Practice: Driving a Linear Programming Solver

(a) The sheet. Two changing cells hold the decision variables: one for $a$, units of product A per week, one for $b$, units of B. A third cell holds the objective as a formula, $=30a+20b$ referring to those two cells. A formula, never a typed number, so it recomputes as Solver varies the plan. Three further formula cells hold the left-hand side of each constraint: $=2a+b$ for machine-hours, $=a+b$ for labour-hours, $=a$ for the sales limit. Each appears in the dialog as a row naming that cell, the relation $\le$, and a bound of 100, 80 and 40 respectively. Because the direction is chosen explicitly in the dialog as Max, no negation is needed: unlike an array interface that only minimises, Solver takes the maximisation as stated. That is the conversion this route avoids. (b) The two settings. The solving method. It must be set to the linear option. The default is a nonlinear method, which may terminate at a point it cannot certify as globally optimal and may report a different answer on different runs from different starting cells. For a genuinely linear model the linear method is both correct and faster, and it certifies optimality. The nonnegativity option. Unconstrained variables are not assumed nonnegative, so this must either be checked or entered as explicit $\ge 0$ constraint rows. Left off, Solver may return a negative quantity of one product, arithmetically optimal, physically meaningless, and easy to miss because the cell simply shows a negative number. (c) The answer. Solver reports an optimal solution: make 20 units of A and 60 units of B, for a contribution of £1,800 per week. Checking against the model as written: - Machine-hours: $2(20)+60=100\le 100$ ✓ binding
- Labour-hours: $20+60=80\le 80$ ✓ binding
- Sales limit: $20\le 40$ ✓ slack of 20
- Objective: $30(20)+20(60)=1800$ ✓ Both resources bind and demand does not, so the plan is capacity-limited rather than demand-limited. (d) Two errors giving a larger answer. A constraint cell referring to the wrong cells. If the machine-hours formula reads $=2a$, the reference to $b$ lost in an edit, the model believes B consumes no machine time, and the reported contribution rises above 1,800. Finding it: read each constraint formula back as a sentence and check it against the description, or substitute the reported plan into the constraints by hand, where the true machine-hour usage would exceed 100. A missing constraint row. If the labour-hours row was never added to the dialog, the model is unconstrained in labour and again returns more than 1,800. Finding it: count the rows in the dialog against the restrictions in the description, and check the reported plan against every stated restriction rather than the ones on the sheet. Both produce a clean optimal status, because each yields a valid program. The check that catches them is verification against the original description, not anything inside the tool. (e) The two entry routes. A spreadsheet makes reference errors easier: a formula can point at the wrong cell and still compute, and the sheet displays values rather than the relationships between them, so the error is invisible on screen. It makes direction errors harder, since the objective sense is an explicit choice in the dialog rather than a convention to remember. An array interface inverts this. It makes convention errors easier, forgetting that the routine only minimises, or that bounds are not assumed, and makes reference errors harder, since the arrays are written out in one place and their shapes must agree or the call fails.

(a) The model, and where each part goes. Let $x$ and $y$ be tonnes of feed X and feed Y.

linprog solves a minimisation with inequalities as $Ax\le b$, equalities in a separate array, and bounds given separately. So: the objective is already a minimisation and needs no negation; the two $\ge$ rows must be rewritten as $\le$ by negating them; the limit on $x$ is either a third inequality row or an upper bound, and appears exactly once either way; there are no equalities; nonnegativity is stated as a lower bound of zero, not assumed. (b) The arrays. Column 1 is $x$, tonnes of feed X per batch; column 2 is $y$, tonnes of feed Y.

with bounds $\ell =(0,0{)}^{\mathsf{T}}$ and $u=(8,\mathrm{\infty }{)}^{\mathsf{T}}$, taking the availability limit as an upper bound. Row 1 is the protein requirement, row 2 the fibre requirement. The rewriting to record: both constraint rows were negated to turn $\ge$ into $\le$. Multiplying an inequality by $-1$ reverses its direction, so $40x+30y\ge 240$ becomes $-40x-30y\le -240$. The objective was not negated, since the model already minimises. Negating it out of habit would invert the question. (c) The result, and reversing. The solver reports an optimum found, at $x=(4,8/3)$, that is, $4$ and approximately $2.667$, with objective value $1360$. The only conversions applied were to the constraint rows, which do not affect how the point or the objective value are read. The objective was not negated, so the reported value stands as it is. In the dairy's terms: blend 4 tonnes of feed X with 8/3 tonnes of feed Y, at a cost of £1,360 per batch. The fractional quantity is ordinary. Nothing in a linear program makes its optimum integral, and a tonnage is divisible, so $8/3$ tonnes is an implementable answer rather than something to round. Rounding it to $2.7$ would raise the cost; rounding to $2.6$ would breach the fibre requirement. (d) Verification against the original model. - Protein: $40(4)+30(8/3)=160+80=240\ge 240$ ✓, binding
- Fibre: $20(4)+60(8/3)=80+160=240\ge 240$ ✓, binding
- Availability: $4\le 8$ ✓, slack of 4 tonnes
- Nonnegativity: both positive ✓
- Objective: $180(4)+240(8/3)=720+640=1360$ ✓ matches Both nutrient requirements bind and the availability limit does not. This is the covering shape behaving as expected: every unit of slack above a floor is paid for and not required, so the cheapest blend meets its minima exactly. The feed X limit is not scarce at this solution, so negotiating more of it would buy nothing. What would change the cost is relaxing either nutrient floor. Both are active, so each has a price attached, and a sensitivity run would report what a kilogram of either requirement is worth. (e) What the exit flag says. It establishes that the routine terminated having found a point it certifies as optimal for the arrays it received, as opposed to stopping at an iteration limit, or concluding the constraints were unsatisfiable. It does not establish that those arrays are the dairy's model. The negations in part (b) are exactly where that could fail: negating a row's coefficients but not its right-hand side yields a valid program describing a requirement nobody stated, and the flag would report success on it.