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Using '''unit contributions''' d<sub>1</sub> and d<sub>2</sub> respectively, and rearranging results in:<br> | Using '''unit contributions''' d<sub>1</sub> and d<sub>2</sub> respectively, and rearranging results in:<br> | ||
: K<sup>f</sup> = (p<sub>1</sub>-k<sup>v</sup><sub>1</sub>) · x<sub>1</sub> + (p<sub>2</sub>-k<sup>v</sup><sub>2</sub>) · x<sub>2</sub> = d<sub>1</sub> · x<sub>1</sub> + d<sub>2</sub> · x<sub>2</sub><br> | : K<sup>f</sup> = (p<sub>1</sub>-k<sup>v</sup><sub>1</sub>) · x<sub>1</sub> + (p<sub>2</sub>-k<sup>v</sup><sub>2</sub>) · x<sub>2</sub> = d<sub>1</sub> · x<sub>1</sub> + d<sub>2</sub> · x<sub>2</sub><br> | ||
It is obvious that the contribution to be achieved with one product depends on the contribution provided by the other product. In extreme cases, we concentrate on one of the products and set the | It is obvious that the contribution to be achieved with one product depends on the contribution provided by the other product. In extreme cases, we concentrate on one of the products and set the number of units of the other equal to zero. This leads to the '''individual break-even-points''' (or '''break-even-corner-points'''):<br> | ||
: [[Datei:x_hat_1.png]] = K<sup>f</sup>/d<sub>1</sub> (if x<sub>2</sub> = 0)<br> | : [[Datei:x_hat_1.png]] = K<sup>f</sup>/d<sub>1</sub> (if x<sub>2</sub> = 0)<br> |