Optimizations and Programming. Abdelkhalak El Hami
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Название: Optimizations and Programming
Автор: Abdelkhalak El Hami
Издательство: John Wiley & Sons Limited
Жанр: Техническая литература
isbn: 9781119818267
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What is the interval for b2?
We will have
2 − a ≥ 0 and 4 + a ≥ 0.
This gives the interval for the parameter b2: −4 ≤ a ≤ 2.
What is the effect on the minimum value of the objective function?
The effect is:
1.10.2. Effect of modifying c
Next, let us find a condition that guarantees that the basis B will remain optimal under
[1.20]
In general, let us determine the effect of modifying a single variable xi:
There are two cases to analyze depending on whether xi is a basic or non-basic variable.
Case of a non-basic variable
Write ei for the canonical basis vector of ℝn, i.e. ei = (0, 0, . . . , 1, 0, . . . , 0)T with 1 at position i. Then:
The ith component of the vector
where the ri are the components of the final row of the simplex tableau. In other words, we have all the information that we need to compute the stability intervals of the coefficients ci.
EXAMPLE 1.13.– Again, consider the previous example. The final simplex tableau is:
The last row gives us the vector of reduced costs
The optimal solution x = (2, 4, 0, 0, 0)T is the same for every value
1.11. Application to an inventory problem
Consider the problem of an automobile manufacturer that offers two models for sale, a small automobile and a large automobile. Suppose that there is sufficient demand that the manufacturer is certain to sell whatever is produced at the current sales price of 16,000 euros for large automobiles, and 10,000 euros for small ones. The manufacturer’s only problem is the limited supply of two raw materials, rubber and steel. Manufacturing a small automobile requires one unit of rubber and one unit of steel, whereas manufacturing a large automobile requires one unit of rubber and two units of steel. If the manufacturer has 400 units of rubber and 600 units of steel in stock, how many small and large vehicles should be produced from this inventory to maximize the turnover?
Let x be the number of large automobiles manufactured, y the number of small automobiles manufactured and z the resulting turnover. The problem can therefore be expressed in the form:
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1.11.1. Optimal solution
It is relatively easy to find a graphical solution for systems like this with only two variables. This allows us to find the optimal solution x = 200 and y = 200, which corresponds to z = 5, 200, 000. In this particular example, the optimal solution is unique, corresponding to a vertex of the region bounded by the constraints.
1.11.2. Sensitivity to variation in stock
Let us observe how the solution of the problem evolves when some of the starting data is modified, for example by increasing the stock of rubber or steel. Suppose that 700 units of steel are in stock instead of 600. The problem now becomes:
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By solving graphically again, we find that the optimal solution is now x = 300 and y = 100, giving z = 5, 800, 000. In other words, increasing the available steel by 100 units increases the turnover by 600,000 euros. We say that the marginal price of a unit of steel is 6,000 euros.
If the amount of steel in stock is increased to 800, the optimal solution becomes x = 400 and y = 0, and the turnover becomes z = 6, 400, 000. Increasing the steel in stock beyond 800 without also changing the rubber in stock no longer affects the optimal solution, because y is constrained to remain positive.
Suppose now that the steel in stock remains fixed at 600, but the rubber in stock is increased by 400 to 500. The new problem is
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Graphical solving shows that the optimal solution is now given by x = 100 and y = 400, which corresponds to z = 5, 600, 000. In other words, increasing the rubber by 100 units changes the turnover by 400,000 euros. We say that the marginal price of a unit of rubber is 4,000 euros.
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