V3,∞ (0,5) = 11 + 0, 666 ∙ 5 + 1, 333 ∙
6
28
3
= 11 + 1, 388 - 12,441 = -0, 053.
Finally
12,666
' (0, 5) =
9, 334
-0, 053
For β2 = 0,99 we obtain
' (0, 99) =
1
1 - 0, 99
5, 5
5, 5
5, 5
ι 1
+ 1 + 0, 99
2, 5 |
1 |
2, 5 |
+ --------- |
, |
1 - 0, 5 ∙ 0,99 |
28
3 -
6 J
5, 5 |
2, 5 |
0 | |||
= 100 |
5, 5 |
+ 0, 502 |
-2, 5 |
+ 1,980 |
0 |
5, 5 |
5 - 6 - |
28 -— L 3-1 |
and the result
νι,∞ (0, 99) = 100 ∙ 5,5 + 0, 502 ∙ 2,5 + 1, 980 ∙ 0 = 550 + 1, 255 + 0 = 551, 255,
ν2,∞ (0, 99) = 100 ∙ 5, 5 - 0, 502 ∙ 2, 5 + 0 = 548, 745,
V3 ∞ (0, 99) = 550 + 0, 502 ∙ 5 + 1, 980 ∙ 28 = 531,938.
63
Finally
551, 255
' (0,99) =
548, 745
531, 938
12
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