2a - Engineering Economics - 1-11-15

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Engineering Economics

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Decision Making

We make decisions for many reasons, but usually one of the biggest factors is:

How do we decide the best option?

$

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Considerations

When to buy or sell? Repair or rebuild?Market conditions (interest rates-loans)Rate of return on investment (ROI) - (interest rate-savings)Source of financingPay-back periodProduct risksProject risks

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Time Value of Money

Time value of money refers to the change in the value of money over time and is influenced by the following three factors: Interest rate Time Investment or Loan Amount

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Cash Terminology

P0 (PW) = value of the money at the present (Present Worth)Pt= value of the money at end of time “t”

n = number of interest periodsi = interest rate per periodF = future value of a present sum of moneyA = Annuity, a series of consecutive, equal, end-of-period amounts of moneyG = gradient, uniform increase each year

Interest Rate

Interest Rate can be defined in two typical ways*:

1.“…money paid for the use of borrowed money.”.2.“…the return obtainable by the productive

investment of capital.”.

Definition 2 is primarily used in Engineering Economics and is another way of saying “return on investment (ROI)”.

* Principles of Engineering Economy, 5th Ed., Grant and Ireson

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Cash Flow Diagrams

Graphical depiction of cash into and out of an account

P = P0= initial purchase

0 1 2 3 4

A1=annual income

F=salvage

A2=annual cost of operation

Income

Expense

End of yearaccrual

F= future income

F= future cost

+

-

Monthly?

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Time Value of Money (example)

Suppose you deposit $1,000,000 in the bank at an annual interest rate, i, of 8%

P0 = $1,000,000

F1 = P0(1+i)

F2 = F1(1+i)

0 1 2 3 4

$1,080,000 $1,166,400 $1,360,500

F3 = F2(1+i)F4 = F3(1+i)

F4 = F3(1+i)= F2(1+i)(1+i)=F1(1+i)(1+i)(1+i)=P0(1+i)(1+i)(1+i)(1+i)=P0(1+i)4

In general, Fn = P0(1+i)n

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Compound Interest

Compound interest includes interest on the interest earned in pervious periodsThe nominal interest rate per year is the period interest rate times the number of periods per yearThe annual interest rate when considering the time value of money for rates quoted for a period (say one month) is called the effective interest rate.

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Compound Interest (cont.)

Visa Card statement lists interest as 2% per monthNominal Interest Rate =

2%/mo x 12 mo = 24% Effective Interest Rate =

ieff = (1+i)n -1 = (1.02)12 – 1 =

0.268 = 26.8%

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Compound Interest (cont.)

i = (1 + ) - 1 rt

t

wherei = effective interest rate per period (year)r = nominal interest rate per period (year)t = number of compounding periods (per year)

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Effective Interest Rate Example

A bank quotes the interest on a credit card balance as 1.5% per month or a nominal annual interest of 18%. What is the effective interest?

Thus, the effective annual interest rate is 19.6%

i = (1 + ) - 1 = (1+.015)12 –1 = 0.1960.1812

12

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Equivalence

Different sums of money at different times can be equal in economic valueFor example, at i = 8%, $1,000,000 today is equivalent to $1,360,500 at the end of 4 years

P0 = $1,000,000

P1 = P0(1+i)

P2 = P1(1+i)

0 1 2 3 4

$1,080,000 $1,166,400 $1,360,500

P3 = P2(1+i)P4 = P3(1+i)

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“New School” vs “Old School”

New School: F = P (1+i)n = $1,000,000(1.08)4

= $1,360,500

Old School: F = P(F/P, i, n) = $1,000,000(F/P,8%,4)

= $1,000,000(1.360) = $1,360,000

See ANGEL for interest rate tables

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Interest Factors

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Future Value of Money

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

Example: if you have $1000 today, what is it worth at the end of 7 years at 10% annual effective interest?

F = 1000(1+.10)7 = $1000 (F/P,10%,7)= $1000 (1.949) = $1949

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Present Value of Future Money

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

Example: if you need $1950 at the end of 7 years, how much will you need to invest today at 10% annual effective interest?

P = 1950(1+.10)-7 = 1950 (P/F,10%,7) = 1950(0.5132) = $1000.74

NOTE: that’s minus n

Annuity

(P/A,i%,n)= [(1+i)n]-1

i(1+i)n

Annuity “A” = P(A/P,i%,n) = P [i(1+i)n ]

[ (1+i)n]-1]

Present Value of an Annuity (fixed payments:

P = A {[(1+i)n]-1}

i(1+i)n

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Equivalence Example

Consider four loan repayment plans for P = $5,000, i = 0.15 per year, and n = 5 years plan 1 - defer all payments until the end plan 2 - pay interest each year and principal

at the end plan 3 - pay interest plus $1,000 each year plan 4 - make equal payments each year

Which is the best option?

