Zadanie. Dla dziesięciu miast 1,2, ... 9, 10 jest znany czas przejazdu tij między odpowiednie miastami i i j. |
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Która droga z miasta 1 do miasta 10 będzie najkrótsza? |
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Rozwiązanie. |
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1. Narysujemy graf sieci dróg. Dla tego połączmy lukami (strzałkami) wierzchołki (miasta), między którymi istnieją drogi. |
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2. Uporządkujemy jego metodą Fulkerson'a. Dla tego wyznaczymy wierzchołki, z których tylko wychodzą łuki. Z wykresu widać, |
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że to będzie wierzchołek 1. On tworzy I-ą grupę wierzchołków. Dalej śkreślamy z grafu wierzchołek 1 i luki, które zaczynają się w |
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wierzchołku 1 (luki 1-2, 1-3, 1-4, onie zaznaczony na wykresie grubszymi liniami). W pozostałym grafie z wierzchołków 2 i 3 |
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również tylko wychodzą luki. Wierzchołki 2 i 3 odniosą się do II-ej grupy. Jeżeli kontynuować ten proces przekształcenia grafu, |
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to będziemy mieć 9 grup wierzchołków, które narysowany na wykresie niżej. |
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3. Wykorzystamy metody programowania dynamicznego dla znalezienia minimalnej odległości między punktami 1 a 10, i również 2 a 10, 3 a 10, ... , 9 a 10. |
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Zaczynamy od ostatniego etapu XIX. Liczymy, że odległość z miasta 10 do miasta 10 równa się 0. Na przedostatnim etapie VIII (wierzchołek 9) z |
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wierzchołka 9 wychodzi tylko jedna luka. To oznacza, że minimalna odległość między miastami 9 i 10 równa się 2 jednostki (0+2=2). |
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Komórkę z optymalnym marszrutem przejazdu zaznaczamy kolorem. |
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Na VII etapie z wierzchołka 8 wychodzi 2 luki. Jeżeli będziemy jechać kierunkiem 8-10, to odległość będzie 0+5=5 jednostek, jeżeli trasą 8-9-10 |
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– wtedy 0+2+6=8 jednostek. Porównując trasy 8-10 i 8-9-10 wyberamy minimalną odległość z 8 do 10: to jest 5 jednostek i odpowiada kierunku 8-10. |
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Dalej potrzebnie zrobić optymalizację dla pozostałych etapów. |
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4. Obliczymy czas optymistyczny dla ścieżki 1-2-5-8-10: topt=10 i ścieżki 1-3-7-10: topt=12. Dla ścieżki 2-5-8-10: topt=9 |
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Obliczmy czas pesymistyczny dla ścieżki 1-2-5-8-10: tpes=24 i ścieżki 1-3-7-10: tpes=19. Dla ścieżki 2-5-8-10: tpes=20 Czas oczekiwany |
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dla ścieżki 1-2-5-8-10: toczekiwane = 14,33 dla ścieżki 1-3-7-10 : toczekiwane = 13,83 dla ścieżki 2-5-8-10: toczekiwane = 12,17.
