Define variables y_{j} as the number of times the j-th pattern is cut. In addition, denote by v_{1}, v_{2}, & v_{3}the number of 8 ft, 5 ft, and 3 ft rods that are purchased. Min z = 1y_{1} + 1y_{2} + 1.5y_{3} + 1.5y_{4} + 0.5y_{5} + 0.5y_{6} + 1y_{7} +

(demand constraints) y_{1}, y_{2}, …, y_{8}; v_{1}, v_{2}, v_{3} ≥ 0 and integer. Optimal solution: = 2, = 0, = 0, = 18, = 5, = 0, and = 0, as well as = 53, = 0, & = 1. No rods are left over, the demand is exactly satisfied. 2-dimensional cutting stock problems: same formulation, cutting plans are more difficult to set up.

x_{1}, x_{2} 0. plus: “if food 1 is in the diet, then food 2 should not be included. Define logical variables y_{1} (and y_{2}) as one, if food 1 (food 2) is included in the diet, and zero otherwise.

y_{1}

y_{2}

OK?

0

0

0

1

1

0

1

1

No

A formulation that allows the first three cases & prohibits the last case is y_{1} + y_{2} 1.

y_{1}, y_{2} = 0 or 1. Adding these constraints to the formulation is not sufficient, though, as it allows the continuous variables x_{1} & x_{2}to change independent of y_{1} and y_{2}. We need linking constraints. Here, x_{1} My_{1} &

x_{2} My_{2}, with M >> 0 (but not too large, scaling). Validity: If y_{1} = 1 (the food is included in the diet), then the constraint reads x_{1} M, &, given that M is sufficiently large, the constraint is redundant. On the other hand, if y_{1} = 0 (the food is not included in the diet), the constraint reads x_{1} ≤ 0, &, since x_{1} ≥ 0, x_{1} = 0 follows. In other words, if a food is not in the diet, its quantity is zero. Different additional (conditional) constraint: “if food 1 is included in the diet, then food 2 must be included in the diet as well.”

y_{1}

y_{2}

OK?

0

0

0

1

1

0

No

1

1

Here, the additional constraint y_{1} y_{2} together with the linking constraints will do. 4.2.3 Land Use Two choices for a parcel of land: harvest or protect (but not both). Define variables y_{1}= 1, if we harvest & 0 otherwise, & y_{2} = 1, if we protect, & 0 otherwise.

y_{1}

y_{2}

OK?

0

0

0

1

1

0

1

1

No

Again, the additional constraint is y_{1} + y_{2} ≤ 1. (There are no other variables, so that there is nothing to link). Allow 3 options: Harvest (y_{1}), build a sanctuary (y_{2}), or allow the building of a municipal well (y_{3}).

y_{1}

y_{2}

y_{3}

OK?

0

0

0

0

0

1

0

1

0

1

0

0

0

1

1

1

0

1

No

1

1

0

No

1

1

1

No

Formulate:y_{1} + y_{2} ≤ 1 (eliminates the solutions in the last two rows of the decision table), & the constraint

y_{1} + y_{3} ≤ 1 (eliminates the solution in the third row from the bottom of the table).