Green Electric Energy: Engineering Economics

ECE 333 Green Electric Energy Lecture 14 Engineering Economics

Professor Tom Overbye Department of Electrical and Computer Engineering

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])

Energy Economic Concepts (From Prof. Gross) •

The economic evaluation of a renewable energy resource requires 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

Time Value of Money • • • •

Basic notion: a dollar today is not the same as a dollar in a year We represent the time value of money by the standard approach of discounted cash flows The notation is P = principal i = interest value The convention we use is that payments occur at the end of each period (e.o.p.)

Simple Example loan P for 1 year repay P + iP = P ( 1 + i ) at the end of 1 year year 0 P year 1 P (1 + i) loan P for n years year 0 year 1 year 2 year M 3 . year n

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

repay/reborrow repay/reborrow repay/reborrow M

( 1 + i )n P

repay

Compound Interest e.o.p.

amount owed

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

P(1+i ) n-1 i

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

M

interest for next period

amount owed for next period

M

n-1

P(1+i ) n-1

n

P(1+i ) n

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)

Terminology/Overview •

Cash flow diagrams Incoming cash flow 0

1

2

3

4

Outward cash flows Present

•

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)

Cash Flows • • •

A cash flow is a transfer of an amount A t from one entity to another at e.o.p. time t A cash-flow set { A0 , A1 , A2 ,..., An} corresponds to the set { 0 ,1, 2,..., n} of times The convention for cash flows is Ex. I take out a loan + inflow − outflow 0 1 2 3 4 I make equal repayments

•

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

Cash Flows •

{ A , A , A ,..., A }

Given a cash-flow set

0

1

2

we

n

define the future worth F n of the cash flow set at e.o.y. n as

A0

n

∑ A ( 1 + i)

Fn = A1

t =0

1

2

t

At

A2

... 0

n− t

A n-2

A n-1

An

n–2

n–2

n

... t

Cash Flows •

Note that each cash flow A t in the set contributes differently to F n

A0 A1

→ →

A2 → M At → M An →

A0 ( 1 + i )

A1 ( 1 + i )

n n−1

A2 ( 1 + i ) M n− t At ( 1 + i ) M An

n− 2

Discount Rate • •

•

The interest rate i is typically referred to as the discount 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 rate that can be earned from the best investment alternative A postulated savings of $ 10,000 in a project in 5 years is worth at present

P = F5 β

5

= 10,000 ( 1 + d )

−5

Discount Rate • • •

For d = 0.1, P = $ 6,201, while for d = 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

Cash Flow Example • Find the future worth of cash flows at the end of year 5 $ 3,000

d = %12 0

$ 2,000

1

2

3

4

F5=? 5 year

$ 2,000

Answer: outflow of $2571

Cash Flows •

Single-payment compound amount factor-

•

Single-payment present worth factor-

( 1+ i)

β @( 1 + i )

•

−1

n

β

n

= ( 1+ i)

−n

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

Cash Flows •

We define the present worth P of the cash flow set as

P =

•

n

∑A β t =0

Note that

P =

n

=

∑ A ( 1 + i) t =0

=

t

t

n

∑ A ( 1 + i) t =0

−t

t

−t

t

n

∑ A ( 1 + i ) (11 4+ i4) 2( 14+4i) 3 t =0

−t

−n

n

t

1

Cash Flows cont’d

n

P = ( 1 + i ) ∑ At ( 1 + i ) 142 4 3 t =0 1 4 42 4 43 n β F −n

n −t

n

= β n Fn or equivalently

Fn = ( 1 + i ) P n

This is the lump equivalent sum at the end of n periods. F is the future worth, and P is the present worth

Example 1 •

• •

Consider a loan of $4,000 at 8% interest to be repaid in two 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 year 0 present dollars, but the repayments are in year 1 and year 4 future dollars

Example 2 •

Given that

P = $1,000

•

and

i = .12

Then we say that for the cost of money of 12%, P and F are equivalent in the sense that $1,000 today has the same worth as $1,762.34 in 5 years

P ( 1 + i)

5

= $1,000( 1 + .12)

5

= $1,762.34 = F

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 evaluate P

rate at which money can be freely lent or borrowed

P = $ 1,000 ( 1 + .1) + $ 2,000( 1 + .1) 142 43 142 43 2 β β = $ 909.9 + $ 1,652.09 = $ 2,561.98 −1

