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Linear and Integer Programming (ADM II) - Institut für Mathematik

Linear and Integer Programming (ADM II) - Institut für Mathematik

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For the blending process certain quality constraints have to be met. High quality<br />

gasoline has to consist of at least 70% of heavy naphtha <strong>and</strong> at least 20% of cracked<br />

oil. High quality fuel oil may not contain any residuum, while the low quality may<br />

contain at most 15% of the residuum. For marketing reasons the refinery decides to<br />

produce at least half as many of the low quality gasoline as of the high quality gas.<br />

Finally, the selling prices per barrel are $365 for the high quality gasoline, $260 for<br />

the low quality, $200 for high quality fuel oil, <strong>and</strong> $140 for the low quality fuel oil.<br />

The lube oil still earns $45 per barrel. One barrel of crude oil costs the refinery $90<br />

<strong>and</strong> for the planning period 1000 barrels will be delivered.<br />

Naturally, the refinery wants to maximize its profit. Set up a linear program that<br />

models the problem <strong>and</strong> solve it using zimpl <strong>and</strong> cplex or soplex.<br />

Remarks (to think about): What happens if you drop the prerequisite of the 1000<br />

barrels of crude oil? Why? How much lube oil is produced? Why?<br />

Exercise 31. 8 points<br />

Consider the following linear program:<br />

minimize −2x1 − x2<br />

subject to x1 − 2x2 ≤ 2<br />

x1 − x2 ≤ 2<br />

x1 + x2 ≤ 6<br />

−4x1 + x2 ≤ 1<br />

x1, x2 ≥ 0<br />

Suppose the right-h<strong>and</strong> side vector is changed to (1, 3, 4, 1) ⊤ <strong>and</strong> the new constraint<br />

2x1 − x2 ≤ 2 is added.<br />

Show how the basis corresponding to your original optimal solution, which was<br />

(hopefully) the point (x1, x2) = (4, 2), can be used to initialise the dual simplex<br />

method <strong>and</strong> solve the new linear program with this approach.<br />

Exercise 32. 8 points<br />

Solve the following linear program with the dual simplex method:<br />

minimize x1 + 2x2<br />

subject to x1 + x2 ≥ 3<br />

x2 ≥ 2<br />

−x1 + x2 ≤ 3<br />

x1 − 2x2 ≤ 1<br />

x1 − x2 ≤ 3<br />

x1, x2 ≥ 0

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