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5. Na podstawie optymalizacji można zrobić następujące wnioski: |
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5.1. Optymalny kierunek z 1 w 10: 1-2-5-8-10 lub 1-3-7-10. Czas przejazdu 13 jednostek. |
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5.2. Optymalne kierunki: z 2 w 10: 2-5-8-10 (11); z 3 w 10: 3-7-10 (8); z 4 w 10: 4-6-7-10 (13); z 5 w 10: 5-8-10 (10); z 6 w 10: 6-7-10 (8); z 7 w 10: 7-10 (5); |
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z 8 w 10: 8-10 (5); z 9 w 10: 9-10 (2). |
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5.3. Oczekiwany czas przejazdu od 1 do 10 wynosi 13,83 jednostek czasowych, od 2 do 10 – 12,17 jednostek. |
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6. Korzystając z empirycznych wzorów dla wartości oczekiwanej i wariancji i przypuszczając że dla czasu przejazdu występuję rozkład normalny, |
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obliczamy parametry rozkładu i badamy prawdopodobieństwa przejazdu od p.1 do p.10 w określionych granicach czasu. |
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Zadanie. |
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1 |
Narysujemy graf sieci dróg. |
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tpl |
topt |
tpes |
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t12 |
2 |
1 |
4 |
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t13 |
5 |
4 |
7 |
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t14 |
9 |
7 |
11 |
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t24 |
4 |
2 |
8 |
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t25 |
1 |
1 |
5 |
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Solver |
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t26 |
9 |
8 |
9 |
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t35 |
5 |
5 |
5 |
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t36 |
9 |
8 |
11 |
2 |
Uporządkujemy graf metodą Fulkersona. |
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xij - zmienne decyzyjne binarne |
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t37 |
3 |
3 |
4 |
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I |
II |
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III |
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IV |
V |
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VI |
VII |
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VIII |
IX |
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x12 |
1 |
t12 |
2 |
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tij - obciążenia grafu |
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t45 |
5 |
3 |
8 |
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x13 |
0 |
t13 |
5 |
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Funkcja celu |
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t46 |
5 |
2 |
8 |
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x14 |
0 |
t14 |
9 |
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Z= |
13,0000000222407 |
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t56 |
4 |
3 |
5 |
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x24 |
0 |
t24 |
4 |
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Ograniczenia |
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t58 |
5 |
3 |
7 |
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x25 |
1 |
t25 |
1 |
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t67 |
3 |
3 |
6 |
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x26 |
0 |
t26 |
9 |
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1: |
1 |
= |
1 |
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t68 |
5 |
4 |
6 |
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x35 |
0 |
t35 |
5 |
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2: |
1 |
= |
1 |
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t78 |
9 |
7 |
13 |
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x36 |
0 |
t36 |
9 |
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3: |
0 |
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0 |
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t79 |
8 |
7 |
15 |
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x37 |
0 |
t37 |
3 |
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4: |
0 |
= |
0 |
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t710 |
5 |
5 |
8 |
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x45 |
0 |
t45 |
5 |
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5: |
1 |
= |
1 |
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t89 |
6 |
5 |
8 |
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x46 |
0 |
t46 |
5 |
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6: |
0 |
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0 |
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t810 |
5 |
5 |
8 |
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x56 |
0 |
t56 |
4 |
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7: |
0 |
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0 |
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t910 |
2 |
1 |
6 |
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x58 |
1 |
t58 |
5 |
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8: |
1 |
= |
1 |
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3 |
Metodami programowania dynamicznego znajdziemy minimalny czas przejazdu |
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x67 |
0 |
t67 |
3 |
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9: |
0 |
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0 |
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1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
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x68 |
0 |
t68 |
5 |
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10: |
1 |
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1 |
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x78 |
0 |
t78 |
9 |
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1_2 |
