Practice: Contour Lines and the Direction of Improvement

(a) $c=(4,2)$. Contours $4x_1+2x_2=k$ have slope $-4/2=-2$. Maximising, so travel along $(4,2)$: right and up.

(b) $c=(4,2)$, identical. Contours identical, slope $-2$. Minimising, so travel against $(4,2)$: left and down.

(c) $c=(3,-1)$. Contours $3x_1-x_2=k$ have slope $-3/(-1)=3$. Maximising, so travel along $(3,-1)$: right and down.

What is the same. (a) and (b) have exactly the same contour family. The same lines, in the same places, with the same slope. A drawing of one is a drawing of the other.

What differs. (a) and (b) travel in opposite directions along that identical family. Nothing in the drawing distinguishes them; the difference lives entirely in whether the problem maximises or minimises.

What (c) adds. A negative coefficient means improvement involves decreasing a variable. A learner following the habit 'push away from the origin' gets (a) right, (b) wrong, and (c) wrong, and the drawing looks correct in all three cases.

The reliable statement. $c$ points towards increase in every case. Maximise: travel along $c$. Minimise: travel against $c$. Read the sign of each coefficient to know what improving means for that variable.

What is correct. The contours do slope down to the right: $c=(2,3)$ gives slope $-2/3$. And moving up and to the right, along $c$, does increase the objective. Both statements are true.

The error. The problem is a minimisation. Having correctly identified the direction of increase, the student travelled that way. Minimising means travelling against $c$, down and to the left, until the contour last touches the region. The student has found the point that maximises the objective over the region.

The deeper error: where the direction came from. The student says the slope told them which way the objective increases. It cannot. The family of parallel lines for $c=(2,3)$ and for $c=(-2,-3)$ is identical, same lines, same slope, opposite directions of increase. Orientation is not recoverable from a family of parallel lines. It comes from the coefficient vector, which must be written down separately.

Why nothing looked wrong. The contours were right, the region was right, the sliding was executed correctly, and the answer landed on a genuine corner of the region with a genuine objective value. The output is a well-formed answer to the wrong question, and the drawing offers no signal at all.

What to do instead. Write $c=(2,3)$ explicitly. State the rule in words before sliding: minimising, so travel against $c$. If any doubt remains about a particular direction $d$, compute $c^{\mathsf{T}}d$ and read the sign, for $d$ pointing down-left, say $(-1,-1)$, the product is , confirming the objective falls that way.

$c^{\mathsf{T}}d=-4$. Moving along $d$ changes the objective at a rate of $-4$ per unit step: each unit of movement lowers the objective by $4$. Since this is a minimisation, the direction improves the objective, and the algorithm should move along it, as far as feasibility permits.

If it were $+4$. The objective would rise along $d$ at the same rate, so for a minimisation the direction is not worth taking. The algorithm would look at a different edge.

If it were $0$. The objective would not change along $d$ at all. Moving that way reaches different points of exactly equal value. If the algorithm is at an optimal point, such a direction is the signature of multiple optima; the point reached is another optimal solution.

What it corresponds to in the picture. Exactly the sliding of the contour. In two variables, $c^{\mathsf{T}}d$ is the rate at which the contour value changes as you walk along $d$: negative means walking downhill across the contours, positive means uphill, and zero means walking along a contour without crossing any.

Why this is the point of the unit. In forty variables there is no picture. What survives is that single number, and the whole geometric account has been compressed into its sign. A learner who acquired 'push the contour away from the origin' has nothing to carry here; a learner who acquired 'the sign of $c^{\mathsf{T}}d$ decides' has the same competence working unchanged. This quantity is the reduced cost, and the sign convention that makes the optimality test intelligible is precisely this one.