General Knowledge: Economics

Richard B. Pool, Ph.D., P.E.Distinguished Professor Emeritus of Civil and Environmental Engineering

University of South Carolina

A Review of Interest Formulas with Examples• Compound Amount / Future Worth• Present Worth• Series Compound Amount

- Sinking Fund• Capital Recovery

- Series Present Worth• Comparison of Alternatives Problems:

- P.W., Annual Cost, and Benefit-Cost Ratio Approaches• Depreciation Methods• Rate of Return After Taxes and Depreciation Problem• Retirement and Replacement Problem• Inflation Consideration Problem

Course Outline

Compound Amount

F = future worth

P = present worth

F = P(l+i)ni = period interest rate

n = number of periods

Basic compound interest equation (Future Worth)

Example 1

Assume

P = $2,500

i = 6%

How much will it be in 5 years? (n = 5)

F = $2,500(1+.06) = $3,345.565

Example 1 (Continued)

How long would it require to triple your input?

3 x 2,500 = 7,500

7,500 = 2,500(1+.06) n

3 = 1.06n

log 3 = nlog1.06

n = log 3/log 1.06 = .47712/.02530587 = 18.85 yrs

Present Worth, P

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

The equation for Present Worth, P, is obviously a form of the previous Future Worth equation:

Series Compound AmountA = uniform amount deposited at the end of each

interest period. F = future worth.

A A A A A A A

F

F = A (1+i)-1n

i

A = 200

i = 0.035

n = 10

Example 2

Assume $200 is deposited at the end of each 6 months for 5 years at 7% per annum compounded semi-annually. How much would one have after 5 years?

F = 200 1.035 -1.035

10=$2346.28

(Note that the sum of the deposits is $2000.00)

Sinking Fund

This comes from the previous formula and is

Used when one wants to know required deposits to reach a goal.

A = F i

(1+ i) - 1n

Capital Recovery

A A A A A A A

P

P = Present w orth

and A = uniform insta llm ents

A = Pi(1+i)

(1+ i) - 1

n

n

The ‘uniform installment’(A) required to ‘recover’ a ‘capital outlay’(P) is given by:

Example 3

Assume a mortgage of $150,000 must be paid off in monthly installments for 20 years @ 8% per annum compounded monthly

P = 150,000 i = .08/12 n = 240

What is the monthly payment?

A = 150,000.08/12(1 + .08/12)

(1 +.08/12) -1240

240

= $1,254.66

Series Present Worth

The Series Present Worth equation is a form of the previous Capital Recovery equation:

Used in determining the P.W. of a series of uniform future payments

P = A(1 + i) - 1

n

i(1 + i)n

Example 4(Comparison of Alternatives Present Worth

Approach)

A firm is considering a pair of options before purchasing certain equipment.

Use ‘series present worth’ to determine a choice.

1st CostLifeSalvageInterestAnnual O&M

Option A Option B$85,00010 years$20,00012%$12,000

$100,00010 years$25,00012%$9,000

Find P.W. of eachApproach

P = 85,000 – 20,000(1.12) + 12,000

= 85,000 – 6,439.46 + 67,802.68= 146,363.22

A)

B) P = 100,000 – 25,000(1.12) + 9,000

= 100,000 – 8,049.33 + 50,852.01= 142,802.68

-10

-10

B is less expensive – therefore best choice

Example 4 (Continued)

1.12 - 1.12(1.12)

10

10

1.12 - 1.12(1.12)

10

10

Example 5(Comparison of Alternatives Annual Cost

Approach)

Find the annual cost of each of the options of Example 4

O ption A

1st C ost - (85,000 - 20,000).12(1.12)

1.12 - 1= 11,503.97

10

10

Interest on salvage 20,000(.12) = 2,400.00O &M = 12,000.00

Annual C ost = $25,903.97

Choose B – Lower annual cost

O ption B

1st Cost - (100,000 - 25,000) .12(1.12)

1.12 - 1= 13,273.81

10

10

Interest on salvage 25,000(.12) = 3,000.00O &M = 9,000.00

Annual C ost = $25,273.81

Example 5(Comparison of Alternatives Annual Cost

Approach)

Shortcut to get A.C. if P.W. is known use Capital Recovery equation

A) 146,363.22.12(1.12)

1.12 - 1

10

= $25,903.97

B) 142,802.68 .12(1.12)10

= $25,273.81

10

1.12 - 110

Example 5 (Continued)

Example 6(Comparison of Alternatives Benefit–Cost Ratio)

Determine the advisability of an investment using benefit-cost ratio. Three alternatives are available for a new parcel “assorter.”