2_4 |
3_5 |
4_5 |
5_6 |
6_7 |
7_8 |
8_9 |
9_10 |
10_10 |
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x79 |
0 |
t79 |
8 |
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13 |
17 |
15 |
15 |
12 |
8 |
14 |
8 |
2 |
0 |
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x710 |
0 |
t710 |
5 |
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1_3 |
2_5 |
3_6 |
4_6 |
5_8 |
6_8 |
7_9 |
8_10 |
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x89 |
0 |
t89 |
6 |
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13 |
11 |
17 |
13 |
10 |
10 |
10 |
5 |
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x810 |
1 |
t810 |
5 |
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1_4 |
2_6 |
3_7 |
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7_10 |
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x910 |
0 |
t910 |
2 |
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22 |
17 |
8 |
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5 |
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4, 5 |
Obliczamy optymistyczny, pesymistyczny, oczekiwany czas przejazdu |
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trasa |
1_2 |
2_5 |
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5_8 |
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8_10 |
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10_10 |
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topt |
10 |
9 |
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8 |
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5 |
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0 |
toczekiwane= |
14,33 |
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tpes |
24 |
20 |
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15 |
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8 |
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0 |
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trasa |
1_3 |
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3_7 |
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7_10 |
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10_10 |
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topt |
12 |
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8 |
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5 |
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0 |
toczekiwane= |
13,83 |
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tpes |
19 |
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12 |
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8 |
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0 |
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trasa |
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2_5 |
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5_8 |
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8_10 |
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10_10 |
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topt |
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9 |
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8 |
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5 |
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0 |
toczekiwane= |
12,17 |
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tpes |
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20 |
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15 |
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8 |
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0 |
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6 |
Oblicz prawdopodobieństwo że czas przejazdu będzie zawierać się: a) między topt a 2topt; b) między 0,9toczek a 1,1 toczek. |
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Przypuszczając, że dla badania losowości czasu przejazdu można stosować rozkład normalny, |
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obliczymy parametry rozkładu |
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- gęstość prawdopodobieństwa dla rozkładu normalnego |
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t |
f(t) |
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Korzystniejszy (minimalny czas) kierunek jest 1-3-7-10. Dla tego |
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1 |
9,16666666666667 |
0,000114711622084 |
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toczekiwane= |
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13,83 |
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2 |
9,45833333333333 |
0,000302224870601 |
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Obliczymy wariancje s2 według wzoru: |
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3 |
9,75 |
0,000748013738611 |
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4 |
10,0416666666667 |
0,001739184049114 |
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5 |
10,3333333333333 |
0,003798727210233 |
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Wtedy |
s2 = |
1,36111111111111 |
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6 |
10,625 |
0,007794482144221 |
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s = |
1,16666666666667 |
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7 |
10,9166666666667 |
0,015024257565916 |
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W naszym przypadku |
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8 |
11,2083333333333 |
0,027205415859143 |
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9 |
11,5 |
0,046277971297018 |
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Obliczamy: |
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10 |
11,7916666666667 |
0,073951987565581 |
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a) prawdopodobieństwo że czas przejazdu będzie między t=12 a t=15; |
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11 |
12,0833333333333 |
0,111015081999335 |
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topt = |
12 |
2topt = |
15 |