−2

Net Present Value (NPV) for Example 3 •

Next, suppose that this investment requires $ 2,400 now. At 10%, the investment has a net present value (NPV ) of NPV = $ 2,561.98 – $ 2,400 = $ 161.98 $ 1,000

$ 2,561.98 0

1

$ 2,400 NPV=$ 161.98 { 0

$ 2,000 2 year

1

2

Uniform Cash Flow Set • • • •

Consider the cash-flow set { A0 , A1 , A2 ,..., An}

At = A

with

t = 1, 2,..., n

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 P =

n

∑ At β t =1

t

n

= A ∑ β t = Aβ 1 + β + β 2 + ... + β n− 1  t =1

Uniform Cash Flow Set, cont. •

Now, for 0 < β < 1 , we have the identity ∞

∑β •

It follows that

j

j =0

1 = 1−β

∞

∑β

j

j=0

1 + β + ... + β n−1 =

∞

j n 2 n −1  β − β 1 + β + β + ... + β +... ∑  j =0

∞

= ( 1−β n ) ∑ β 1− β = 1− β

n

j =0

j

Uniform Cash Flow Set, cont. • •

Therefore

But

 1 − β n P = A β   1− β 

β = (1 + d)

−1

and so

1 d 1−β = 1− = = βd 1+ d 1+d n 1 − β • We write P = A d 1− βn the equal payment series present • We call d value function

Uniform Cash Flows, PVF

1−β d

n

1 − (1 + d ) = d

−n

(1 + d ) − 1 = d (1 + d )n n

this is also written as (P|A,d%,n)

In the News: States Not Meeting Renewable Energy Goals • USA Today had an article last week that discussed progress states •

were making on meeting their renewable portfolio standards; 35 states have such goals There is no central clearing house on compliance, so the article just discussed several examples – – –

NJ would like to have 1000MW of offshore wind by 2012, but will miss that by about a year CA has a 20% goal by 2010, but won’t achieve that until 2013 or 2014 AZ planned to get 0.3% from solar by 2009, but probably won’t achieve that until 2011

www.usatoday.com/money/industries/energy/2009-10-08-altenergy_N.htm

California and Solar • In California Gov Schwarzenegger signed two solar energy bills this week that will help make solar energy more cost effective for consumers (with solar). –

–

AB 920 requires utility companies to pay customers for any surplus electricity they produce from solar or wind (previously they had to give it back any surplus for free) AB 32 creates an above-market tariff, called a “feed-in-tariff” that requires the state’s utilities to buy solar electricity from 1.5 to 3.0 MW units at rates above what they would pay for conventional sources

Equivalence •

We consider two cash-flow sets

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

and

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

under a given discount rate d

•

We say

{A }

are equivalent cash-flow sets if

and { A } their future worths (or their net worths at any point in a t

time) are identical.

b t

Equivalence, Example •

Consider the two cash-flow sets under d = 7%

8,200.40

2000 2000 2000 2000 2000

0

1

2

3

4 a

5

6

7

0

1 b

2

Equivalence, cont. •

7

We compute

P a = 2000 ⋅ ∑ β

t

= 7162.33

t =3

and

•

P b = 8200.40 ⋅ β 2 = 7162.33

Therefore, { A a } and { A b} are equivalent cash flow sets t t under d = 7%

Example •

Consider the set of cash flows illustrated below $ 400 $ 300

$ 200

$ 200 3 0

1

4

2 $ 300

5

6

7

8

d = 6%

Example, cont. •

We compute F 8 at t = 8 for d = 6% F8 = 300 ( 1 + .06)

7

− 300( 1 + .06)

200 ( 1 + .06) + 400( 1 + .06) = $ 951.56 4

•

We next compute P

P = 300 ( 1 + .06)

−1

+ 200 ( 1 + .06 ) = $ 597.04

•

2

− 300( 1 + .06)

−4

5

+ + 200

−3

+ 400( 1 + .06)

−6

+ 200( 1 + .06)

We check that for d = 6% 8 F8 = 597.04 ( 1 + .06) = $ 951.56

−8

Motor Purchase Example • •

•

We consider the purchase of two 100-hp motors – a and b – to be used over a 20-year period; the discount rate is 10% The relative merits of a and b are motor

costs ( $ )

load ( kW )

a

2,400

79.0

b

2,900

77.5

The motor is used 1,600 hours per year and electricity costs are constant at 0.08 $/kWh

Motor Purchase Example, motor a •

We evaluate yearly energy costs over 20 years for motor a A a = ( 79.0 kW ) ( 1600 h / yr) ( .08 $ / kWh) = $ 10,112/ yr 0

1

Aa

2

Aa

...

t

19

20

Aa

Aa

...

