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Dynamisk programmering II - Matematik og optimering

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Hvis vi indfører:<br />

v (k1,...,kN<br />

⎛<br />

v<br />

⎜<br />

)<br />

(n) = ⎜<br />

⎝<br />

(k1,...,kN )<br />

1 (n)<br />

v (k1,...,kN )<br />

2 (n)<br />

.<br />

v (k1,...,kN<br />

⎞<br />

⎟<br />

⎠<br />

)<br />

N (n)<br />

⎛ ⎞ ⎛<br />

q k1<br />

1<br />

q k2<br />

2<br />

q (k1,...,kN<br />

⎜ ⎟<br />

) ⎜ ⎟<br />

= ⎜ . ⎟<br />

⎝ . ⎠ =<br />

⎜<br />

⎝<br />

⎛<br />

q kN<br />

N<br />

p k1<br />

11<br />

p k2<br />

21<br />

P (k1,...,kN<br />

⎜<br />

) ⎜<br />

= ⎜ .<br />

⎝ .<br />

p kN<br />

N1<br />

da f˚as alts˚a sammenhængen:<br />

Dias 29/36 — Henrik Holm (IGM) — <strong>Dynamisk</strong> <strong>pr<strong>og</strong>rammering</strong> <strong>II</strong> — 31. marts 2011<br />

p k1<br />

11<br />

p k2<br />

21<br />

k1 r11 k2 r21 p kN kN<br />

N1rN1 + pk1<br />

12<br />

+ pk2<br />

22<br />

p k1<br />

12 ··· p k1<br />

1N<br />

p k2<br />

22 ··· p k2<br />

2N<br />

.<br />

. ..<br />

p kN<br />

N2 ··· p kN<br />

NN<br />

k1 r12 k2 r22 kN + pkN<br />

N2rN2 ⎞<br />

⎟<br />

. ⎟<br />

. ⎠<br />

k1 ⎞<br />

+ ··· + pk1<br />

1Nr1N k2<br />

+ ··· + pk2<br />

2Nr ⎟<br />

2N ⎟<br />

.<br />

⎟<br />

.<br />

⎠<br />

kN<br />

+ ··· + pkN<br />

NNrNN v (k1,...,kN ) (n) = q (k1,...,kN ) + P (k1,...,kN ) v (k1,...,kN ) (n − 1)<br />

Man kan fx nu succesivt udregne:<br />

v (1,1) <br />

0<br />

(0) =<br />

0<br />

v (1,1) <br />

68 0.7 0.3 0 68<br />

(1) = +<br />

=<br />

28 0.6 0.4 0 28<br />

v (1,1) <br />

68 0.7 0.3 68 124<br />

(2) = +<br />

=<br />

28 0.6 0.4 28 80<br />

v (1,1) <br />

68 0.7 0.3 124 179<br />

(3) = +<br />

=<br />

28 0.6 0.4 80 134<br />

Dias 31/36 — Henrik Holm (IGM) — <strong>Dynamisk</strong> <strong>pr<strong>og</strong>rammering</strong> <strong>II</strong> — 31. marts 2011<br />

.<br />

For strategien (k1,k2) = (1,1) (dvs. man vælger konsekvent den dyreste<br />

reklamekampagne <strong>og</strong> den dyreste markedsanalyse) f˚as:<br />

v (1,1) <br />

v<br />

(n) =<br />

(1,1)<br />

1 (n)<br />

v (1,1)<br />

<br />

2 (n)<br />

q (1,1) <br />

1<br />

q1 =<br />

q 1 <br />

1<br />

p11r =<br />

2<br />

1 11 + p1 12r 1 12<br />

p 1 21r 1 21 + p1 22r 1 <br />

=<br />

22<br />

P (1,1) <br />

1<br />

p11 p<br />

=<br />

1 <br />

12 0.7 0.3<br />

=<br />

0.6 0.4<br />

p 1 21<br />

p 1 22<br />

Dias 30/36 — Henrik Holm (IGM) — <strong>Dynamisk</strong> <strong>pr<strong>og</strong>rammering</strong> <strong>II</strong> — 31. marts 2011<br />

<br />

0.7 · 80 + 0.3 · 40<br />

=<br />

0.6 · 40 + 0.4 · 10<br />

Tilstand Valg p k ij r k<br />

ij q k i k j = 1 j = 2 j = 1 j = 2<br />

i<br />

1<br />

1<br />

2<br />

0.7<br />

0.5<br />

0.3<br />

0.5<br />

80<br />

100<br />

40<br />

60<br />

68<br />

80<br />

2<br />

1<br />

2<br />

0.6<br />

0.33<br />

0.4<br />

0.67<br />

40<br />

60<br />

10<br />

21<br />

28<br />

34<br />

Gevinsten g (k1,...,kN ) for strategien (k1,...,kN) er den asymptotiske<br />

gennemsnitlige ˚arlige indtægt – som faktisk er uafhængig af i – alts˚a:<br />

1<br />

n v (k1,...,kN )<br />

i (n) → g (k1,...,kN )<br />

for n → ∞<br />

For strategien (k1,k2) = (1,1) finder vi:<br />

1<br />

10 v (1,1) (10) = 1<br />

<br />

561 56.1<br />

=<br />

10 517 51.7<br />

1<br />

100 v (1,1) (100) = 1<br />

<br />

5481 54.8<br />

=<br />

100 5437 54.4<br />

1<br />

1000 v (1,1) (1000) = 1<br />

<br />

54681 54.7<br />

=<br />

1000 54637 54.6<br />

1<br />

10000 v (1,1) (10000) = 1<br />

<br />

546681 54.7<br />

=<br />

10000 546637 54.7<br />

Gevinsten for denne strategi er alts˚a g (1,1) = 54.7 (kr/˚ar i gennemsnit).<br />

Dias 32/36 — Henrik Holm (IGM) — <strong>Dynamisk</strong> <strong>pr<strong>og</strong>rammering</strong> <strong>II</strong> — 31. marts 2011<br />

<br />

68<br />

28

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