Engineering Economy Chapter 2x

CHAPTER 2

Factors: How Time and Interest Affect Money

ENGINEERING ECONOMY, Sixth Edition

by Blank and Tarquin

GrawHill

Authored by Don Smith, Texas A&M University 2004

1. Foundations: Overview1. F/P and P/F Factors

2. P/A and A/P Factors

3. F/A and A/F Factors

4. Interpolate Factor Values

5. P/G and A/G Factors

6. Geometric Gradient

7. Calculate i

8. Calculate “n”

9. Spreadsheets

CHAPTER 2: Section 1

F/P and P/F Factors

2.1 Basic Derivations: F/P factor

F/P Factor To find F given P

………….

To Find F given P

Compound forward in time

2.1 Derivation by Recursion: F/P factor

F1 = P(1+i)

F2 = F1(1+i)…..but:

F2 = P(1+i)(1+i) = P(1+i)2

F3 =F2(1+i) =P(1+i)2 (1+i) = P(1+i)3

In general:

Fn = P(1+i)n

Fn = P(F/P,i%,n)

2.1 Present Worth Factor from F/P

Since Fn = P(1+i)n

We solve for P in terms of FN

P = F{ 1/ (1+i)n} = F(1+i)-n

Thus:P = F(P/F,i%,n) where

(P/F,i%,n) = (1+i)-n

Thus, the two factors are:

1. F = P(1+i)n finds the future worth of P;

2. P = F(1+i)-n finds the present worth from F

2.1 P/F factor – discounting back in time

Discounting back from the future

………….

P/F factor brings a single future sum back to a specific point in time.

P/A and A/P Factors

2.2 Example- F/P Analysis

Example: P= $1,000;n=3;i=10%What is the future value, F?

0 1 2 3P=$1,000

F = ??

i=10%/year

F3 = $1,000[F/P,10%,3] = $1,000[1.10]3

= $1,000[1.3310] = $1,331.00

2.2 Example – P/F Analysis

Assume F = $100,000, 9 years from now. What is the present worth of this amount now if i =15%?

0 1 2 3 8 9

…………

$100,000

i = 15%/yr

P0 = $100,000(P/F, 15%,9) = $100,000(1/(1.15)9)

= $100,000(0.2843) = $28,430 at time t = 0

2.2 Uniform Series Present Worth and Capital Recovery Factors

Annuity Cash Flow

$A per period

P = ??

………….. n 1 2 3 .. ..

Desire an expression for the present worth – P of a stream of equal, end of period cash flows - A

0 1 2 3 n-1 n

A = given

P = ??

Write a Present worth expression

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

Term inside the brackets is a geometric progression.

Mult. This equation by 1/(1+i) to yield a second equation

The second equation

To isolate an expression for P in terms of A, subtract Eq [1] from Eq. [2]. Note that numerous terms will drop out.

1 1 1 1..

1 (1 ) (1 ) (1 ) (1 )n n

i i i i i

Setting up the subtraction

1 1 1 1..

(1 ) (1 ) (1 ) (1 )n nP A

i i i i

1 (1 ) (1 )n

iP Ai i i

2 3 4 1

1 1 1 1 1...

(1 ) (1 ) (1 ) (1 ) (1 ) (1 )n n

i i i i i i

Simplifying Eq. [3] further

1 (1 ) (1 )n

iP Ai i i

(1 )nA

(1 ) 1 0

iP A for i

This expression will convert an annuity cash flow to an equivalent present worth amount one period to the left of the first annuity cash flow.

(1 ) 1 0

iP A for i

/ %, P A i n factor

2.2 Capital Recovery Factor A/P, i%, n

Given the P/A factor

(1 ) 1 0

iP A for i

(1 ) 1

i iA P

Solve for A in terms of P

Yielding….

A/P,i%,n factor

The present worth point of an annuity cash flow is always one period to the left of the first A amount

F/A and A/F Factors

2.3 F/A and A/F Derivations

Annuity Cash Flow

………….. N

$A per period

Find $A given the Future amt. - $F

Take advantage of what we already haveRecall:

(1 )nP F

(1 ) 1

i iA P

Substitute “P” and simplify!

