Linear optimization: Mathematical formulations of problems presented in the course 1: Unterschied zwischen den Versionen
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| − | + | =='''Mathematical formulations of problems presented in the course'''== | |
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| + | Example: | ||
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| + | A company makes two products (X and Y) using two machines (A and B). Each unit of X that is produced requires 50 minutes processing time on machine A and 30 minutes processing time on machine B. Each unit of Y that is produced requires 24 minutes processing time on machine A and 33 minutes processing time on machine B. | ||
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| + | At the start of the current week there are 30 units of X and 90 units of Y in stock. Available processing time on machine A is forecast to be 40 hours and on machine B is forecast to be 35 hours. | ||
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| + | The demand for X in the current week is forecast to be 75 units and for Y is forecast to be 95 units. Company policy is to maximise the combined sum of the units of X and the units of Y in stock at the end of the week. | ||
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| + | x = number of units of X produced in the current week | ||
| + | y = number of units of Y produced in the current week | ||
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| + | Constraints: | ||
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| + | <math>50x + 24y \leq 40</math> machine A time | ||
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| + | <math>30x + 33y \leq 35</math> machine B time | ||
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| + | <math>x \geq (75 - 30)</math> | ||
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| + | i.e. <math>x \geq 45</math> so production of <math>X \geq demand (75) - initial stock (30)</math> , which ensures we meet demand | ||
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| + | <math>y \geq 95 - 90</math> | ||
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| + | i.e. <math>y \geq 5</math> so production of <math>Y \geq demand (95) - initial stock (90)</math> , which ensures we meet demand | ||
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| + | objective function: <math>max (x+30-75) + (y+90-95) = (x+y-50)</math> | ||
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| + | i.e. to maximise the number of units left in stock at the end of the week | ||
Version vom 19. Juni 2013, 14:45 Uhr
Mathematical formulations of problems presented in the course
Example:
A company makes two products (X and Y) using two machines (A and B). Each unit of X that is produced requires 50 minutes processing time on machine A and 30 minutes processing time on machine B. Each unit of Y that is produced requires 24 minutes processing time on machine A and 33 minutes processing time on machine B.
At the start of the current week there are 30 units of X and 90 units of Y in stock. Available processing time on machine A is forecast to be 40 hours and on machine B is forecast to be 35 hours.
The demand for X in the current week is forecast to be 75 units and for Y is forecast to be 95 units. Company policy is to maximise the combined sum of the units of X and the units of Y in stock at the end of the week.
x = number of units of X produced in the current week y = number of units of Y produced in the current week
Constraints:
Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): 50x + 24y \leq 40
machine A time
Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): 30x + 33y \leq 35
machine B time
Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): x \geq (75 - 30)
i.e. Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): x \geq 45
so production of Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): X \geq demand (75) - initial stock (30) , which ensures we meet demand
Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): y \geq 95 - 90
i.e. Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): y \geq 5
so production of Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): Y \geq demand (95) - initial stock (90) , which ensures we meet demand
objective function: Fehler beim Parsen (http://mathoid.testme.wmflabs.org Serverantwort ist ungültiges JSON.): max (x+30-75) + (y+90-95) = (x+y-50)
i.e. to maximise the number of units left in stock at the end of the week
