math cad ror solution using a biquadratic bypass method

MathCAD - ROR Solution Using a Biquadratic Bypass Method.xmcd

Example 7A person invested $10k in a project. At the end of three years, he received $9k, and atthe end of six years he received $6k. What rate of return would the investor receive onthe project?

Reference"Engineering Fundamentals Examination Review", 2nd Ed. Donald G. Newnan & Bruce E. LarockISBN 0 471-01900-3.Page 436.

Nomenclature

i = Interest rate per interest periodn = Number of interest periodsP = A present sum of money (Present value, PV)F = A sum of money n periods from the present time that is equivalent to P with interest,i. (Future Value, FV).A = The end-of-period payment or disbursement in a "uniform series" continuing for n-periods, the entire seres being equivalent to P at interest rate, i.G = Uniform "period-by-period" increase in cash flows; the arithmetic gradient.

SolutionThe problem will be solved using a Block solver, first, then with an exact equation derivein the appendix.

PWcost PWbenefit= 1( )

P F1 PF i n1 F2 PF i n2 = 2( )

10 k 9 k PF i 3( ) 6 k PF i 6( )= 3( )

Where PF i n( )1

1 i( )n

=P

F= 4( )

Express equation 3 in terms of equation 4

n 3 6( )T

where ORIGIN 1

Julio C. Banks, P.E. page 1 of 6


Initial estimate of the interest rate: i 12 %

0.1 0.91

1 i( )n1

0.61

1 i( )n2

=

The block solver is as follows:

Given

10.9

1 i( )n1

0.6

1 i( )n2

= i 0

ROR Find i( )

ROR 10.41 %

The alternate (second) method is based on the derivation given in the appendix

The solution requires equation 3 to be represented in the by quadratic form. The choice of the exponents was initially recognized when in their numerical form first.

x2

2 B1 x B0 0=

Let n2 2 n1=

10.9

1 i( )n1

0.6

1 i( )2 n1

= 5( )

Multiply Eq. 5 by

1 i( )2 n1

1 i( )2 n1

1 i( )2 n1 0.9

1 i( )n1

1 i( )2 n1 0.6

1 i( )2 n1

=



1 i( )2 n1

0.9 1 i( )n1 0.6=

1 i( )2 n1

0.9 1 i( )n1 0.6 0=

Let x 1 i( )n1

=

and x2

1 i( )2 n1

=

2 B1 0.9= ⇒ B10.92

0.45

B0 0.6

ΛB0

B12

2.963

x B1 1 1 Λ 1.346

Since the root-solution sought is the variable w, then one must produce a post-processsolution

x 1 i( )n1

=

Solve for i

i3x 1 10.41 % QED



Appendix

Derive the solution of biquadratic equations

An equation of the form,

w2 p

B1 wp Bo 0= A 1( )

Is a by quadratic equation which can be solved using the standard quadratic equation

x2

2 B1 x Bo 0= A 2( )

Equation A-1 can be transformed into Eq. A-2 using the bypass variables,

x wp

= A 3( )

x2

w2 p

= A 4( )

Substitute equations A-3 and A-4 into A-1 to arrive at the equivalent quadratic equationA-2. The solution of Eq. A-2 produces two (2) roots which in most cases of physicalrelevance such root is positive, such as length, area, volume, etc. The two (2) rootssolution of Eq. A-2 is

A 5( )x B1 11 Λ

1 Λ

=

Where ΛB0

B12

= A 6( )

Since the root-solution sought is the variable w, then one must produce a post-processsolution

w x

1

p= A 7( )



Example 1 i( )2 n1

0.9 1 i( )n1 0.6 0= 1( )

Let x 1 i( )n1

= 2( )

3( )and x2

1 i( )2 n1

=

Substitute equations 2 and 3 into 1 to arrive at the equivalent quadratic equation A-2.

x2

2 B1 x Bo 0= A 2( )

The solution of Eq. A-2 produces two (2) roots which in most cases of physical relevancsuch root is positive, such as length, area, volume, etc. The two (2) roots solution of Eq.A-2 is

x B1 11 Λ

1 Λ

= A 5( )

Where ΛB0

B12

= A 6( )

The solution sought must be the positive root of equation A-2 which is

x B1 1 1 Λ 1.346

2 B1 0.9= ⇒ B10.92

0.45

B0 0.6



ΛB0

B12

2.963

x B1 1 1 Λ 1.346

Since the root-solution sought is the variable w, then one must produce a post-processsolution. w. from the bypass variable, x.

Bypass root-solution: x 1 i( )n1

=

Solve for i

i3x 1 10.41 %

Substantiation of the symbolic solutionUse a block solver to validate the biquadratic bypass method

Given

10.9

1 i( )n1

0.6

1 i( )n2

= i 0

ROR Find i( )

ROR 10.41 %

QED: The numerical results of the block solver exactly matches thesymbolic bypass solution method derived in this appendix.