2.3 A/F Factor

By substitution we see:

Simplifying we have:Which is the (A/F,i%,n) factor

1 (1 )

(1 ) (1 ) 1

i iA F

(1 ) 1nA

2.3 F/A factor from the A/F Factor

Given:

Solve for F in terms of A

(1 ) 1n

2.3 F/A and A/F Derivations

Annuity Cash Flow

………….. N

$A per period

Find $F given the $A amounts

2.3 Example 2.5

Formosa Plastics has major fabrication plants in Texas and Hong Kong. It is desired to know the future worth of $1,000,000 invested at the end of each year for 8 years, starting one year from now. The interest rate is assumed to be 14% per year.

2.3 Example 2.5

•A = $1,000,000/yr; n = 8 yrs, i = 14%/yr

•F8 = ??

2.3 Example 2.5

Solution:The cash flow diagram shows the annual payments starting at the end of year 1 and ending in the year the future worth is desired. Cash flows are indicated in $1000 units. The F value in 8 years is

F = l000(F/A,14%,8) = 1000( 13.23218) = $13,232.80 = 13.232 million 8 years from now.

2.3 Example 2.6

How much money must Carol deposit every year starting, l year from now at 5.5% per year in order to accumulate $6000 seven years from now?

2.3 Example 2.6

Solution

The cash How diagram from Carol's perspective fits the A/F factor.

A= $6000 (A/F,5.5%,7) = 6000(0.12096) = $725.76 per yearThe A/F factor Value 0f 0.12096 was computed using the A/F factor formula

Interpolation in Interest Tables

2.4 Interpolation of Factors

• All texts on Engineering economy will provide tabulated values of the various interest factors usually at the end of the text in an appendix

• Refer to the back of your text for those tables.

2.4 Interpolation of Factors

• Typical Format for Tabulated Interest Tables

2.4 Interpolation (Estimation Process)

• At times, a set of interest tables may not have the exact interest factor needed for an analysis

• One may be forced to interpolate between two tabulated values

• Linear Interpolation is not exact because:

• The functional relationships of the interest factors are non-linear functions

• Hence from 2-5% error may be present with interpolation.

2.4 An Example

• Assume you need the value of the A/P factor for i = 7.3% and n = 10 years.

• 7.3% is most likely not a tabulated value in most interest tables

• So, one must work with i = 7% and i = 8% for n fixed at 10

• Proceed as follows:

2.4 Basic Setup for Interpolation

•Work with the following basic relationships

2. 4 i = 7.3% using the A/P factor

• For 7% we would observe:

COMPOUND PRESENT SINKING COMPOUND CAPITAL

N AMT. FACTOR WORTH FUND AMOUNT RECOVERY

F/P P/F A/F F/A A/P

10 1.9672 0.5083 0.0724 13.8164 0.14238

A/P,7%,10) = 0.14238

2. 4 i = 7.3% using the A/P factor

• For i = 8% we observe:

COMPOUND PRESENT SINKING COMPOUND CAPITAL

N AMT. FACTOR WORTH FUND AMOUNT RECOVERY

F/P P/F A/F F/A A/P

10 2.1589 0.4632 0.0690 14.4866 0.14903

(A/P,8%,10) = 0.14903

2. 4 Estimating for i = 7.3%

• Form the following relationships

2.4 Final Estimated Factor Value

• Observe for i increasing from 7% to 8% the A/P factors also increases.

• One then adds the estimated increment to the 7% known value to yield:

2.4. The Exact Value for 7.3%

• Using a previously programmed spreadsheet model the exact value for 7.3% is:

CHAPTER 2 Section 5

P/G and A/G Factors

2.5 Arithmetic Gradient Factors

• In applications, the annuity cash flow pattern is not the only type of pattern encountered

•Two other types of end of period patterns are common

•The Linear or arithmetic gradient

•The geometric (% per period) gradient

•This section presents the Arithmetic Gradient

• An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a constant amount over n time periods.

•A linear gradient is always comprised of TWO components:

•The Two Components are:

•The Gradient component

•The base annuity component

•The objective is to find a closed form expression for the Present Worth of an arithmetic gradient

2.5 Linear Gradient Example

Assume the following:

0 1 2 3 n-1 N

A1+(n-2)G

A1+(n-1)G

This represents a positive, increasing arithmetic gradient

2.5 Example: Linear Gradient

• Typical Negative, Increasing Gradient: G=$50

The Base Annuity= $1500

2.5 Example: Linear Gradient

• Desire to find the Present Worth of this cash flow

The Base Annuity= $1500

• The “G” amount is the constant arithmetic change from one time period to the next.