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p = |
0,783303179199215 |
(narysowane na wykresie) |
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12 |
12,375 |
0,156556358904875 |
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b) prawdopodobieństwo że czas przejazdu będzie między 0,9toczek a 1,1toczek; |
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13 |
12,6666666666667 |
0,207403478159265 |
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0,9toczek = |
12,45 |
2topt = |
15,2166666666667 |
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p = |
0,764264849023151 |
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14 |
12,9583333333333 |
0,258117798989831 |
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7. Wnioski: |
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15 |
13,25 |
0,301770280083684 |
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1. Optymalny kierunek z 1 w 10: 1-2-5-8-10 lub 1-3-7-10 . Ego długość 13. |
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16 |
13,5416666666667 |
0,331429814402442 |
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2. Optymalne kierunki: z 2 w 10: 2-5-8-10 (11); z 3 w 10: 3-7-10 (8); z 4 w 10: 4-6-7-10 (13); |
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17 |
13,8333333333333 |
0,341950526058371 |
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z 5 w 10: 5-8-10 (10); z 6 w 10: 6-7-10 (8); z 7 w 10: 7-10 (5); |
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18 |
14,125 |
0,331429814402443 |
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z 8 w 10: 8-10 (5); z 9 w 10: 9-10 (2). |
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19 |
14,4166666666667 |
0,301770280083687 |
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3. Oczekiwany czas z 1 w 10: 1-2-5-8-10 --- 14,33 , dla 1-3-7-10 --- 13,83. Kierunek 1-3-7-10 jest lepszy. |
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20 |
14,7083333333333 |
0,258117798989834 |
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Oczekiwany czas z 2 w 10: 2-5-8-10 --- 12,17. |
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21 |
15 |
0,207403478159268 |
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4. |
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22 |
15,2916666666667 |
0,156556358904878 |
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a) prawdopodobieństwo że czas przejazdu będzie między topt a 2topt; |
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p = |
0,783 |
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23 |
15,5833333333333 |
0,111015081999338 |
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b) prawdopodobieństwo że czas przejazdu będzie między 0,9toczek a 1,1toczek; |
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p= |
0,764 |
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24 |
15,875 |
0,073951987565583 |
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25 |
16,1666666666667 |
0,046277971297019 |
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26 |
16,4583333333333 |
0,027205415859144 |
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27 |
16,75 |
0,015024257565916 |
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28 |
17,0416666666667 |
0,007794482144221 |
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29 |
17,3333333333333 |
0,003798727210233 |
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30 |
17,625 |
0,001739184049114 |
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31 |
17,9166666666667 |
0,000748013738611 |
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32 |
18,2083333333333 |
0,000302224870601 |
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33 |
18,5 |
0,000114711622084 |
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34 |
18,7916666666667 |
4,09016884638904E-05 |
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Zadanie. |
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1 |
Narysujemy graf sieci dróg. |
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tpl |
topt |
tpes |
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t12 |
2 |
1 |
4 |
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t13 |
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7 |
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t14 |
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11 |
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t24 |
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2 |
8 |
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t25 |
1 |
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5 |
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Solver |
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t26 |
9 |
8 |
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t35 |
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5 |
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t36 |
9 |
8 |
11 |
2 |
Uporządkujemy graf metodą Fulkersona. |
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xij - zmienne decyzyjne binarne |
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t37 |
3 |
3 |
4 |
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I |
II |
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III |
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IV |
V |
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VI |
VII |
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VIII |
IX |
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x12 |
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t12 |
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tij - obciążenia grafu |
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t45 |
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3 |
8 |
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x13 |
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t13 |
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Funkcja celu |
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t46 |
