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Engineering Economics Chapter 3 Interest and Equivalence

Engineering Economics - Islamic University of Gazasite.iugaza.edu.ps/.../files/2014/01/Chapter-3-engineering-Economics.… · Engineering Economics Chapter 3 Interest and Equivalence

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Engineering Economics

Chapter 3

Interest and Equivalence

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.

2

3

Example 3-1

Purchase 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 3-2

A 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 Flows

0 (Now) +$1000

1 -580

2 -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

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Interest

• Simple Interest

• Compound interest

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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 n

Or F = P(1+ i n)

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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) = $2000

Amount due at the end of loan = P + P i n = 5000 + 2000

= $7000

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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.

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Example 3-4

… 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 interest per 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

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Single Payment Compound Amount Formula

If 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 periods

P = a present sum of money

F = 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.

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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 Solution

Single Payment Present Worth Formula

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

P = F (P/F,i,n) , i = 5% and n = 4 periods

From tables in Appendix B, (P/F,i,n) = 0.8227

P = 800 x 0.8227 = $658.16

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F = 800

P = ?

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

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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.

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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.

…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

Total

Paymentint*b b+c

1 5,000 400 5,400 1,000 1,400

2 4,000 320 4,320 1,000 1,320

3 3,000 240 3,240 1,000 1,240

4 2,000 160 2,160 1,000 1,160

5 1,000 80 1,080 1,000 1,080

SUM 1,200 5,000 6,200

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…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

Total

Paymentint*b b+c

1 5,000 400 5,400 0 400

2 5,000 400 5,400 0 400

3 5,000 400 5,400 0 400

4 5,000 400 5,400 0 400

5 5,000 400 5,400 5,000 5,400

SUM 2,000 5,000 7,000

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… 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

Total

Paymentint*b b+c

1 5,000 400 5,400 852 1,252

2 4,148 332 4,480 920 1,252

3 3,227 258 3,485 994 1,252

4 2,233 179 2,412 1,074 1,252

5 1,160 93 1,252 1,160 1,252

SUM 1,261 5,000 6,261

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… 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

Total

Paymentint*b b+c

1 5,000 400 5,400 0 0

2 5,400 432 5,832 0 0

3 5,832 467 6,299 0 0

4 6,299 504 6,802 0 0

5 6,802 544 7,347 5,000 7,347

SUM 2,347 5,000 7,347

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The four plans were

Year Plan 1 Plan 2 Plan 3 Plan 4

1 $1400 $400 $1252 0

2 1320 400 1252 0

3 1240 400 1252 0

4 1160 400 1252 0

5 1080 5400 1252 7347

Total $6200 $7000 $6260 $7347

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.

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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.

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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 Solution

Single Payment Present Worth Formula

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

P = F (P/F,i,n) , i = 5% and n = 4 periods

From tables in Appendix B, (P/F,i,n) = 0.8227

P = 800 x 0.8227 = $658.16

F = 800

P = ?

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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

Example 3-8

$400

0 1 2 3 4 5

$600

Alternate Solution

P = F(1+i)-n

P = 400(1+0.12)-3

+ 600(1+0.12)-5

P = $625.17

P

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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

Example 3-8 (Interest Compounded monthly)

$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

P

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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.

Points of view

Year Cash flow

0 - P 1 0 2 0 3 +400 4 0 5 +600

Year Cash flow

0 + P 1 0 2 0 3 -400 4 0 5 -600

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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