•The “G” amount may be positive or negative!

•The present worth point is always one time period to the left of the first cash flow in the series or,

•Two periods to the left of the first gradient cash flow!

Focus Only on the gradient Component

(n-2)G

(n-1)G

0 1 2 3 n-1 N

“0” G

Removed Base annuity

2.5 Present Worth Point…

The Present worth point of a linear gradient is always: 2 periods to the left of the “1G” point or,

1 period to the left of the very first cash flow in the gradient series.

DO NOT FORGET THIS!

0 1 2 3 4 5 6 7

$100$200

The Present Worth Point of the Gradient

2.5 Gradient Component

$100$200

0 1 2 3 4 5 6 7

•The Gradient Component

0 1 2 3 4 5 6 7

Base Annuity – A = $100

•PW of the Base Annuity is at t = 0

•PWBASE Annuity=$100(P/A,i%,7)

2.5 Present Worth: Linear Gradient

The present worth of a linear gradient is the present worth of the two components: 1. The Present Worth of the Gradient

Component and, 2. The Present Worth of the Base

Annuity flow Requires 2 separate calculations!

2.5 Present Worth: Gradient Component

The PW of the Base Annuity is simply the Base Annuity –A{P/A, i%, n} factorWhat is needed is a present worth expression for the gradient component cash flow.We need to derive a closed form expression for the gradient component.

2.5 Present Worth: Gradient Component

General CF Diagram – Gradient Part Only

0 1 2 3 4 ……….. n-1 n

3G(n-2)G

(n-1)G

We want the PW at time t = 0 (2 periods to the left of 1G)

2.5 To Begin- Derivation of P/G,i%,n

Next Step:

Factor out G and re-write as …..

( / , %,2) 2 ( / , %,3) ...

[(n-2)G](P/F,i%,n-1) + [(n-1)G](P/F,i%,n)

P G P F i G P F i

2.5 Factoring G out…. P/G factor

P/F,i%,2 P/F,i%,P=G ( )+2( ) +...(3 P/F,in-1)( %,n)

What is inside of the { }’s?

2.5 Replace (P/F’s) with closed-form

2 3 n-1 n

1 2 n-2 n-1P=G ...

(1+i) (1+i) (1+i) (1+i)

Multiply both sides by (1+i)

2.5 Mult. Both Sides By (n+1)…..

11 2 n-2 n-1

1 2 n-2 n-1P(1+i) =G ...

(1+i) (1+i) (1+i) (1+i)

•We have 2 equations [1] and [2].

•Next, subtract [1] from [2] and work with the resultant equation.

2 3 n-1 n

1 2 n-2 n-1P=G ...

(1+i) (1+i) (1+i) (1+i)

2.5 Subtracting [1] from [2]…..

11 2 n-2 n-1

1 2 n-2 n-1P(1+i) =G ...

(1+i) (1+i) (1+i) (1+i)

(1 ) 1

(1 ) (1 )

2.5 The P/G factor for i and N

( / , %, )P G i N factor

(1 ) 1

(1 ) (1 )

2.5 Further Simplification on P/G

(1 ) 1( / , %, )

i iNP G i N

Remember, the present worth point of any linear gradient is 2 periods to the left of the 1-G cash flow or, 1 period to the left of the “0-G” cash flow.

P=G(P/G,i,n)

2.5 Extension – The A/G factor

Some authors also include the derivation of the A/G factor.A/G converts a linear gradient to an equivalent annuity cash flow.Remember, at this point one is only working with gradient componentThere still remains the annuity component that you must also handle separately!

2.5 The A/G Factor

Convert G to an equivalent A

( / , , )( / , , )A G P G i n A P i n

How to do it…………

2.5 A/G factor using A/P with P/G

The results follow…..

