Engineering Economics ECIV 5245

Chapter 3

Interest and Equivalence

2

Cash Flow Diagrams (CFD)

Used to model the positive and negative cash flows. At each time at which cash flow will occur, a vertical

arrow is added, point down for costs and up for revenues.

Cash flow are drawn to relative scale Rent and insurance are beginning-of-period cash

flows; i.e. just put an arrow in where it occurs. O&M, salvages, and revenues are assumed to be end-

of-period cash flows.

3

Example 3-1Purchase a new $30,000 mixing machine. The machine

may be paid for in one of two ways A. Pay the full price now minus a 3% discount B. Pay $5000 now, $8000 at the end of 1st yr, and $6000 at

end of each following year

List the alternatives in the form of a table of cash flows

4

Continue … Example 3-1

End of year

0 (now)

1

2

3

4

5

Pay in Full Now Pay over 5 Yrs

-$29,100 -$5000

0 -$8000

0 -$6000

0 -$6000

0 -$6000

0 -$6000

Cash flow table:

5

Example 3-2A man borrowed $1000 from a bank at 8% interest.

At the end of 1st yr: Pay half of the $1000 principal amount plus the interest.

At the end of 2nd yr: Pay the remaining half of the principal amount plus the interest for the second year.

Compute the borrower’s cash flow

End of Year Cash Flows0 (Now) +$1000

1 -5802 -540

6

Time Value of Money If monetary consequences occur in a short period of

time → Simply add the various sums of money What if time span is greater? $100 cash today vs. $100 cash a year from now? Money is rented. The rent is called the interest If you put $100 in the bank today, and interest rate is

9% → $109 a year from now

7

Interest

Simple Interest Compound interest

8

Simple Interest

Interest that is computed only on the original sum and not on accrued interest. e.g. if you loaned someone the amount of P at a simple

interest rate of i for a period of n years: Total interest earned = P × i × n = P i n The amount of money due after n years:

F = P + P i nOr F = P(1+ i n)

9

Example 3-3You loaned a friend $5000 for 5 years at a simple

interest rate of 8% per year.How much interest you receive from the loan?How much will your friend pay you at the end of 5 yrs.

Total interest earned = P i n = (5000)(0.08)(5) = $2000Amount due at the end of loan = P + P i n = 5000 + 2000

= $7000

10

Compound Interest This is the interest normally used in real life Interest on top of interest Next year’s interest is calculated based on the unpaid

balance due, which includes the unpaid interest from the preceding period.

11

Example 3-4

12

… Compound InterestCompound interest is interest that is charged on the original

sum and un-paid interest.

You put $500 in a bank for 3 years at 6% compound interestper year.

At the end of year 1 you have (1.06) 500 = $530. At the end of year 2 you have (1.06) 530 = $561.80. At the end of year 3 you have (1.06) $561.80 = $595.51.

Note: $595.51 = (1.06) 561.80

= (1.06) (1.06) 530 = (1.06) (1.06) (1.06) 500 = 500 (1.06)3

13

Single Payment Compound Amount FormulaIf you put P in the bank now at an interest rate of i% for n years, the future amount you will have after n years is given by

F = P (1+ i )n

i = interest rate per interest period (stated as decimal)n = number of interest periodsP = a present sum of moneyF = A future sum of money

The term (1+i)n is called the single payment compound factor.F = P (1+i)n = P (F/P,i,n)

Also P = F (1+i)-n = F (P/F,i,n)The factor (F/P,i,n) is used to compute F, given P, and given i and n.The factor (P/F,i,n) is used to compute P, given F, and given i and n.

14

Present ValueExample 3-6If you want to have $800 in savings at the end of four years, and 5% interest is paid annually, how much do you need to put into the savings account today?We solve F = P (1+i)n for P with i = 0.05, n = 4, F = $800.

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

P = 800/(1.05)4 = 800 (1.05)-4 = 800 (0.8227) = $658.16.

Alternate SolutionSingle Payment Present Worth Formula

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

P = F (P/F,i,n) , i = 5% and n = 4 periodsFrom tables in Appendix B, (P/F,i,n) = 0.8227

P = 800 x 0.8227 = $658.16

F = 800

P = ?

15

Factors in the Book (page 573 in 9-th edition)

16

Present ValueExample: You borrowed $5,000 from a bank at 8%

interest rate and you have to pay it back in 5 years. The debt can be repaid in many ways.

Plan A: At end of each year pay $1,000 principal plus interest due.

Plan B: Pay interest due at end of each year and principal at end of five years.

Plan C: Pay in five end-of-year payments.

Plan D: Pay principal and interest in one payment at end of five years.

17

…Example (cont’d)You borrowed $5,000 from a bank at 8% interest rate and you have to pay

it back in 5 years.Plan A: At end of each year pay $1,000 principal

plus interest due.a b c d e f

Year Amnt.Owed

Int. Owed Total OwedPrincip.Payment

TotalPaymentint*b b+c

1 5,000 400 5,400 1,000 1,4002 4,000 320 4,320 1,000 1,3203 3,000 240 3,240 1,000 1,2404 2,000 160 2,160 1,000 1,1605 1,000 80 1,080 1,000 1,080

SUM 1,200 5,000 6,200

18

…Example (cont'd)You borrowed $5,000 from a bank at 8% interest rate and you have to pay

it back in 5 years.

