ゲーム理論BASIC 演習45 -ナッシュ交渉解-
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ゲーム理論BASIC 演習45 -ナッシュ交渉解- https://youtu.be/sv2X1LSning
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ゲーム理論BASIC 演習45 -ナッシュ交渉解-
- 3. ަবͷఆٛ ఆٛɿަব ࣮ݱՄೳू߹ɿ ަবͷෆҰகɿ ͷ ʹΑͬͯఆٛ͞ΕΔ Ծఆ w ۭ࣮࣍ؒݩ ͷίϯύΫτͰತͳ෦ू߹ w w ͷ ͕ଘࡏͯ͠ U d = (d1 , d2 ) (U, d) U ℜ2 d ∈ U U u = (u1 , u2 ) ui di (i = 1, 2) 0 u2 u1 (d1 , d2 ) = (20, 20) U 100 100
- 4. ަবղ͕ຬ͖ͨͭ͢ͷެཧ ύϨʔτ࠷దੑ ͯ͢ͷ ʹରͯ͠ Ͱ গͳ͘ͱͭͷ ʹରͯ͠ Ͱ͋Δ ͷ ଘࡏ͠ͳ͍ ରশੑ ަব ͕ରশͳΒ ˎަব͕ରশͰ͋Δͱ ͳΒ ਖ਼࣍ม͔Βͷಠཱੑ ަব ͕ަব ͔Βޮ༻ͷਖ਼࣍ม ʹΑͬͯಘΒΕΔͱ͖ ແؔͳ݁Ռ͔Βͷಠཱੑ ̎ͭͷަব ͱ ʹ͓͍ͯ ͱ͢Δ ͠ ͳΒ Ͱ͋Δ i = 1, 2 ti ≥ fi (U, d) i ti fi (U, d) U t = (t1 , t2 ) (U, d) f1 (U, d) = f2 (U, d) d1 = d2 (u1 , u2 ) ∈ U (u2 , u1 ) ∈ U (U′  , d′  ) (U, d) ui′  = αi ui + βi , αi 0 (i = 1,2) fi (U′  , d′  ) = αi fi (U, d) + βi , αi 0 (i = 1,2) (U, d) (T, d) T ⊂ U f(U, d) ∈ T f(T, d) = f(U, d)
- 5. ަবղ͕ຬ͖ͨͭ͢ͷެཧ ఆཧ ަব ͷ ͱ͢Δ ͭͷެཧΛຬͨ͢ަবͷղ ͕Ұҙʹଘࡏ͠ ͯ͢ͷަব ʹରͯ͠ ͷղͰ͋Δ ͜ͷղΛφογϡަবղͱ͍͏ ·ͨ Λφογϡੵͱ͍͏ (U, d) B f : B → ℜ2 (U, d) f(U, d) max u∈U:u≥d (u1 − d1 )(u2 − d2 ) (u1 − d1 )(u2 − d2 )
- 7. ਓͷట͕͢ྗڠΔ͜ͱͰ ສԁΛखʹೖΕΔ͜ͱ͕Ͱ͖Δ ͍ͳ͠ྗڠͷͰ͋Ε ਓԿखʹೖΕΒΕͣ ͏ਓສԁΛಠΓͰखʹೖΕΔ͜ͱ͕Ͱ͖Δ ֤ట ϓϨΠϠʔ ҎԼͷΑ͏ͳޮ༻ؔΛͭͱ͢Δɿ ϓϨΠϠʔϦεΫճආతͳޮ༻ؔ ϓϨΠϠʔϦεΫதཱతͳޮ༻ؔ ͜ͷͱ͖φογϡަবղΛٻΊΑ ҎԼͷφογϡੵ࠷େԽΛղ͚Α͍ u1(x1) = x1 u2(x2) = x2 max (u1,u2)≥(0,100) (u1 − 0)(u2 − 100) s . t . x1 + x2 = 400 x1, x2 ≥ 0 ղ max 0≤x1≤400 ( x1 − 0)(400 − x1 − 100)
- 8. ͱ͢Δͱ Ұ֊݅ΑΓ ࠷େԽΛߟ͍͑ͯΔͷͰ ͕ղͱͳΔ Ώ͑ʹ ޮ༻Ͱߟ͑Δͱ max 0≤x1≤400 ( x1 − 0)(300 − x1) f(x1) = ( x1 − 0)(300 − x1) f′  (x1) = {( x1 − 0)(300 − x1)}′  = 1 2 x− 1 2 1 (300 − x1) + x1(−1) = 3 2 x−1 2 1 (100 − x1) x1 = 100 (x1, x2) = (100, 300) (u1, u2) = ( x1, x2) = (10, 300) ղ ʜ ʜ