(1 ) 1

(1 ) (1 )

(A/P,i,n)

(1 ) 1

2.5 Resultant A/G factor

(A/G,i,n) = 1

(1 ) 1n

(1 ) 1

(1 ) (1 )

(A/P,i,n)

(1 ) 1

2.5 Gradient Example

• Consider the following cash flow

0 1 2 3 4 5

$100$200

$300$400

Find the present worth if i = 10%/yr; n = 5 yrs

Present Worth Point is here!And the G amt. = $100/period

2.5 Gradient Example- Base Annuity

• First, The Base Annuity of $100/period

0 1 2 3 4 5

A = +$100

•PW(10%) of the base annuity = $100(P/A,10%,5)

•PWBase = $100(3.7908)= $379.08

•Not Finished: We need the PW of the gradient component and then add that value to the $379.08 amount

2.5 Focus on the Gradient Component

0 1 2 3 4 5

$0$100

$200$300

We desire the PW of the Gradient Component at t = 0

PG@t=0 = G( P/G,10%,5 ) = $100( P/G,10%,5 )

2.5 The Set Up

0 1 2 3 4 5

$0$100

$200$300

PG@t=0 = G(P/G,10%,5) = $100(P/G,10%,5)

G (1 ) 1P=

i (1 ) (1 )

Could substitute n=5, i=10% and G = $100 into the P/G closed form to get the value of the factor.

2.5 PW of the Gradient Component

PG@t=0 = G(P/G,10%,5) = $100(P/G,10%,5)

Calculating or looking up the P/G,10%,5 factor yields the following:Pt=0 = $100(6.8618) = $686.18 for

the gradient PW

P/G,10%,5)

G (1 ) 1P=

i (1 ) (1 )

Sub. G=$100;i=0.10;n=5

6.8618

2.5 Gradient Example: Final Result

• PW(10%)Base Annuity = $379.08

•PW(10%)Gradient Component= $686.18

•Total PW(10%) = $379.08 + $686.18

•Equals $1065.26

•Note: The two sums occur at t =0 and can be added together – concept of equivalence

2.5 Example Summarized

0 1 2 3 4 5

$100$200

$300$400

$500This Cash Flow…

Is equivalent to $1065.26 at time 0 if the interest rate is 10% per year!

2.5 Shifted Gradient Example: i = 10%

• Consider the following Cash Flow

0 1 2 3 4 5 6 7

1. This is a “shifted” negative, decreasing gradient.

2. The PW point in time is at t = 3 (not t = o)

$600$550

$500$450

2.5 Shifted Gradient Example

0 1 2 3 4 5 6 7

$600$550

$500$450

•The PW @ t = 0 requires getting the PW @ t =3;

•Then using the P/F factor move PW3 back to t=0

2.5 Shifted Gradient Example

•The base annuity is a $600 cash flow for 3 time periods

0 1 2 3 4 5 6 7

$600$550

$500$450

2.5 Shifted Gradient Example: Base Annuity

• PW of the Base Annuity: 2 Steps

A = -$600

0 1 2 3 4 5 6 7

P3=-600(P/A,10%,4)

P0=P3(P/F,10%,3)

P0= [-600(P/A,10%,4)](P/F,10%,3)3.1699 0.7513

P 0-base annuity = -$1428.93

2.5 Shifted Gradient Example: Gradient

• PW of Gradient Component: G = -$50

0 1 2 3 4 5 6 7

P0=P3(P/F,10%,3)

0G1G 2G 3G

P3-Grad = +50(P/G,10%,4)

P0-grad = {+50(P/G,10%,4)}(P/F,10%,3)4.3781 0.7513

=-$164.46

Geometric Gradient

2.6 Geometric Gradients

• An arithmetic (linear) gradient changes by a fixed dollar amount each time period.

•A GEOMETRIC gradient changes by a fixed percentage each time period.

•We define a UNIFORM RATE OF CHANGE (%) for each time period

•Define “g” as the constant rate of change in decimal form by which amounts increase or decrease from one period to the next

2.6 Geometric Gradients: Increasing

• Typical Geometric Gradient Profile

•Let A1 = the first cash flow in the series

0 1 2 3 4 …….. n-1 n

A1 A1(1+g)A1(1+g)2

A1(1+g)3

A1(1+g)n-1

2.6 Geometric Gradients: Decreasing

• Typical Geometric Gradient Profile

•Let A1 = the first cash flow in the series

A1(1-g)

A1(1-g)2

A1(1-g)3

A1(1-g)n-1

0 1 2 3 4 …….. n-1 n

2.6 Geometric Gradients: Derivation

• First Major Point to Remember:

•A1 does NOT define a Base Annuity;

•There is no BASE ANNUITY for a Geometric Gradient!