...

Aa

$2400

•

20

We the+present P evaluate = 2, 400 10,112worth = $a88, 489 ∑ ( 1.1of) motor a

t =1

−t

Motor Purchase Example, motor b •

We evaluate yearly energy costs over 20 years for motor b

A b = ( 77.5 kW ) ( 1600 h / yr) ( .08 $ / kWh) = $ 9, 920 / yr 0

1

Ab

2

Ab

...

t

19

20

Ab

Ab

...

...

Ab

$2900

•

20

WePnext present worth b = evaluate 2, 900 + the 9, 920 = motor $ 87, 354 ) of ∑ ( 1.1 b

t =1

−t

Motor Purchase Example, cont. • Comparing the present worths of motors a and b gives the following:

P a − P b = 88, 489 − 87, 354 = $ 1,135

• Therefore, the purchase of motor b results in a savings of $ 1,135 compared to motor a due to the use of the smaller load motor under the specified 10% discount rate

• What if it were a generator instead of a motor?

Infinite Horizon Cash-Flow Sets •

Consider a uniform cash-flow set with n → ∞

{ At = A : t = 0, 1, 2, ... } •

Then,

P = A

n 1 − β ( )

d

n →∞

1 A d

For an infinite horizon uniform cash-flow set

A = d P

Infinite Horizon Cash-Flow Sets, cont. •

We may view d as the capital recovery factor with the following interpretation: For an initial investment of P, dP = A is the annual amount recovered in terms of returns on investment

Internal Rate of Return • •

We consider a cash-flow set

{A

t

= A : t = 0, 1, 2, ...

}

The value of d for which

P =

n

t A β ∑ t =0 t =0

•

is called the internal rate of return (IRR) The IRR is a measure of how fast we recover an investment or stated differently, the speed with which the returns recover an investment

Internal Rate of Return Example

•

Consider the following cash-flow set $6,000 $6,000 $6,000 $6,000

$6,000

0 1

$30,000

2

3

4

8

Internal Rate of Return • The present value 1− β 8 P = − 30,000 + 6,000 =0 d

•

has the (non-obvious) solution of d equal to about 12%. The interpretation is that under a 12% discount rate, the present value of the cash flow set is 0 and so 12% is the IRR for the given cash- flow set –

The investment makes sense as long as other investments yield less than 12%.

Internal Rate of Return •

Consider an infinite horizon simple investment

I

0

A

A

1

2

A • Therefore d = I

•

...

A

n

ratio of annual return to initial investment

For I = $ 1,000 and A = $ 200, d = 20% and we interpret that the returns capture 20% of the investment each year or equivalently that we have a simple payback period of 5 years

Efficient Refrigerator Example • •

A more efficient refrigerator incurs an investment of additional $ 1,000 but provides $ 200 of energy savings annually For a lifetime of 10 years, the IRR is computed from the solution of

1− β 0 = − 1,000 + 200 d or

1− β d

10

= 5

10

The solution of this equation requires either an iterative approach or a value looked up from a table

Efficient Refrigerator Example, cont. •IRR tables show that

1 − β 10 = 5.02 d d = 15%

and so the IRR is approximately 15% If the refrigerator has an expected lifetime of 15 years this value becomes

1 − β 15 = 5.00 d d = 18.4%

As was mentioned earlier, the value is 20% if it lasts forever

Impacts of Inflation • Inflation is a general increase in the level of prices in an • •

economy; equivalently, we may view inflation as a general decline in the value of the purchasing power of money Inflation is measured using prices: different products may have distinct escalation rates Typically, indices such as the CPI – the consumer price index – use a market basket of goods and services as a proxy for the entire U.S. economy

–

reference basis is the year 1967 with the price of $ 100 for the basket (L 0); in the year 1990, the same basket cost $ 374 (L 23 )

Figuring Average Rate of Inflation •

Calculate average inflation rate e from 1967 to 1990

374 = 3.74 ( 1 + e) = 100 ln ( 3.74 ) ln ( 1 + e ) = → e = 0.059% 23 23

Current (1/2009) basket value is about 632.

Source: http://en.wikipedia.org/wiki/File:US_Historical_Inflation.svg