Deluxe R egular Econom y

1st C ostAnnual Incom eO &M (Annual)SalvageM AR R 15%

$220,00079,00038,00016,000

$125,00043,00013,0006,900

$75,00028,0008,0003,000

Life of each = 10 years

Solution - U se Annual C ost

Example 6 (Continued) Solution

Deluxe (220000 - 16000).15(1.15)

1.15 - 1 10

10

= 43047.42 B /C = 79000/43047.42 = 1.84

+ 16000(.15)

Regular (125000 - 6900).15(1.15)

1.15 - 1 10

10+ 6900(.15)

=23531.67 B /C = 43000/23521.67 = 1.83

Econom y (75000 - 3000)

= 14796.15 B /C = 28000/14796.15 = 1.89

.15(1.15)

1.15 - 1 10

10+ 3000(.15)

Depreciation Methods – Annual ValuesThree Methods of Depreciation:

* Straight Line (Example 7a)* Sum of Digits (Example 7b) * Double Declining Balance (Example 7c)

Example 7a (Straight Line Depreciation Method)

1st Cost $80,000 Salvage $10,000 Life 15 yrs

Straight Line Depreciation = (80,000-10,000)/15 = $4,666.67/yr

Depreciation Methods (Continued)Example 7b

(Sum Of Digits Method)

1st Cost = $10,000 Salvage = $1,000Life = 4 years 4+3+2+1 = 10 sum of digits !

Changes each year:1st Year’s Depreciation = 4/10 (10,000 – 1000) = 36002nd Year’s Depreciation = 3/10 (10,000 – 1000) = 27003rd Year’s Depreciation = 2/10 (10,000 – 1000) = 18004th Year’s Depreciation = 1/10 (10,000 – 1000) = 900

Total Depreciation = $9000

Depreciation Methods (Continued)Example 7c

(Double Declining Balance Method)

* Using values from previous example (7b:Sum of Digits):

Depreciation = (Cost - Depr. to date)

1st Yr = (10,000 - 0) = 5,000

All OK here ifsalvage = $625.00

2nd Yr = (10,000 - 5,000) = 2,500

(10,000 - 7,500) = 1,250

(10,000 - 8,750) = 625

3rd Yr =

4th Yr =

24

24

24

24

2N

N=Life

Total = $9,375

Eq’n

Example 8(Rate of Return Problem)

A firm wishes to invest in a project if it can make a MARR of 12% after taxes.Assume:

Equipment cost = $110,000Salvage after 4 years = $15,000Annual O&M costs are $7,500.Annual income expected at $50,000Taxes @40%Use Double Declining Balance Depreciation

Depreciation - 1st Year2nd "3rd "4th "

- 2/4 (110,000)= 55,000- 2/4 (55,000) = 27,500- 2/4 (27,500) = 13,750 - Use 12,500- 2/4 (13,750) = 6,875 - Use Zero

Note - Total depreciation cannot exceed $95,000

Example 8 (Continued)

Year

0

1

2

3

4

Income

50,000

50,000

50,000

50,000

Depr.

55,000

27,500

12,500

0

I-D

-5,000

22,500

37,500

50,000

Taxes

-2,000

9,000

15,000

20,000

O&M

7,500

7,500

7,500

7,500

Net

44,500

33,500

27,500

22,500

Other

-110,000

+15,000

Example 8 (Continued)

The firm failed to make 12% by a very narrow margin becausethe sum is negative.

Present Worth

First Cost

1st Year

2nd "

3rd "

4th "

44,500(1.12)

33,500(1.12)

27,500(1.12)

(22,500 + 15,000)(1.12)

-110,000.00

39,732.14

26,705.99

19,573.96

23,831.93

-155.98Sum

-1

-2

-3

-4

Example 8 (Continued)

Example 9(Retirement and Replacement Problem)

A warehouse presently in service is believed to be sufficient for 10 more years with an estimated selling price at that time of $150,000. O.&M. costs have been about $10,000/year, and property taxes and insurance $6,000/year. The warehouse could be sold now for $195,000. Rental space can be obtained for $30,000/year, with O.&M. costs plus fire insurance for $6,000/year. Use 12% and make a ten year study of annual costs to compare options.

A - Continue with present warehouse

A = (195,000 - 150,000)

= 18,000.00= 16,000.00

$41,964.29

= 7,964.29

150,000 x .12O.&M., taxes, insurance

Sum

Example 9 (Continued)

1.12 - 1.12(1.12)

10

10

The comparison favors disposing of current warehouse and renting space.

B - R enta l Space

= 30,000.00= 6 ,000.00

$36,000.00

R entO .&M . + Insurance

Sum

Example 9 (Continued)

Example 10

(Inflation Consideration Problem)

Assume inflation to occur at 3% / year for the next four years. A borrower gets a loan of $5,000 for four years at 8%. How much does the lender really make if payments are in annual installments?

A = 5,000 .08 x 1.081.08 - 1

= 1,509.60 yearly payment

Value of payments in current dollars.

year 1 – 1,509.60(1.03) = 1,465.63

year 2 – 1,509.60(1.06) = 1,422.94

year 3 – 1,509.60(1.09) = 1,381.50

year 4 – 1,509.60(1.12) = 1,341.26

-1

-2

-3

-4

4

Example 10 (Continued)

Lender really makes only 4.85% Inflation always favors the borrower.

Try 6% P = 1,465.63(1.06) + 1,422.94(1.06) + 1,381.50(1.06) + 1,341.26(1.06) = 4,871.42

Try 5% P = 1,465.63(1.05) + 1,422.94(1.05) + 1,381.50(1.05) + 1,341.26(1.05) = 4,983.34

Extrapolate to get $5,000.00

.06 - (.01) = .04855,000 – 4,871.424,983.34 – 4,871.42

-2

-2

-4

-4

-1 -3

-3

-1

Example 10 (Continued)