•The objective is to determine the Present Worth one period to the left of the A1 cash flow point in time

•Remember: The PW point in time is one period to the left of the first cash flow – A1!

2.6 Geometric Gradients: Derivation

• For a Geometric Gradient the following parameters are required:

•The interest rate per period – i

•The constant rate of change – g

•No. of time periods – n

•The starting cash flow – A1

2.6 Geometric Gradients: Starting

• Pg = The Aj’s time the respective (P/F,i,j) factor

•Write a general present worth relationship to find Pg….

2 11 1 1 1

(1 ) (1 ) (1 )...

(1 ) (1 ) (1 ) (1 )

A A g A g A gP

i i i i

Now, factor out the A1 value and rewrite as..

1 (1 ) (1 ) (1 )...

(1 ) (1 ) (1 ) (1 )

g g gP A

i i i i

(1+g)Multuply both sides by to create another equation

(1+i)1 2 1

(1+g) (1+g) 1 (1 ) (1 ) (1 )...

(1+i) (1+i) (1 ) (1 ) (1 ) (1 )

g g gP A

i i i i

Subtract (1) from (2) and the result is…..

1+g (1 ) 11

1+i (1 ) 1

Solve for Pg and simplify to yield….

P Ai g

2.6 Geometric Gradient P/A factor

• This is the (P/A,g,i,n) factor and is valid if g is not equal to i.

P Ai g

2.6 Geometric Gradient P/A factor

•Note: If g = i we have a division by “0” – undefined.

•For g = i we can derive the closed form PW factor for this special case.

•We substitute i for g into the Pg

relationship to yield:

2.6 Geometric Gradient: i = g Case

1 1 1 1P =A ...

(1+i) (1+i) (1+i) (1+i)

For the case i = g

2.6 Geometric Gradients: Summary

•Pg = A1(P/A,g,i,n)

g not = to i

Case: g = i

2.6 Geometric Gradient: Notes

•The geometric gradient requires knowledge of:

•A1, i, n, and g

•There exist an infinite number of combinations for i, n, and g: Hence one will not find tabulated tables for the (P/A, g,i,n) factor.

2.6 Geometric Gradient: Notes

•You have to calculated either from the closed form for each problem or apply a pre-programmed spreadsheet model to find the needed factor value

•No spreadsheet built-in function for this factor!

2.6 Geometric Gradient: Example

•Assume maintenance costs for a particular activity will be $1700 one year from now.

•Assume an annual increase of 11% per year over a 6-year time period.

2.6 Geometric Gradient: Example

•If the interest rate is 8% per year, determine the present worth of the future expenses at time t = 0.

•First, draw a cash flow diagram to represent the model.

2.6 Geometric Gradient Example (+g)

•g = +11% per period; A1 = $1700; i = 8%/yr

0 1 2 3 4 5 6 7

$1700 $1700(1.11)1

$1700(1.11)2

$1700(1.11)3

$1700(1.11)5

PW(8%) = ??

2.6 Solution

303: Use "g" 667: use f-bar

Geometric Gradients"E" or g or f-bar = 11%

i= 8%N= 7

P/A,g,i,n factor is…… 7.04732

First Amt= 1,700.00$ P. Value = 11,980.44$

• P = $1700(P/A,11%,8%,7)•Need to calculate the P/A factor from the closed-form expression for a geometric gradient.•From a spreadsheet we see:

2.6 Geometric Gradient ( -g )

• Consider the following problem with a negative growth rate – g.

0 1 2 3 4

g = -10%/yr; i = 8%; n = 4

A1 = $1000 $900

$810$729

We simply apply a “g” value = -0.10

2.6 Geometric Gradient (-g value)

• Evaluate:

For a negative g value = -0.10

303: Use "g" 667: use f-bar

Geometric Gradients"E" or g or f-bar = -10%

i= 8%N= 4

P/A,g,i,n factor is…… 2.87637

First Amt= 1,000.00$ P. Value = 2,876.37$

Determination of an Unknown Interest Rate

2.7 When the i – rate is unknown

• A class of problems may deal with all of the parameters know except the interest rate.

