Dynamisk programmering II - Matematik og optimering
Dynamisk programmering II - Matematik og optimering
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