- 9. ʿ
- 11. ެཧ͔ΒٻΊΔ ͱ͍͏݅ͷͱͰ࣮ݱՄೳू߹Λਤࣔ͢Δ u1(x1) = x1 u2(x2) = x2 x1 + x2 = 400 x1, x2 ≥ 0 u2 1 + u2 = 400 ղ 0 u2 u1 400 (d1, d2) = (0, 100) 20
- 12. ެཧ͔ΒٻΊΔ ͱ͍͏݅ͷͱͰ࣮ݱՄೳू߹Λਤࣔ͢Δ u1(x1) = x1 u2(x2) = x2 x1 + x2 = 400 x1, x2 ≥ 0 u2 1 + u2 = 400 ղ 0 u2 u1 400 (d1, d2) = (0, 100) 20 0 u′  2 u′  1 3 4 (d′  1, d′  2) = (0, 0) 1 ਖ਼ΞϑΟϯมΛߟ͑Δ ͜ͷม͍ͯͭʹ#4*$ୈճࢀর u′  1 = 1 20 u1 u′  2 = 1 400 (u2 − 100) 4u′  1 2 + 4u′  2 = 3
- 13. ެཧ͔ΒٻΊΔ ͱ͍͏݅ͷͱͰ࣮ݱՄೳू߹Λਤࣔ͢Δ u1(x1) = x1 u2(x2) = x2 x1 + x2 = 400 x1, x2 ≥ 0 u2 1 + u2 = 400 ղ 0 u2 u1 400 (d1, d2) = (0, 100) 20 0 u′  2 u′  1 3 4 (d′  1, d′  2) = (0, 0) 1 ( 1 2 , 1 2) ରশੑɾύϨʔτ࠷దੑ 4u′  1 2 + 4u′  2 = 3 ਖ਼ΞϑΟϯมΛߟ͑Δ ͜ͷม͍ͯͭʹ#4*$ୈճࢀর u′  1 = 1 20 u1 u′  2 = 1 400 (u2 − 100)
- 14. ެཧ͔ΒٻΊΔ ͱ͍͏݅ͷͱͰ࣮ݱՄೳू߹Λਤࣔ͢Δ u1(x1) = x1 u2(x2) = x2 x1 + x2 = 400 x1, x2 ≥ 0 u2 1 + u2 = 400 ղ 0 u2 u1 400 (d1, d2) = (0, 100) 20 0 u′  2 u′  1 3 4 (d′  1, d′  2) = (0, 0) 1 ( 1 2 , 1 2) ແؔͳ݁Ռ͔Βͷಠཱੑ ແؔͳ݁Ռ͔Βͷಠཱੑ ̎ͭͷަব ͱ ʹ͓͍ͯ ͱ͢Δ ͠ ͳΒ Ͱ͋Δ (U, d) (T, d) T ⊂ U f(U, d) ∈ T f(T, d) = f(U, d) 4u′  1 2 + 4u′  2 = 3 ਖ਼ΞϑΟϯมΛߟ͑Δ ͜ͷม͍ͯͭʹ#4*$ୈճࢀর u′  1 = 1 20 u1 u′  2 = 1 400 (u2 − 100)
- 15. ެཧ͔ΒٻΊΔ ͱ͍͏݅ͷͱͰ࣮ݱՄೳू߹Λਤࣔ͢Δ u1(x1) = x1 u2(x2) = x2 x1 + x2 = 400 x1, x2 ≥ 0 u2 1 + u2 = 400 ղ 0 u2 u1 400 (d1, d2) = (0, 100) 20 0 u′  2 u′  1 3 4 (d′  1, d′  2) = (0, 0) 1 ( 1 2 , 1 2) ແؔͳ݁Ռ͔Βͷಠཱੑ ແؔͳ݁Ռ͔Βͷಠཱੑ ̎ͭͷަব ͱ ʹ͓͍ͯ ͱ͢Δ ͠ ͳΒ Ͱ͋Δ (U, d) (T, d) T ⊂ U f(U, d) ∈ T f(T, d) = f(U, d) 4u′  1 2 + 4u′  2 = 3 ਖ਼ΞϑΟϯม͔Βͷಠཱੑ 1 2 = 1 20 u1 1 2 = 1 400 (u2 − 100) ⇔ u1 = 10 u2 = 300 (10, 300)