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Alle optimale løsninger<br />

Optimale basisløsninger:<br />

x ∗ = (x 1 , x 2 ) = (4, 3)<br />

x ∗ = (2, 6)<br />

Alle optimale løsninger:<br />

x ∗ = w 1 (4, 3) + w 2 (2, 6)<br />

for alle w 1 , w 2 med w 1 + w 2 = 1, w 1 ≥ 0, w 2 ≥ 0<br />

4.6. Kunstige variabler<br />

Strålebehandling (fra s. 44)<br />

min. Z = .4x 1 + .5x 2<br />

uht. .3x 1 + .1x 2 ≤ 2.7<br />

.5x 1 + .5x 2 = 6<br />

.6x 1 + .4x 2 ≥ 6<br />

x 1 ≥ 0, x 2 ≥ 0<br />

slack x 3 , surplus x 5 , kunstige x 4 og x 6<br />

alle x-variabler ikkenegative<br />

.3x 1 + .1x 2 + x 3 = 2.7<br />

.5x 1 + .5x 2 + x 4 = 6<br />

.6x 1 + .4x 2 − x 5 + x 6 = 6<br />

4. Simplexmetoden – p. 13/31<br />

4. Simplexmetoden – p. 14/31<br />

Tofasemetoden<br />

I fase 1 minimeres summen af de kunstige variabler:<br />

min. Z = x 4 + x 6 eller<br />

max. −Z uht. −Z + x 4 + x 6 = 0<br />

(husk at etablere et legitimt Simplextableau)<br />

Når Z = 0 har vi en basis af ikkekunstige variabler<br />

I fase 2 optimeres den rigtige målfunktion:<br />

min. Z = .4x 1 + .5x 2 eller<br />

max. −Z uht. −Z + .4x 1 + .5x 2 = 0<br />

<strong>Fase</strong> 1: legitimt tableau<br />

Z x 1 x 2 x 3 x 4 x 5 x 6<br />

−1 1 1<br />

3/10 1/10 1 27/10<br />

1/2 1/2 → 1 6<br />

3/5 2/5 −1 → 1 6<br />

−1 −11/10 −9/10 1 −12<br />

⇒ 3/10 1/10 1 0 27/10<br />

1/2 1/2 1 0 6<br />

3/5 2/5 −1 1 6<br />

4. Simplexmetoden – p. 15/31<br />

4. Simplexmetoden – p. 16/31

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