CE 203 EEA Chap 3

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CE 203 EEA Chap 3
Interest and Equivalence
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Time Value of Money

• The use of money has value
• Somebody will pay you to use your money
• You will pay others to use their money
• Interest is Rent for the use of money
• When decisions involve cash flows over a
  considerable length of time, economic analysis
  should include the effects of
• Interest
• Inflation

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Pre-Course EEA Assessment

• 1000 in savings account at 10 for 1 year?
• 1000 per year for 30 years at 10?
• For typical investment plan, 1000 per year for
  30 years at 10 yields 164,494
• (see formula for sinking fund)

4
Your time value of money?

• Would you rather have 100 now or
• 100 a year from now?
• 110 a year from now?
• 120 a year from now?
• 150 two years from now?

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Simple Interest

• Interest computed only on the original sum of
  money or principal
• Total interest earned I P x i x n where
• P principal or present sum of money
• i interest rate per period
• n number of periods
• Example 1000 borrowed at 8 for two years,
  simple interest
• I 1000 x .08 x 2 160

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Simple InterestFuture Value, F, of a Loan, P

• F P P x in P (1 in) P 1 i(period)
• Example 1000 borrowed at 8 for five years,
  simple interest
• F 1000 (1 .08(5)) 1400

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Compound Interest

• Interest on original sum on interest
• Example 1000 _at_ 10 per year, F
• F1 (first year) 1000 1000(0.1)
• or 1000 (1 0.1) 1100
• F2 (second year) 1100 1100(0.1)
• or 1000(10.1)(10.1) 1210
• or 1000(10.1)2
• F3 1000(10.1)3
• Fn 1000(10.1)n
• General Fn P(1i)n

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Compound Interest

• Interest computed on the original sum and any
  unpaid interest

Period Beginning Balance Interest for Period Ending Balance
1 P iP P(1 i)
2 P(1 i) iP(1 i) P(1 i)2
3 P(1 i)2 iP(1 i)2 P(1 i)3
n P(1 i)n-1 iP(1 i)n-1 P(1 i)n
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Compound Interest

• Future Value, F, P (1 i)n
• P principal or present sum of money
• i interest rate per period
• n number of periods (years, months, )
• Example 1000 borrowed at 8 for five years,
  compound interest, all of principal and interest
  repaid in 5 years
• F 1000 (1 .08)5 1469.33
• F P (F/P, i, n) 1000 (F/P, .08, 5)
  1469.00 (see EEA p. 576)

Note Simple interest yield would be 1400
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Compound Interest

• Total interest earned In P (1 i)n - P
• P principal or present sum of money
• i interest rate
• n number of periods
• Example 1000 borrowed at 8 for five years,
  compound interest
• I 1000 (1 .08)5 - 1000 469.33

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Simple vs. Compound Interest

• Future value, F, for P 1000 at 8

Periods F/P, simple i F/P, compound i
1 1080 1080
2 1160 1166
3 1240 1260
4 1320 1360
5 1400 1469
10 1800 2159
15 2200 3172
20 2600 4661
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Specification of Interest Rate, i

• 1) 8 - assumed to mean per year and
  compounded annually
• 2) 8 compounded quarterly - 2 per each 3
  months, compounded every 3 months
• 3) 8 compounded monthly 2/3 each month,
  compounded every month
• 4) 8 .08 in equations NB1000

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In-class Example

• Would you rather have 5000 today or 35,000 in
  25 years if the interest rate was 8 compounded
  yearly/ monthly
• 1) yearly?
• F P(F/P, .08, 25) 5000 (6.848) 34, 240
• 2) monthly?
• F P (F/P, .006667, 300) 36,700

