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Transportation, blending and scheduling models; ratios that are linear after all; and what infeasible means.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to recognise the production, diet, transportation and scheduling shapes and predict how many variables and rows each needs, turn a ratio or a percentage requirement into a linear row by multiplying out, name the requirements that need a binary variable instead, and say what an infeasible result means and what could repair it.
You can write a small model and put it into standard form. This lesson widens the range of problems you can reach with the same four questions, and marks the edge where the language stops.
A transportation model ships from sources to destinations, with one variable per route, supply rows and demand rows. An assignment model is a transportation model where everything is one unit. A blending model mixes inputs to meet composition requirements, and its rows are percentages multiplied out.
Most linear programs met in practice are one of a handful of shapes. Production mix: one variable per product, one row per resource. Diet or blending: one variable per ingredient, one row per nutrient or specification, mostly $\ge$. Transportation: one variable per route, one row per supply and one per demand. Scheduling with overlapping shifts: one variable per shift pattern, one row per hour covered. Recognising the shape tells you how many variables and rows to expect before you write any of them.
Two pieces of English look non-linear and are not. A ratio — chairs at most $30\%$ of output — multiplies out to $0.7x - 0.3y \le 0$. A maximum bounded above — $\max(x, y) \le 10$ — splits into two rows. One piece of English looks harmless and is not: a cost paid once, which jumps and needs a binary.
And a model may have no feasible plan at all. That is an answer about the constraints, not a failure of the solver, and the repair is always to enlarge the feasible region.
Another way: steps
Another way: picture
A transportation model drawn as a grid: shops down the side, warehouses across the top, one cell per route. Each row must add to at least its demand and each column to at most its supply, and every cell belongs to exactly one row and one column.
A ratio is abandoned as non-linear before it is multiplied out, and a jump is written as if it were a rate. The third error is treating infeasibility as a bug: the model is re-run, the tolerances are loosened, and the contradiction stays exactly where it was.
'Ingredient A is at most $30\%$ of the blend' reads as $\dfrac{a}{a + b} \le 0.3$, which is a ratio of variables.
As written, outside the language.
Multiply both sides by $a + b$, which is positive: $a \le 0.3a + 0.3b$.
Clear the denominator.
Collect: $0.7a - 0.3b \le 0$. Linear, and it says exactly what the original did.
Linear after all.
The English says $a \ge 2b$.
Read the comparison.
Collect it into standard shape: $a - 2b \ge 0$, or $-a + 2b \le 0$.
A plan ships to three shops from two warehouses. Shop A receives $8$ and $6$; shop B receives $5$ and $2$; shop C receives $3$ and $9$. Fill in what each shop receives in total.
| From warehouse 1 | From warehouse 2 | Received in total | |
|---|---|---|---|
| Shop A | 8 | 6 | |
| Shop B | 5 | 2 | |
| Shop C | 3 | 9 |
A workshop with $30$ hours of labour is being modelled. Which of these requirements needs a binary variable rather than a linear constraint?
Goods move from $3$ warehouses to $6$ shops, and the plan decides how much goes along each route. How many decision variables is that?
Answer:
You are given a described decision with $5$ products. Put the steps of formulating it into the order you do them.
Number the steps in order (write the number in the box):
A workshop makes $x$ chairs and $y$ tables and has $50$ hours of labour. Match each requirement to the constraint that says it.
| $3x + 2y \le 50$ | $x + 2y \ge 8$ | $x = 2y$ | $0.7x - 0.3y \le 0$ | |
|---|---|---|---|---|
| At most $50$ hours of labour | ||||
| At least $8$ units of protein | ||||
| Twice as many chairs as tables | ||||
| Chairs are at most $30\%$ of the output |
A diet model comes back infeasible: no plan satisfies every constraint. Select every action that could make it feasible.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A plan ships to three shops from two warehouses. Shop A receives $2$ and $7$; shop B receives $5$ and $7$; shop C receives $4$ and $7$. Fill in what each shop receives in total.
| From warehouse 1 | From warehouse 2 | Received in total | |
|---|---|---|---|
| Shop A | 2 | 7 | |
| Shop B | 5 | 7 | |
| Shop C | 4 | 7 |
You can recognise the standard model shapes, turn ratio requirements into linear rows, and read an infeasible result as a statement about the constraints. Say in your own words why running the solver again cannot repair infeasibility.
7. Your turn: 'at least twice as much A as B' as a linear constraint, step 2