Green Electric Energy Lecture 14

8/14/2019 Green Electric Energy Lecture 14

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Lecture 14

Engineering Economics

Professor Tom OverbyeDepartment of Electrical and

Computer Engineering

ECE 333

Green Electric Energy


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Announcements

Be reading Chapter 5

Homework 6 is due Oct 22. It is 4.9, 5.2, 5.4, 5.6, 5.11

Wind Farm field trip will be on Nov 5 from about 8am to 4pm turn in forms to sign up.

Campus Wind Turbine Project: Let your voice be heard! If you would like to email your student senators

with your wind turbine views some folks to talk with are Bradley Tran ([email protected]), Bobby

Gregg ([email protected]), Melanie Cornell ([email protected])
mailto:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]:[email protected]


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The economic evaluation of a renewable energy resourcerequires a meaningful quantification of cost elements

fixed costs

variable costs We use engineering economics notions for this purpose

since they provide the means to compare on a consistent

basis

two different projects; or, the costs with and without a given project

Energy Economic Concepts(From Prof. Gross)


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loanP for 1 yearrepayP+ iP = P(1 + i) at the end of 1 year

year0 P

year 1 P(1 + i)

loanP forn years

year0 P

year 1 (1 + i) P repay/reborrowyear 2 (1 + i)2P repay/reborrow

year 3 (1 + i)3P repay/reborrow

.

yearn (1 + i)nP repay

Simple Example

M M M


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Compound Interest

e.o.p. amount owed interest for

next period

amount owed for next period

0 P Pi P + Pi = P(1+i)

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

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

3 P(1+i)3 P(1+i)3 i P(1+i)3 + P(1+i)3 i = P(1+i)4

n-1 P(1+i)n-1 P(1+i)n-1 i P(1+i)n-1 + P(1+i)n-1 i = P(1+i)n

n P(1+i)n

M M

The value in the last column for the e.o.p. (k-1) provides the value in

the first column for the e.o.p.k(e.o.p. is end of period)


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Terminology/Overview

Cash flow diagrams

Common conversion factors Present Value- (P|A,i%,n) and (P|F,i%,n) Future Value- (F|A,i%,n) and (F|P,i%,n)

Capital Recovery Factor- (A|P,i%,n)

Incoming cash flow

Outward cash flows

0 1 2 3 4

Present


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Cash Flows

A cash flow is a transfer of an amount At from one

entity to another at e.o.p. time t

A cash-flow set corresponds to the set

of times

The convention for cash flows is + inflow

outflow

Each cash flow has (1) amount, (2) time, and (3) sign

{ }1 2, , , ...,0 n A A A A{ },1,2,...,0 n

I take out a loan

I make equal repayments

0 1 2 3 4

Ex.


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Cash Flows

Given a cash-flow set we

define the future worthFnof the cash flow set at e.o.y.

n as

{ }1 2, , , ...,0 n A A A A

( )1n n tn tt 0

F A i

== +

0 1 2 t n 2 nn 2

A0 A1 A2 At An-2 An-1 An

. . . . . .


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Cash Flows

Note that each cash flow At in the set contributesdifferently toF

n

( )( )( )( )

1

1 1

2

2 2

1

1

1

1

n

0 0

n

n

n t

t t

n n

A A i

A A i

A A i

A A i

A A

+ + + +

M M

M M


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Discount Rate

The interest rate i is typically referred to as thediscount rate and is denoted by d

In converting a future amount F to a present worth

P we can view the discount rate as the interest ratethat can be earned from the best investment

alternative

A postulated savings of $10,000 in a project in 5

years is worth at present

( ) 555 10,000 1 P F d = = +


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Discount Rate

Ford = 0.1, P = $6,201,while ford = 0.2, P = $4,019

In general, the lower the discount factor, the

higher the present worth The present worth of a set of costs under a given

discount rate is called the life-cycle costs


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Cash Flow Example

0 1 3

$ 2,000

$ 3,000

F5=?

year

2 4 5

$ 2,000

Find the future worth of cash flows at the end of year 5

Answer: outflow of $2571

d = %12


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Cash Flows

Single-payment compound amount factor-

Single-payment present worth factor-

n is typically in years and need not be an integer value

( ) 11 i +@ ( )1 nn i = +( )1 ni


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Cash Flows

We define thepresent worthPof the cash flow set as

Note that

( )1n n ttt tt 0 t 0

P A A i

= == = +

( )

( ) ( ) ( )

1

1 1 1

1

nt

t

t 0

n

t n nt

t 0

P A i

A i i i

=

=

= +

= + + + 1 4 4 2 4 4 3


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Cash Flows

or equivalently

( ) ( )1 1n

n

nn n t

t

t 0

F

P i A i

== + +

142 431 4 42 4 43

( )1 nnF i P+This is the lump equivalent sum at the end of n periods.

Fis thefuture worth, andPis thepresent worth

nnF

contd


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Example 1

Consider a loan of$4,000 at 8% interest to be repaid intwo installments

$ 1,000 and interest at the e.o.y. (end of year) 1

$ 3,000 and interest at the e.o.y. 4

The cash flows are e.o.y. 1: 1000 + 4000 (.08) = $ 1,320

e.o.y. 4: 3000 (1 + .08 )

3

= $ 3,779.14 Note that the loan is made in year0 presentdollars,

but the repayments are in year 1 and

year 4future dollars


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Example 2

Given that

Then we say that for the cost of money of 12%,PandFare equivalent in the sense that $1,000 todayhas the same worth as $1,762.34 in 5 years

1, 000 .12 P $ i =and

( ) ( )5 51 1, 000 1 .12 1, 762.34 P i $ $ F = + = =


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Example 3

Consider an investment that returns

$1,000 at the e.o.y. 1

$2,000 at the e.o.y. 2

i = 10%

We evaluateP

rate at which

money can befreely lent or

borrowed

( ) ( )1 22

1, 000 1 .1 2, 000 1 .1

909.9 1, 652.09 2, 561.98

P $ $

$ $ $

= + + +

= + =142 43 142 43


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Net Present Value (NPV) for Example 3

Next, suppose that this investment requires $ 2,400now. At 10%, the investment has a net present value

(NPV) of

NPV = $ 2,561.98 $ 2,400 = $ 161.98

0 1 2

$ 2,561.98$ 1,000 $ 2,000

year

NPV=$ 161.98 {

$ 2,400

0 1 2


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Uniform Cash Flow Set

Consider the cash-flow set with

Such a set is called an equal payment cash flow set Each payment can be thought of as a future value and

the results will be the same

We compute the present worth

1,2,...,t A A t n={ }1 2, , , ...,0 n A A A A

2 1

1 1

1 ...n n

t t n

t

t t

P A A A = =

= = + + + +


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Uniform Cash Flow Set, cont.

Now, for , we have the identity

It follows that

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Green Electric Energy Lecture 14

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