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Equivalence (cont.)

Even though each payment plan is different and the amount paid is different, each is equivalent to $5,000 at i = 0.15 and n = 5

Plan 1Plan 2Plan 3Plan 4

$10,0578,7507,2507,458

Total Amount Repaid

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Amortization Tablen Interest

PaymentPrincipal Payment

Total Payment

1 $750.00 $741.58 $1491.58

2 $638.76 $852.81 $1491.58

3 $510.84 $980.74 $1491.58

4 $363.73 $1127.85 $1491.58

5 $194.55 $1297.02 $1491.58

Total $2457.89 $5000.00 $7457.89

Assignment 3

Recreate the previous amortization table using Microsoft Excel. Use Excel for calculating the yearly payment, interest, principal and total payments.

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Assignment 3A

Recreate the previous amortization table using Microsoft Excel. Use Excel for calculating the monthly payment, interest, principal and total payments.

Answers: A = $118.95; i (total) = $2,136.98; Principal Total = $5,000; Total Paid = $7,136.98

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Amortization Schedule

See example mortgage spreadsheet

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Equivalence: Varying Interest Rate and Time

Amount Invested = $1,000,000 (one time)

Annual Interest Rate 8% = 0.08

Compounding Period No. Periods Value at End of 4 Years

1 year 4 $1,360,488.96

1 month 48 $1,375,666.10

1 week 208 $1,376,789.17

1 day 1460 $1,377,079.48

1 hour 35040 $1,377,125.75

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Amount Invested (one time) = $1,000,000Annual Interest Rate = Varies

Compounding Period = Monthly

RateValue at End of

4 Years

5% $1,220,895.36

8% $1,375,666.10

10% $1,489,354.10

15% $1,815,354.85

25% $2,690,496.60

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Amount Invested (per month) = $ 24,413.00

Annual Interest Rate = 8% = 0.08

Compounding Period = Monthly

Value at 4 years $1,375,670.48

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Assignment 4

Recreate the previous three slides using Excel. Put them all on one spreadsheet. Upload to the specified Drop Box.

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Assignment 5 - Credit Card Debt

Assume you have $5,000 in credit card debtAnnual interest rate = 12%How much will your monthly payment be if you payback over 12, 24, 36 and 48 months?What is the total amount repaid?How much total interest is paid?If you pay back only $100 per month, how long will it take you to repay your debt?

Assignment 5 Answers

12 months: A = $444.24; Total Payment = $5,330.88; Total i = $330.8824 months: A = $235.37; Total Payment = $5,648.88; Total i = $648.8836 months: A = $166.07; Total Payment = $5,978.52; Total i = $978.5248 months: A = $131.67; Total Payment = $6,320.16; Total i = $1,320.16Number of months = 69.7 +/-

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ExampleHow much can you take a loan for in 2 years if you only want to repay $400 each year for 5 years starting 3 years from now? i=10%

0 1 2 3 4 5 6 7

P2 = ?

F3=$400F4=$400

F5=$400F6=$400

F7=$400

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Example (cont’d)

0 1 2 3 4 5 6 7

P0 = ?

•Shift time axis, so that P2 is now P0

•Calculate the PV of each $400 payment.•$400 five years from now is $248.37 today P= = F(1+i)-n

P=F(P/F,10%,5)=400(1+.1)-

5=$248.37

0 1 2 3 4 5

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Example (cont’d)To solve the entire problem move all FV to PV:

P= (P/F,10%,5)+ (P/F,10%,4)+ (P/F,10%,3)+ (P/F,10%,2)+ (P/F,10%,1)

P=$400[(1+.1)-5+ (1+.1)-4+(1+.1)-3+ (1+.1)-2

+(1+.1)-1]

=$1516.31

So you can take a $1516 loan.

This method derives the Annuity Equation!

P= A(P/A,I%,n)= A[(1+i)n-1]= 400[(1.15-1)] = $1516.31

i(1+i)n.1(1.1)5

Watch this!

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Example (cont’d)

How much could you take a loan out for now, if you want to start paying back $400 per year at the end of year 3?

0 1 2 3 4 5 6 7

P0 = ?

F3=$400F4=$400

F5=$400F6=$400

F7=$400

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Example (cont’d)

We can’t apply the annuity equation, since we derived it as if the PV is only one year before the first payment (at t=2) and we want it at t=0 (3 years before the first payment).So we have to add another step.First we find the PV at t=2From previous calcs PV=$1516.31

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Example (cont’d)

Then we have to move the $1516.31 to t=0The $1516.31 now becomes a future value

and we solve for P0.

