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# 数学代写|运筹学代写Operations Research代考|MATH3830 Diet Problems Revisited

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## 数学代写|运筹学代写Operations Research代考|Diet Problems Revisited

First, recall the standard diet problem in linear programming. To simplify matters, consider only two foodstuffs and a single nutrient. The quantities of the two foods are defined as $x_1$ and $x_2$, respectively, and at least five units of the nutrient are required in the diet. We assume that the problem has been formulated as follows:
$\mathrm{P}: \operatorname{Min} z=3 x_1+4 x_2$
s.t. $x_1+2 x_2 \geq 5$
$x_1, \quad x_2 \geq 0$.
Suppose now that the additional requirement is that if food 1 is included in the diet (in any quantity), then food 2 should not be. (One reason for this may be incompatibilities due to taste such as ice cream and mustard, or unfortunate side effects of incompatible foods, such as water and green apples, or, worse, yoghurt and yeast.) This is a conditional constraint of the type “if food 1 is included, then food 2 should not be.” We first must define logical zero-one variables, one for each foodstuff. These new variables $y_1$ (and $y_2$ ) are defined as being one, if food 1 (food 2 ) is included in the diet, and zero otherwise. We will need these variables in addition to the variables $x_1$ and $x_2$ that denote the quantities of the two foods that are included in the diet.

For instance, the first row in Table $5.6$ includes neither of the two foods, and while it may leave us hungry, it does not violate the condition. Here, we find that only the solution that has $y_1$ and $y_2$ both equal to one (the case in which both foods are in the diet) is prohibited. We can now use the exclusion constraints introduced earlier.

Eliminating this solution from consideration is achieved by writing the constraint $y_1+y_2 \leq 1$

## 数学代写|运筹学代写Operations Research代考|Land Use

The next example deals with land use. Suppose that a land owner owns a parcel of land that he has to decide what to do with. He has narrowed down his decisions to two: sell stumpage, i.e., harvest the land, or build an animal sanctuary, but not both. Here, we will formulate only this aspect of the model and ignore all other considerations. We will need a decision variable for each possible decision, so that we define $y_1=1$, if we decide to harvest the parcel, and 0 otherwise, and $y_2=1$, if we decide to build an animal sanctuary, and 0 otherwise. The decision table for this problem is then shown in Table $5.8$.

Once formalized as done here, we see that the situation is the same as in the diet problem by not allowing both options at the same time, which is modeled as $y_1$ $+y_2 \leq 1$. Since this problem does not appear to need quantitative variables $x_1$ and $x_2$ as was the case in the diet problem, we do not need linking variables, as there is nothing to link.

Things may get much more complicated when more options exist. Suppose now that for the parcel in question, three choices have been identified: Harvest (decision variable $y_1$ ), build a sanctuary (decision variable $y_2$ ), or allow the building of a municipal well (decision variable $y_3$ ). As in the previous land use example, it is not possible to harvest and have a sanctuary at the same time in the parcel in questions. Furthermore, the parcel cannot be harvested if there is a municipal well on the parcel, while we could very well have a well and a sanctuary on the same parcel. The decision table for this extended problem is shown in Table 5.9.

## 数学代写|运筹学代写Operations Research代考|Diet Problems Revisited

$\mathrm{P}: \operatorname{Min} z=3 x_1+4 x_2$

$x_1, \quad x_2 \geq 0$.

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