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2 |
8 |
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x14 |
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t14 |
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Z= |
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t56 |
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5 |
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x24 |
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t24 |
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Ograniczenia |
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t58 |
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7 |
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x25 |
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t25 |
1 |
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t67 |
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6 |
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x26 |
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t26 |
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1: |
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t68 |
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6 |
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x35 |
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t35 |
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t78 |
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7 |
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x36 |
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t36 |
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t79 |
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x37 |
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t37 |
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t710 |
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8 |
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x45 |
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t45 |
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t89 |
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x46 |
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t46 |
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t810 |
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x56 |
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t56 |
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1 |
6 |
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x58 |
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t58 |
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3 |
Metodami programowania dynamicznego znajdziemy minimalny czas przejazdu |
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x67 |
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t67 |
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x68 |
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t68 |
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x78 |
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t78 |
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1_2 |
2_4 |
3_5 |
4_5 |
5_6 |
6_7 |
7_8 |
8_9 |
9_10 |
10_10 |
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x79 |
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t79 |
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x710 |
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t710 |
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x89 |
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t89 |
6 |
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x810 |
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t810 |
5 |
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x910 |
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t910 |
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4, 5 |
Obliczamy optymistyczny, pesymistyczny, oczekiwany czas przejazdu |
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trasa |
1_2 |
2_5 |
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5_8 |
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8_10 |
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10_10 |
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topt |
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toczekiwane= |
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tpes |
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trasa |
1_3 |
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3_7 |
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7_10 |
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10_10 |
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topt |
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toczekiwane= |
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tpes |
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trasa |
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2_5 |
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5_8 |
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8_10 |
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10_10 |
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topt |
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toczekiwane= |
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tpes |
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6 |
Oblicz prawdopodobieństwo że czas przejazdu będzie zawierać się: a) między topt a 2topt; b) między 0,9toczek a 1,1 toczek. |
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Przypuszczając, że dla badania losowości czasu przejazdu można stosować rozkład normalny, |
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obliczymy parametry rozkładu |
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- gęstość prawdopodobieństwa dla rozkładu normalnego |
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Korzystniejszy (minimalny czas) kierunek jest 1-3-7-10. Dla tego |
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toczekiwane= |
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Obliczymy wariancje s2 według wzoru: |
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Wtedy |
