Читать книгу Optimizations and Programming - Bouchaib Radi, Ghias Kharmanda, Michel Ledoux - Страница 42

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.

If the rubber in stock is increased to 600, the optimal solution becomes x = 0 and y = 600, and the turnover becomes z = 6, 000, 000. Increasing the amount of rubber in stock beyond 600, without also increasing the amount of steel in stock, no longer affects the optimal solution because x is constrained to remain positive.