Recipe

Find the allowable range for one LP coefficient, cost or resource

Use when: you know an LP's optimal vertex and need the exact interval ONE cost coefficient, or ONE resource (right-hand-side) value, can move through before today's optimal vertex changes

Goal: the exact interval that coefficient or resource level can move through while today's optimal vertex -- and its shadow price -- stays valid

Taught in Sensitivity to resource changes · extends Find the allowable range for one cost coefficient

Quick recipe

Perturb the value in question
Substitute the shift into whatever stays fixed
Impose feasibility (or optimality)
Read the tightest bracket, then add the base value back

1
Write the value you're ranging as its current value plus an unknown shift, $\Delta$ . Everything else -- every other coefficient, every other constraint -- holds still.

Example

Ranging Meadowbrook's protein requirement $b_1=10$ : write it as $10+\Delta$ , holding $b_2=15$ and both costs fixed.
- ranging a cost coefficient $c_j$ — the feasible region itself never moves; only which vertex is cheapest can change (Section 6's own case).
- ranging a resource level $b_i$ — the feasible region's own boundary moves, carrying the optimal vertex along the intersection of the SAME two constraints.
Why: isolating one unknown, with everything else fixed, is what turns "how far can this move" into a single answerable question.
2
For a cost coefficient, substitute into the objective at every candidate vertex (Section 6, Step 2). For a resource level, substitute the shifted right-hand side into the SAME two constraints that are tight at the optimum, and solve that pair of equations for the vertex's coordinates as functions of $\Delta$ .

Example

$x_1+2x_2=10+\Delta$ , $3x_1+x_2=15$ . From the second, $x_2=15-3x_1$ ; substituting: $x_1+2(15-3x_1)=10+\Delta \Rightarrow -5x_1=\Delta-20 \Rightarrow x_1=4-\tfrac{\Delta}{5}$ , then $x_2=3+\tfrac{3\Delta}{5}$ .

Why: the same two boundary lines stay tight until one of the vertex's own coordinates is forced out of the feasible region, so re-solving that pair as functions of $\Delta$ tracks the vertex exactly as it slides.
3
For a cost coefficient, impose that the current vertex's cost stays at or below every other vertex's cost (Section 6, Step 3). For a resource level, impose that every coordinate of the SLIDING vertex stays nonnegative -- each coordinate's own formula in $\Delta$ gives one inequality.

Example

$x_1=4-\tfrac{\Delta}{5}\ge0 \Rightarrow \Delta\le20$ . $x_2=3+\tfrac{3\Delta}{5}\ge0 \Rightarrow \Delta\ge-5$ .
- a coordinate's own coefficient on $\Delta$ is negative — increasing $\Delta$ eventually drives it to $0$ , setting the UPPER edge.
- a coordinate's own coefficient on $\Delta$ is positive — decreasing $\Delta$ eventually drives it to $0$ , setting the LOWER edge.
Why: the vertex stops being a genuine basic feasible solution the moment either coordinate would need to go negative -- Chapter 7's own feasibility requirement, applied to a moving point instead of a fixed one.
4
Keep the tightest lower bound and the tightest upper bound among every inequality found, then add the original value back to convert the $\Delta$ -range into the actual allowable range.

Example

$\Delta\in[-5,20]$ ; adding back $b_1=10$ : $b_1\in[10-5,\ 10+20]=[5,\ 30].$

Why: the question was about $b_1$ itself, not about the bookkeeping shift used to answer it.

Common mistakes

Trying to hold the vertex's coordinates fixed while only the constraint moves -- the whole point is that the vertex itself slides.
Forgetting to re-derive which two constraints stay tight; the recipe assumes the SAME two stay active, which is true only within the allowable range, not automatically forever.
Mixing up the cost case and the resource case mid-calculation -- a resource shift never touches the objective, and a cost shift never touches a constraint's right-hand side.
Stopping at the $\Delta$ -range and reporting it as if it were the resource level itself.