s2 = |
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s = |
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W naszym przypadku |
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Obliczamy: |
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a) prawdopodobieństwo że czas przejazdu będzie między t=12 a t=15; |
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topt = |
12 |
2topt = |
15 |
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p = |
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(narysowane na wykresie) |
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b) prawdopodobieństwo że czas przejazdu będzie między 0,9toczek a 1,1toczek; |
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0,9toczek = |
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2topt = |
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p = |
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Zestaw 1-5. Dla dziesięciu miast 1,2, ... 9, 10 wiadomy czas przejazdu tij między odpowiednie miastami i i j . Która droga z miasta 1 |
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do miasta 10 będzie najkrótsza? |
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Potrzebnie: 1. Narysować graf sieci dróg. 2. Uporządkować jego metodą Falkersona. 3. Metodami programowania dynamicznego |
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znaleźć minimalny czas przejazdu między punktami 1 a 10, i również 2 a 10, 3 a 10, ... , 9 a 10. 4. Oblicz oczekiwany czas |
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przejazdu od miasta 1 do miast 10 i od miasta 2 do miasta 10. 5. Oblicz prawdopodobieństwo że czas przejazdu będzie zawierać |
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się: a) między tpes a 1,5tpes; b) między 0,75toczek a 1,25 toczek. 6. Zrobić wnioski. |
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Zestaw 6-10. Dla dziesięciu miast 1,2, ... 9, 10 taksówkarzu wiadomy płacy za czas przyjazdu tij między odpowiednie |
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miastami i i j . Którą drogą z miasta 1 do miasta 10 powinien taksówkarz przewozić pasażerów, żeby otrzymać |
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największy zysk? Drogi przejezdne tylko w jednym kierunku. |
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Potrzebnie: 1. Narysować graf sieci dróg. 2. Uporządkować jego metodą Falkersona. 3. Metodami programowania |
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dynamicznego znaleźć maksymalny czas przejazdu (i maksymalny zysk) od przewozów między punktami 1 a 10, i również 2 a |
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10, 3 a 10, ... ,9 a 10. 4. Oblicz oczekiwany czas przejazdu od miasta 1 do miast 10 i od miasta 2 do miasta 10. |
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5. Oblicz prawdopodobieństwo że czas przejazdu będzie zawierać się: a) między topt a tpes; b) między 0,8toczek a 1,2toczek. |
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6. Zrobić wnioski. |
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W tablicach są zapisane macierzy zbieżności wierzchołków. Liczba w komórce oznacza połączenie pomiędzy i-tym a j-ym |
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punktem (opłatę, czas), komunikat "nie" - że połączenie nie istnieje. |
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Z prawej strony w macierzach zapisane, na ile procent zmniejsza się (zwiększa się) odpowiednia wartość dla czasu |
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optymistycznego (pesymistycznego). |
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Zestaw 1, 6 |
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Zestaw 1, 6 Granice dolne |
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Zestaw 1, 6 Granice górne |
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Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
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Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
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Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
1 |
0 |
9 |
5 |
7 |
4 |
nie |
nie |
nie |
nie |
nie |
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1 |
0 |
6 |
1 |
3 |
2 |
nie |
nie |
nie |
nie |
nie |
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1 |
0 |
10 |
6 |
10 |
7 |
nie |
nie |
nie |
nie |
nie |
2 |
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0 |
nie |
nie |
7 |
nie |
nie |
2 |
6 |
nie |
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2 |
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0 |
nie |
nie |
5 |
nie |
nie |
1 |
4 |
nie |
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2 |
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0 |
nie |
nie |
10 |
nie |
nie |
3 |
9 |
nie |
3 |
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0 |
5 |
nie |
9 |
nie |
nie |
nie |
nie |
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3 |
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0 |
2 |
nie |
7 |
nie |
nie |
nie |
nie |
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3 |
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0 |
8 |
nie |
12 |
nie |
nie |
nie |
nie |
4 |
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0 |
3 |
6 |
nie |
nie |
nie |
8 |
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4 |
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0 |
2 |
3 |
nie |
nie |
nie |
6 |
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4 |
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0 |
4 |
8 |
nie |
nie |
nie |
11 |
5 |
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0 |
nie |
9 |
nie |
5 |
nie |
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5 |
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0 |
nie |
5 |
nie |
2 |
nie |
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5 |
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0 |
nie |
12 |
nie |
6 |
nie |
6 |