P0= F(1+i)-n= $1516.31(1.1)-2= $1253.15

If you want the money now but don’t want to start paying back until the end of year 3, then you get less money now.

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Useful Economic Formulas

(F/P,i%,n) = (1+i)n Future value of present money

F = P(1+i)n

(P/F,i%,n)= (1+i)-n Present value of future money

P = F(1+i)-n

(P/A,i%,n)= [(1+i)n]-1 Present value of fixed payments

i(1+i)n

P = A {[(1+i)n]-1} i(1+i)n

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Projects of Finite Life

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Example

Your client wants to build a commercial building and will have $1,000,000 in the bank by the end of 2006. Assume that construction will be completed by the end of 2009, when all money will be due the contractor.

If the money is collecting 6.5% annual interest until then, what should the final budget cost of the building be?

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Future Value of Money

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

F= 1,000,000(1+.065)3= (1,000,000,6.5%,3) =

$1,207,950

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Background

Engineering economics is often used as a basis for decision making.Decisions can be to determine project feasibility, determine long term project costs, or to determine which alternative to choose.Alternatives having the same economic life are evaluated on the basis of their present worth.Economic life is defined as the time in which you want to recover your investment. It does not have to be the actual life of the alternative.

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Example 1Suppose a developer wishes to construct a building as office rental property. The building will cost $5 million. Rental income is forecast at $0.75 million

annually. Maintenance cost will average $0.15 million

annually. The developer plans to sell the building in 5

years. How much will the selling price need to be for a $2 million profit if i = 8%? The construction time frame is 1 year.

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Example 1 (cont’d)

Notice that the economic life is five years even though the building will last longer. The economic life is often how long you want to take to recover your investment.

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Example 1 (cont.)

0 1 2 3 4 5$5 m

$0.15 m

$0.75 m F = ?

1 2 3 4 5$5 m

$0.15 m

$0.75 m

P0 = ?

Step 2: time shift & find the PV of the income and expenses

Step 1: draw the cash flow diagram

0 1 2 3 4

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Example 1 (cont.)

P0 = -5.0 + (0.75 - 0.15) (P/A,8,4)=-5 + (0.60((1+.08)4-1 )) 0.08(1+0.08)4

P0 = -5.0 + 0.60 (3.3121) = -$3.013 m

Step 3: change time scale back and transfer P1 to P0

0 1

P0 3.013P0 = P1 (P/F,8,1)= -3.013(1+0.08)-1

P0 = -3.013 (0.9259) = -$2.789 m (loss)

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Example 1 (cont.)Step 4: calculate selling price (include $2m profit!)

Selling price = F0 + $2m

F = P0 (F/P,8,5) + 2.0 = 2.789(1+.08)5 + 2.0

F = 2.789 (1.4693) + 2.0

F = $6.099 m

0 1 2 3 4 5

2.789

F0

$2m

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Example 2

Suppose the same developer has an option on property elsewhere that is for sale for $3.5 million. A building could be constructed on that site for $6.7 million. The income would be $1.45 m/year and maintenance of $0.20 m/year beginning at the end of year 2. At the end of year 5, the sale price will be $8.5 million. Which alternative is better?

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Example 2 (cont.)

0 1 2 3 4 5

$6.7 m $0.20 m

$1.45 m $8.5 m

0 1 2 3 4$6.7 m

$0.20 m

$1.45 m

P1 = ?

Step 1

$3.5 m

+ F = ?Step 1: draw cash flow diagram

Step 2: find P1 (let F, Profit and $8.5m wait for later)

Sale price

Profit

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Example 2 (cont.)

P1 = -6.7 + 1.25 (P/A,8,4) = -6.7 + 1.25 (1+.08)4-1 .08(1+.08)4 P1 = -6.7 + 1.25 (3.3121)

P1 = -$2.560 m

0 1 2 3 4 5$3.5 m

$8.5 m

$2.560 mP0 = ?

Step 3: find P0 (include $8.5m) + FProfit

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Example 2 (cont.)P0 = -3.5 + P1 (P/F,8,1)

P0 = -3.5 + (-2.560 (0.9259))

P0 = -$5.870 m

Step 4: find F and ProfitF = P0 (F/P,8,5)

F = -$5.870 (1.4693)

F = -$8.62 mProfit = $8.5m - $8.62Profit = -$0.12

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Example SummaryIf we compare the two examples as potential alternatives for investment, we find that the first alternative is the best economic alternative because:

The second alternative is at best a breakeven proposition

Alternative 1 Alternative 2There is $2 mprofit at the end of year 5

This is abreakeven proposition

2a - Engineering Economics - 1-11-15

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