Plan B: Pay interest due at end of each year and principal at end of five years.a b c d e f

Year Amnt.Owed

Int. Owed Total OwedPrincip.Payment

TotalPaymentint*b b+c

1 5,000 400 5,400 0 4002 5,000 400 5,400 0 4003 5,000 400 5,400 0 4004 5,000 400 5,400 0 4005 5,000 400 5,400 5,000 5,400

SUM 2,000 5,000 7,000

19

… Example (cont'd)You borrowed $5,000 from a bank at 8% interest rate and you have to pay

it back in 5 years.

Plan C: Pay in five end-of-year payments.

a b c d e f

Year Amnt.Owed

Int. Owed Total OwedPrincip.Payment

TotalPaymentint*b b+c

1 5,000 400 5,400 852 1,2522 4,148 332 4,480 920 1,2523 3,227 258 3,485 994 1,2524 2,233 179 2,412 1,074 1,2525 1,160 93 1,252 1,160 1,252

SUM 1,261 5,000 6,261

20

… Example (cont'd)You borrowed $5,000 from a bank at 8% interest rate and you have to pay

it back in 5 years.

Plan D: Pay principal and interest in one payment at end of five years.a b c d e f

Year Amnt.Owed

Int. Owed Total OwedPrincip.Payment

TotalPaymentint*b b+c

1 5,000 400 5,400 0 02 5,400 432 5,832 0 03 5,832 467 6,299 0 04 6,299 504 6,802 0 05 6,802 544 7,347 5,000 7,347

SUM 2,347 5,000 7,347

21

The four plans were

$7347$6260$7000$6200Total73471252540010805

0125240011604012524001240301252400132020$1252$400$14001

Plan 4Plan 3Plan 2Plan 1Year

How do we know whether these plans are equivalent or not?→We won’t be able to know by simply looking at the cash flows, therefore some effort should be made.

22

Equivalence In the previous example, four payment plans were

described. The four plans were used to accomplish the task of

repaying a debt of $5000 with interest at 8%. All four plans are equivalent to $5000 now. i.e. all four plans are said to be equivalent to each

other and to $5000 now.

23

Present ValueExample 3-8If you want to have $800 in savings at the end of four years, and 5% interest is paid annually, how much do you need to put into the savings account today?We solve F = P (1+i)n for P with i = 0.05, n = 4, F = $800.

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

P = 800/(1.05)4 = 800 (1.05)-4 = 800 (0.8227) = $658.16.

Alternate SolutionSingle Payment Present Worth Formula

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

P = F (P/F,i,n) , i = 5% and n = 4 periodsFrom tables in Appendix B, (P/F,i,n) = 0.8227

P = 800 x 0.8227 = $658.16

F = 800

P = ?

24

In 3 years, you need $400 to pay a debt. In two more years, you need $600 more to pay a second debt. How much should you put in the bank today to meet these two needs if the bank pays 12% per year?

Interest is compounded yearly

P = 400(P/F,12%,3) + 600(P/F,12%,5)= 400 (0.7118) + 600 (0.5674) = 284.72 + 340.44 = $625.16

$400

0 1 2 3 4 5

$600

Alternate SolutionP = F(1+i)-n

P = 400(1+0.12)-3 + 600(1+0.12)-5

P = $625.17

Example 3-8

P

25

In 3 years, you need $400 to pay a debt. In two more years, you need $600 more to pay a second debt. How much should you put in the bank today to meet these two needs if the bank pays 12% compounded monthly?

Interest is compounded yearly

P = 400(P/F,12%,3) + 600(P/F,12%,5)= 400 (0.7118) + 600 (0.5674) = 284.72 + 340.44 = $625.16

$400

0 1 2 3 4 5

$600

Interest is compounded monthly

P = 400(P/F,12%/12,3*12) + 600(P/F,12%/12,5*12)= 400(P/F,1%,36) + 600(P/F,1%,60)= 400 (0.6989) + 600 (0.5504)

= 279.56 + 330.24 = $609.80

Example 3-8 (Interest Compounded monthly)

P

26

Borrower point of view:You borrow money from the bank to start a business.

Investors point of view:You invest your money in a bank and buy a bond.

Year Cash flow

0 - P1 02 03 +4004 05 +600

Year Cash flow

0 + P1 02 03 -4004 05 -600

Points of view

27

Appendix B in the text book tabulate:

Compound Amount Factor(F/P,i,n) = (1+i)n

Present Worth Factor(P/F,i,n) = (1+i)-n

These terms are in columns 2 and 3, identified as Compound Amount Factor: “Find F Given P: F/P”Present Worth Factor: “Find P Given F: P/F”

Concluding Remarks