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0 |
nie |
nie |
nie |
2 |
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6 |
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0 |
nie |
nie |
nie |
1 |
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6 |
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0 |
nie |
nie |
nie |
3 |
7 |
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0 |
nie |
2 |
6 |
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7 |
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0 |
nie |
1 |
2 |
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7 |
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0 |
nie |
3 |
7 |
8 |
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0 |
4 |
nie |
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8 |
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0 |
1 |
nie |
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8 |
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0 |
6 |
nie |
9 |
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0 |
7 |
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9 |
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0 |
3 |
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9 |
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0 |
8 |
10 |
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0 |
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10 |
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0 |
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10 |
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0 |
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Zestaw 2, 7 |
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Zestaw 2, 7 Granice dolne |
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Zestaw 2, 7 Granice górne |
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Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
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Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
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Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
1 |
0 |
7 |
4 |
9 |
nie |
nie |
6 |
nie |
nie |
nie |
|
1 |
0 |
3 |
1 |
5 |
nie |
nie |
4 |
nie |
nie |
nie |
|
1 |
0 |
8 |
7 |
11 |
nie |
nie |
8 |
nie |
nie |
nie |
2 |
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0 |
nie |
5 |
4 |
nie |
7 |
nie |
nie |
nie |
|
2 |
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0 |
nie |
1 |
1 |
nie |
5 |
nie |
nie |
nie |
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2 |
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0 |
nie |
8 |
6 |
nie |
9 |
nie |
nie |
nie |
3 |
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0 |
nie |
nie |
8 |
nie |
9 |
nie |
nie |
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3 |
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0 |
nie |
nie |
6 |
nie |
5 |
nie |
nie |
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3 |
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0 |
nie |
nie |
10 |
nie |
12 |
nie |
nie |
4 |
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0 |
nie |
nie |
2 |
nie |
nie |
nie |
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4 |
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0 |
nie |
nie |
1 |
nie |
nie |
nie |
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4 |
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0 |
nie |
nie |
3 |
nie |
nie |
nie |
5 |
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0 |
nie |
9 |
nie |
3 |
nie |
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5 |
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0 |
nie |
5 |
nie |
2 |
nie |
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5 |
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0 |
nie |
11 |
nie |
4 |
nie |
6 |
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0 |
3 |
nie |
nie |
5 |
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6 |
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0 |
2 |
nie |
nie |
3 |
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6 |
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0 |
4 |
nie |
nie |
7 |
7 |
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0 |
nie |
4 |
6 |
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7 |
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0 |
nie |
1 |
3 |
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7 |
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0 |
nie |
7 |
8 |
8 |
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0 |
nie |
4 |
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8 |
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0 |
nie |
2 |
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8 |
|
|
|
|
|
|
|
0 |
nie |
7 |
9 |
|
|
|
|
|
|
|
|
0 |
8 |
|
9 |
|
|
|
|
|
|
|
|
0 |
5 |
|
9 |
|
|
|
|
|
|
|
|
0 |
9 |
10 |
|
|
|
|
|
|
|
|
|
0 |
|
10 |
|
|
|
|
|
|
|
|
|
0 |
|
10 |
|
|
|
|
|
|
|
|
|
0 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
|
Zestaw 3, 8 |
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|
|
Zestaw 3, 8 Granice dolne |
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|
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|
|
Zestaw 3, 8 Granice górne |
|
|
|
|
|
|
|
|
|
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
1 |
0 |
4 |
5 |
7 |
3 |
nie |
nie |
nie |
nie |
nie |
|
1 |
0 |
0 |
3 |
5 |
2 |
nie |
nie |
nie |
nie |
nie |
|
1 |
0 |
6 |
7 |
10 |
4 |
nie |
nie |
nie |
nie |
nie |
2 |
|
0 |
nie |
8 |
nie |
3 |
nie |
nie |
nie |
nie |
|
2 |
|
0 |
nie |
6 |
nie |
2 |
nie |
nie |
nie |
nie |
|
2 |
|
0 |
nie |
11 |
nie |
4 |
nie |
nie |
nie |
nie |
3 |
|
|
0 |
nie |
8 |
nie |
nie |
nie |
7 |
nie |
|
3 |
|
|
0 |
nie |
4 |
nie |
nie |
nie |
4 |
nie |
|
3 |
|
|
0 |
nie |
9 |
nie |
nie |
nie |
9 |
nie |
4 |
|
|
|
0 |
nie |
nie |
nie |
9 |
nie |
nie |
|
4 |
|
|
|
0 |
nie |
nie |
nie |
6 |
nie |
nie |
|
4 |
|
|
|
0 |
nie |
nie |
nie |
11 |
nie |
nie |
5 |
|
|
|
|
0 |
nie |
nie |
1 |
5 |
nie |
|
5 |
|
|
|
|
0 |
