Example 11.2a Suppose that three investment projects with the fol- lowing return functions are available: 10.x 1+x fi(x) = x = 0, 1,..., f₂(x)=√x, x = 0, 1,.... f3(x) = 10(1-e), x = 0, 1,..., and that we want to maximize our return when we have 5 to invest. Now, 10x 1+x] yı(x) = x. V₁(x) = f(x) = Because V₂(x) = max [f2(y) + V₁(x − y)} = max Osys Osysx we see that 10(x - y) 1+x=y] V₂(1) = max(10/2, 1} = 5, y2(1) = 0, V₂ (2) = max(20/3, 1+5, √2}=20/3, y2 (2) = 0, V₂(3) = max(30/4, 1+20/3, √2+5, √3)=23/3, y2(3) = 1, V₂(4)= max(40/5, 1+30/4, √2+20/3, √3+5, √4) = 8.5, 32(4) = 1, V₂(5)= max(50/6, 1+8, √√2+7.5, √√3+20/3, √4+5, √5) = 9, 32(5) = 1. Continuing, we have that V3(x) = max [f3(y) + V₂(x−y)} = max (10(1-e¹³)+V₂(x−y)}-
Example 11.2a Suppose that three investment projects with the fol- lowing return functions are available: 10.x 1+x fi(x) = x = 0, 1,..., f₂(x)=√x, x = 0, 1,.... f3(x) = 10(1-e), x = 0, 1,..., and that we want to maximize our return when we have 5 to invest. Now, 10x 1+x] yı(x) = x. V₁(x) = f(x) = Because V₂(x) = max [f2(y) + V₁(x − y)} = max Osys Osysx we see that 10(x - y) 1+x=y] V₂(1) = max(10/2, 1} = 5, y2(1) = 0, V₂ (2) = max(20/3, 1+5, √2}=20/3, y2 (2) = 0, V₂(3) = max(30/4, 1+20/3, √2+5, √3)=23/3, y2(3) = 1, V₂(4)= max(40/5, 1+30/4, √2+20/3, √3+5, √4) = 8.5, 32(4) = 1, V₂(5)= max(50/6, 1+8, √√2+7.5, √√3+20/3, √4+5, √5) = 9, 32(5) = 1. Continuing, we have that V3(x) = max [f3(y) + V₂(x−y)} = max (10(1-e¹³)+V₂(x−y)}-
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
PLEASE LOOK AT THE BOOK EXAMPLE BELOW:
Explain why do you get y2(3) = 1 but y2(2) = 0.
Transcribed Image Text:Example 11.2a Suppose that three investment projects with the fol-
lowing return functions are available:
f₂(x)=√x, x=0, 1,...,
f(x) = 10(1e), x=0, 1,...,
and that we want to maximize our return when we have 5 to invest. Now,
10x
yi(x) = x.
Because
10.x
fi(x) = -, x=0, 1,...,
1+x
V₂(x) =
we see that
Vi(x)= fi(x) =
+x
max [f2(y) + V₁(x − y)} = max
Osysx
0y≤x
10(x - y)
1+x=y
V₂(1) = max{10/2, 1) = 5, y2₂(1) = 0,
V₂ (2) = max(20/3, 1+5, √2)=20/3, y2(2) = 0,
V₂(3) = max(30/4, 1+20/3, √√2+5,√√3)=23/3, y2(3) = 1₁
V₂(4)= max(40/5, 1+30/4, √2+20/3, √3+5, √4)
= 8.5, y2 (4) = 1,
V₂(5)= max(50/6, 1+8, √√2+7.5, √√3+20/3, √√4+5, √5)
=9, y₂(5) = 1.
Continuing, we have that
V3(x) = max {f3(y) + V₂(x −y)} = max (10(1-e³)+V₂(x−y)}-
0y≤x
Transcribed Image Text:Using that
1-e¹.632,
we obtain
1-e².865, 1-e³ = .950,
1-e982, 1-es = .993,
V3(5)= max(9, 6.32 +8.5, 8.65 +23/3,
9.50+20/3, 9.82 +5,9.93) = 16.32,
y3 (5) = 2.
Thus, the maximal sum of returns from investing 5 is 16.32; the optimal
amount to invest in project 3 is y3(5) = 2; the optimal amount to invest
in project 2 is y₂(3) = 1; and the optimal amount to invest in project 1
is yı(2) = 2.
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