質問 13-2. p_1,p_2, Rの間にはどのような関係式が成り立つか。
  1. (2p1+3p2+6(1,1/2)(p1,p2)/10)(1+R)=8p1 (4p1+2p2+4(1,1/2)(p1,p2)/10)(1+R)=6p2
  2. (3p1+2p2+6(1,1/2)(p1,p2))(1+R)=8p1 (4p1+2p2+4(1,1/2)(p1,p2))(1+R)=6p2
  3. (2p1+3p2+6(1,1/2)(p1,p2)/10)(1+R)=8p1 (4p1+2p2+4(1,1/2)(p1,p2)/10)(1+R)=6p2
  4. [(2,3)(p1,p2) + 6(1, 1/2)(p1,p2)]*(1 + R) = p1 [(4,2)(p1,p2) + 4(1, 1/2)(p1,p2)]*(1 + R) = p2
  5. (2p1+3p2+6(1,1/2)/10(1+R)=8p1 (4p1+2p2+4(1,1/2)/10(1+R)=6p2
  6. (3p1+2p2+6(p1+p2/2)/10)(1+R)=8p1(4p1+2p2+4(p1+p2/2)/10)(1+R)=6p2
  7. 2p1+3p2+6(1,1/2)(p1,p2)/10*(1+R)=8p14p1+2p2+4(1,1/2)(p1,p2)/10*(1+R)=6P2
  8. 8p1-(2p1+3P2+6w)/2p1+3p2+6w=R6p2-(4p1+2p2+4w)/4p1+2p2+4w=R
  9. (2p1+3p2+6(1,1/2)(p1,p2)/10)(1+R)=8p1 (4p1+2p2+4(1,1/2)(p1,p2)/10)(1+R)=6p2
  10. (R+1)(26p1+33p2)/80=p1(R+1)(22p1+11p2)/30=p2
  11. (2p1+3p2+6(1p1+(1/2)p2/10)(1+R)=8p1 (4p1+2p2+4(1p1+(1/2)p2/10)(1+R)=6p2
  12. (2p1+3p2+6(1p1+(1/2)p2/10)(1+R)=8p1(4p1+2p2+4(1p1+(1/2)p2/10)(1+R)=6p2
  13. 3p1+2p2+6w=8p1/(1+R) 4p1+2p2+4w=6p1/(1+R)
  14. (2p1+3p2+6(p1+1/2×p2)/10)(1+R)=8p1 4p1+2p2+4(p1+1/2×p2)/10)(1+R)=6p2
  15. (2p1+3p2)*(1+R1)=8p1-(2p1+3p2+6w) (4p1+2p2)*(1+R2)=6p2-(4p1+2p2+4w)
  16. (7p1+(11/2)p2)*(1+R)=8p1+6p2
  17. （1,1/2）・（p1,p2）×（1＋R）＝300＋400