nie |
nie |
1 |
1 |
nie |
|
5 |
|
|
|
|
0 |
nie |
nie |
2 |
6 |
nie |
6 |
|
|
|
|
|
0 |
5 |
2 |
nie |
nie |
|
6 |
|
|
|
|
|
0 |
3 |
1 |
nie |
nie |
|
6 |
|
|
|
|
|
0 |
7 |
3 |
nie |
nie |
7 |
|
|
|
|
|
|
0 |
nie |
nie |
9 |
|
7 |
|
|
|
|
|
|
0 |
nie |
nie |
7 |
|
7 |
|
|
|
|
|
|
0 |
nie |
nie |
11 |
8 |
|
|
|
|
|
|
|
0 |
nie |
8 |
|
8 |
|
|
|
|
|
|
|
0 |
nie |
4 |
|
8 |
|
|
|
|
|
|
|
0 |
nie |
10 |
9 |
|
|
|
|
|
|
|
|
0 |
3 |
|
9 |
|
|
|
|
|
|
|
|
0 |
2 |
|
9 |
|
|
|
|
|
|
|
|
0 |
4 |
10 |
|
|
|
|
|
|
|
|
|
0 |
|
10 |
|
|
|
|
|
|
|
|
|
0 |
|
10 |
|
|
|
|
|
|
|
|
|
0 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
|
Zestaw 4, 9 |
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|
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|
|
Zestaw 4, 9 Granice dolne |
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|
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|
|
|
|
|
Zestaw 4, 9 Granice górne |
|
|
|
|
|
|
|
|
|
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
1 |
0 |
7 |
9 |
nie |
5 |
nie |
nie |
nie |
nie |
nie |
|
1 |
0 |
4 |
6 |
nie |
2 |
nie |
nie |
nie |
nie |
nie |
|
1 |
0 |
10 |
11 |
nie |
8 |
nie |
nie |
nie |
nie |
nie |
2 |
|
0 |
nie |
nie |
9 |
6 |
nie |
nie |
nie |
nie |
|
2 |
|
0 |
nie |
nie |
5 |
4 |
nie |
nie |
nie |
nie |
|
2 |
|
0 |
nie |
nie |
12 |
7 |
nie |
nie |
nie |
nie |
3 |
|
|
0 |
4 |
3 |
nie |
5 |
nie |
nie |
nie |
|
3 |
|
|
0 |
1 |
2 |
nie |
1 |
nie |
nie |
nie |
|
3 |
|
|
0 |
7 |
4 |
nie |
8 |
nie |
nie |
nie |
4 |
|
|
|
0 |
nie |
nie |
1 |
nie |
3 |
nie |
|
4 |
|
|
|
0 |
nie |
nie |
0 |
nie |
2 |
nie |
|
4 |
|
|
|
0 |
nie |
nie |
2 |
nie |
4 |
nie |
5 |
|
|
|
|
0 |
nie |
4 |
3 |
nie |
8 |
|
5 |
|
|
|
|
0 |
nie |
0 |
2 |
nie |
6 |
|
5 |
|
|
|
|
0 |
nie |
5 |
4 |
nie |
10 |
6 |
|
|
|
|
|
0 |
nie |
2 |
nie |
4 |
|
6 |
|
|
|
|
|
0 |
nie |
1 |
nie |
0 |
|
6 |
|
|
|
|
|
0 |
nie |
3 |
nie |
6 |
7 |
|
|
|
|
|
|
0 |
nie |
9 |
8 |
|
7 |
|
|
|
|
|
|
0 |
nie |
5 |
5 |
|
7 |
|
|
|
|
|
|
0 |
nie |
11 |
9 |
8 |
|
|
|
|
|
|
|
0 |
nie |
3 |
|
8 |
|
|
|
|
|
|
|
0 |
nie |
2 |
|
8 |
|
|
|
|
|
|
|
0 |
nie |
4 |
9 |
|
|
|
|
|
|
|
|
0 |
6 |
|
9 |
|
|
|
|
|
|
|
|
0 |
2 |
|
9 |
|
|
|
|
|
|
|
|
0 |
8 |
10 |
|
|
|
|
|
|
|
|
|
0 |
|
10 |
|
|
|
|
|
|
|
|
|
0 |
|
10 |
|
|
|
|
|
|
|
|
|
0 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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|
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|
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|
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|
|
Zestaw 5, 10 |
|
|
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|
|
|
Zestaw 5, 10 Granice dolne |
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|
|
|
|
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|
|
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|
|
Zestaw 5, 10 Granice górne |
|
|
|
|
|
|
|
|
|
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
1 |
0 |
8 |
4 |
7 |
9 |
nie |
nie |
nie |
nie |
nie |
|
1 |
0 |
5 |
1 |
5 |
7 |
nie |
nie |
nie |
nie |
nie |
|
1 |
0 |
11 |
6 |
8 |
10 |
nie |
nie |
nie |
nie |
nie |
2 |
|
0 |
nie |
5 |
nie |
3 |
nie |
nie |
nie |
nie |
|
2 |
|
0 |
nie |
1 |
nie |
2 |
nie |
nie |
nie |
nie |
|
2 |
|
0 |
nie |
8 |
nie |
4 |
nie |
nie |
nie |
nie |
3 |
|
|
0 |
nie |
2 |
nie |
6 |
nie |
nie |
nie |
|
3 |
|
|
0 |
nie |
1 |
nie |
3 |
nie |
nie |
nie |
|
3 |
|
|
0 |
nie |
3 |
nie |
9 |
nie |
nie |
nie |
4 |
|
|
|
0 |
nie |
2 |
nie |
6 |
nie |
nie |
|
4 |
|
|
|
0 |
nie |
1 |
nie |
2 |
nie |
nie |
|
4 |
|
|
|
0 |
nie |
3 |
nie |
7 |
nie |
nie |
5 |
|
|
|
|
0 |
nie |
4 |
5 |
8 |
nie |
|
5 |
|
|
|
|
0 |
nie |
1 |
3 |
4 |
nie |
|
5 |
|
|
|
|
0 |
nie |
7 |
8 |
11 |
nie |
6 |
|
|
|
|
|
0 |
nie |
9 |
nie |
8 |
|
6 |
|
|
|
|
|
0 |
nie |
6 |
nie |
6 |
|
6 |
|
|
|
|
|
0 |
nie |
10 |
nie |
9 |
7 |
|
|
|
|
|
|
0 |
nie |
5 |
nie |
|
7 |
|
|
|
|
|
|
0 |
nie |
1 |
nie |
|
7 |
|
|
|
|
|
|
0 |
nie |
6 |
nie |
8 |
|
|
|
|
|
|
|
0 |
1 |
5 |
|
8 |
|
|
|
|
|
|
|
0 |
1 |
3 |
|
8 |
|
|
|
|
|
|
|
0 |
2 |
6 |
9 |
|
|
|
|
|
|
|
|
0 |
7 |
|
9 |
|
|
|
|
|
|
|
|
0 |
5 |
|
9 |
|
|
|
|
|
|
|
|
0 |
10 |
10 |
|
|
|
|
|
|
|
|
|
0 |
|
10 |
|
|
|
|
|
|
|
|
|
0 |
|
10 |
|
|
|
|
|
|
|
|
|
0 |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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Punkty |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
|
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|
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|
|
1 |
0 |
15,6 |
10,5 |
nie |
12,5 |
nie |
nie |
13,8 |
15,8 |
1,9 |
|
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
2 |
|
0 |
11,3 |
nie |
18 |
nie |
19,2 |
nie |
nie |
nie |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
3 |
|
|
0 |
nie |
2,8 |
11,6 |
nie |
nie |
nie |
2,5 |
|
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|
|
|
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|
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|
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|
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4 |
|
|
|
0 |
10,2 |
nie |
11,3 |
nie |
nie |
nie |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
5 |
|
|
|
|
0 |
8,9 |
nie |
nie |
14,8 |
4 |
|
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|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
6 |
|
|
|
|
|
0 |
nie |
nie |
nie |
nie |
|
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|
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|
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7 |
|
|
|
|
|
|
0 |
15,3 |
19,9 |
nie |
|
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|
|
|
|
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8 |
|
|
|
|
|
|
|
0 |
15,9 |
nie |
|
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|
|
|
|
|
|
|
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9 |
|
|
|
|
|
|
|
|
0 |
14,4 |
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|
10 |
|
|
|
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|
0 |
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