha share of 3/13, about 23.08% of the batch. Clearing is only as good as the total you cleared.
The same move applied to N ÷ B ≥ 2 gives N − 2B ≥ 0, and that is not a rearrangement. Multiplying an inequality by B preserves it only when B is strictly positive, and the question of whether it can vanish is itself a linear programme: add B ≤ 0 to the constraints and ask whether anything is left.
The batch (30, 70, 0) fills 100 litres, is 30% naphtha and is within the butane ceiling, so it is feasible with B = 0. At that point N ÷ B is not a number at all, while N − 2B ≥ 0 reads 30 ≥ 0 and holds comfortably. The cleared row admits points the original requirement says nothing about, so the lab declines the move and says which feasible blend refutes it.
The honest form of the rule is therefore conditional. A ratio requirement may be cleared when the denominator is a quantity the constraints keep strictly positive — and when it is the total of the batch, that is usually evident. When it is one component, it is a claim, and claims on this course get checked.
Interactive
A percentage of a total the decisions themselves add up to
Method
Four steps, of which the third is the one that separates a rearrangement from a new statement.
“30% of the batch” and “30% of the other components” are different requirements, and they differ by more than notation: the first gives a naphtha share of 30% and the second about 23%.
N ≥ 3/10 (N + R + B), not N ≥ 3/10. Keeping the total visible for one line is what makes the next step a calculation rather than a guess.
The total litres of a batch cannot be negative, so multiplying through preserves the inequality. Collect terms into (1 − t) on the component and −t on every other, with 0 on the right.
Compute each component divided by the total at the optimum and compare it with what was required. A share that comes out wildly below its floor is the signature of a proportion written as a quantity.
The clearing is three lines of algebra. The check at the end is what tells you the three lines were about the right total.
Both stated shares are met exactly, which is what happens when one bound forces expensive naphtha in and the other stops cheap butane taking over. The check is not ceremony: it is the only step that would notice the requirement had been written about the wrong total.
For a faded rehearsal, tighten the butane ceiling from 25% to 10% and leave the naphtha floor alone. The supplied first move is the cleared form: only the coefficient t changes, so the row becomes −(1/10)N − (1/10)R + (9/10)B ≤ 0. Predict the new blend and its cost, compute the three achieved shares as exact fractions, and say which of the two bounds is now doing the work.
You should be able to turn a percentage requirement into a linear row with 0 on the right, say why multiplying through was legal, compute the achieved shares at the optimum as exact fractions, and refuse a ratio whose denominator the constraints allow to vanish.
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