•For many application-type problems, this can become a difficult task

•Termed, “rate of return analysis”

•In some cases:

•i can easily be determined

•In others, trial and error must be used

2.7 Example: i unknown

• Assume on can invest $3000 now in a venture in anticipation of gaining $5,000 in five (5) years.

•If these amounts are accurate, what interest rate equates these two cash flows?

• The Cash Flow Diagram is…

0 1 2 3 4 5

$3,000

$5,000

•F = P(1+i)n

•5,000 = 3,000(1+i)5

•(1+i)5 = 5,000/3000 = 1.6667

• Solution:

0 1 2 3 4 5

$3,000

$5,000

•(1+i)5 = 5,000/3000 = 1.6667

•(1+i) = 1.66670.20

•i = 1.1076 – 1 = 0.1076 = 10.76%

2.7 For “i” unknown

• In general, solving for “i” in a time value formulation is not straight forward.

•More often, one will have to resort to some form of trial and error approach as will be shown in future sections.

•A sample spreadsheet model for this problem follows.

2.7 Example of the IRR function

=IRR($D7:$D12)

Determination of Unknown Number of Years

2.8 Unknown Number of Years

• Some problems require knowing the number of time periods required given the other parameters

•Example:

•How long will it take for $1,000 to double in value if the discount rate is 5% per year?

•Draw the cash flow diagram as….

0 1 2 . . . . . . ……. n

P = $1,000

Fn = $2000

i = 5%/year; n is unknown!

• Solving we have…..

0 1 2 . . . . . . ……. n

P = $1,000

Fn = $2000

•Fn=? = 1000(F/P,5%,x): 2000 =

1000(1.05)x

•Solve for “x” in closed form……

• Solving we have…..

•(1.05)x = 2000/1000

•Xln(1.05) =ln(2.000)

•X = ln(1.05)/ln(2.000)

•X = 0.6931/0.0488 = 14.2057 yrs

•With discrete compounding it will take 15 years to amass $2,000 (have a little more that $2,000)

2.8 No. of Years – NPER function

• From Excel one can formulate as:

=NPER(C23,C22,C20,C21)

Spreadsheet Application – Basic Sensitivity Analysis

2.9 Basic Sensitivity Analysis

• Sensitivity analysis is a procedure applied to a formulated problem whereby one can assess the impact of each input parameter relating to the output variable.

•Sensitivity analysis is best performed using a spreadsheet model.

•The procedure is to vary the input parameters within certain ranges and observe the change on the output variable.

• By proper modeling, one can perform “what-if” analysis on one or more of the input parameters and observe any changes in a targeted output (response) variable

•Commercial add-in packages are available that can be linked to Excel to perform such an analysis

•Specifically: Palisade Corporation’s TopRank Excel add-in is most appropriate.

• When you build your own models, devise an approach to permit varying at least one of the input parameters and store the results of each change in the output variable…then plot the results.

•If a small change in one of the input parameters represents a significant change in the output variable then…

•That input variable is “sensitive”

• If an input parameter is deemed “sensitive” then some effort should go into the estimation of that parameter

•Because it does influence the response (output) variable.

•Less sensitive input parameters may not have as much effort required to estimate as those input parameters do not have that much impact on the targeted response variable.

• When you build your own models, devise an approach to permit varying at least one of the input parameters and store the results of each change in the output variable…then plot the results.

•If a small change in one of the input parameters represents a significant change in the output variable then…

•That input variable is “sensitive”

CHAPTER 2: Summary of Important Points

GrawHill

Chapter Summary

• This chapter presents the fundamental time value of money relationships common to most engineering economic analysis calculations

•Derivations have been presented for:

•Present and Future Worth- P/F and F/P

•Annuity Cash flows – P/A, A/P, F/A and A/F

•Gradients – P/G, A,G and P/A,g,i,n

Chapter Summary

• One must master these basic time value of money relationships in order to proceed with more meaningful analysis that can impact decision making.

•These relationships are important to you professionally and in your personal lives.

•Master these concepts!!!

McGrawHill

ENGINEERING ECONOMY, Sixth Edition

Blank and Tarquin

End of Slide Set

Engineering Economy Chapter 2x

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Engineering Economy & Finanace

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