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Four Ways to Repay a Debt
Plan Principal repaid Interest paid Trend on interest earned
1 in equal annual installments on unpaid balance declines
2 at end of loan on unpaid balance constant
3 in equal annual installments in equal annual installments declines at increasing rate
4 at end of loan at end of loan (compounded) increases at increasing rate
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Loan Repayment Plan 1 (5000, 5 years, 8)
Year Owed at begin-ning Annual Interest Total owed at end Principal Paid Total pay-ment
1 5000 400 5400 1000 1400
2 4000 320 4320 1000 1320
3 3000 240 3240 1000 1240
4 2000 160 2160 1000 1160
5 1000 80 1080 1000 1080
1200 5000 6200
Bank loans with yearly payments Principal
repaid in equal installments
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Loan Repayment Plan 2 (5000, 5 years, 8)
Year Owed at begin-ning Annual Interest Total owed at end Princi-pal Paid Total pay-ment
1 5000 400 5400 0 400
2 5000 400 5400 0 400
3 5000 400 5400 0 400
4 5000 400 5400 0 400
5 5000 400 5400 5000 5400
2000 5000 7000
Interest only loans
used in bonds and international loans
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Loan Repayment Plan 3 (5000, 5 years, 8)
Year Owed at begin-ning Annual Interest Total owed at end Princi-pal Paid Total pay-ment
1 5000 400 5400 852 1252
2 4148 331 4479 921 1252
3 3227 258 3485 994 1252
4 2233 178 2411 1074 1252
5 1159 93 1252 1159 1252
2000 5000 6260
Equal annual installments
Auto/home loans (but usually monthly payments)
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Loan Repayment Plan 4 (5000, 5 years, 8)
Year Owed at begin-ning Annual Interest Total owed at end Princi-pal Paid Total pay-ment
1 5000 400 5400 0 0
2 5400 432 5832 0 0
3 5832 467 6299 0 0
4 6299 504 6804 0 0
5 6803 544 7347 5000 7347
2347 5000 7347
Interest and principal repaid at end of loan
Certificates of deposit (CDs) and IRAs
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Loan repayment plans 1-4

• Are all equivalent to 5000 now in terms of time
  value of money,
• May not be equally attractive to loaner or
  borrower
• Have different cash flow diagrams
• Equivalent in nature but different in structure

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Equivalence

• The present sum of money is equivalent to the
  future sum(s) (from our perspective), if..
• ..we are indifferent as to whether we have a
  quantity of money now or the assurance of some
  sum (or series of sums) of money in the future
  (with adequate compensation)

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Equivalence

• Used to make a meaningful engineering economic
  analysis
• Apply by finding equivalent value at a common
  time for all alternatives
• value now or Present Worth
• value at some logical future time or Future
  Worth
• Assume the same time value of money (interest
  rate) for all alternatives

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You borrow 8000 to help pay for senior year at
ISU the bank offers you two repayment plans for
paying off the loan in four years
Year Plan 1 payment Plan 2 payment
1 0 0
2 0 3300
3 0 3300
4 10,400 3300
Total 10,400 9900
Which would you choose?
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Comparing Economic Alternatives

• Technique of Equivalence requires that we
• 1)Determine our time value of money
• 2)Determine a single equivalent value at a
  selected time for Plan 1
• 3) Determine a single equivalent value at the
  same selected time for Plan 2
• 4) Compare the two values

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Technique of Equivalence

• Cash flows can be compared by calculating their
  total Present Worth (now value) or their total
  Future Worth (at some time in the future)
• Future Worth, F P (1 i)n
• Solving the previous equation for P gives
• Present Worth, P F (1 i)-n
• P some present amount of money
• F some future amount of money
• i interest rate per time period
  (appropriate time value of money)
• n number of time periods

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Technique of Equivalence

• To compare payment plans, move all amounts
  to now (Present Worth)
• or to 4 years hence (Future Worth)

Year Plan 1 payment Plan 2 payment
1 0 0
2 0 3300
3 0 3300
4 10,400 3300
Total 10,400 9900
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Technique of Equivalence Present Worth
Assume our time value of money is 7 APR
PW?
Present WorthPayment Plan 1 P F(1 i)-n
1
2
3
4
0
P4 10,400(1 .07)-4 7,934.11 Total PW
7,934.11Or P 10,400 (P/F,
.07, 4) .7629 (10,400) 7934.16
10,400
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Technique of Equivalence Present Worth
Assume our time value of money is 7 APR
PW?
Present WorthPayment Plan 1 P F(1 i)-n
1
2
3
4
0
3300
3300
3300
P2 3300(1 .07)-2 2882.35P3 3300(1
.07)-3 2693.78 P4 3300(1 .07)-4
2517.55 Total PW 8093.68
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Technique of Equivalence Present Worth
Which would you choose?
7934.11
Plan 1
8093.68
1
2
3
4
0
Plan 2
1
2
3
4
10,400
0
3300
3300
3300
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Technique of Equivalence Future Worth
Assume our time value of money is 7 APR
Future WorthPayment Plan 1 F P(1 i)n
1
2
3
4
0
F4 10,400(1 .07)0 10,400 Total FW
10,400.00
10,400
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Technique of Equivalence Future Worth
Assume our time value of money is 7 APR
1
2
3
4
0
Future WorthPayment Plan 2 P F(1 i)-n
3300
3300
3300
F2 3300(1 .07)2 3778.17F3 3300(1
.07)1 3531.00 F4 3300(1 .07)0
3300.00 Total FW 10609.17
Again, larger than Plan 1
FW?
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In-class Example

• You deposit 3000 in an account that earns 5,
  compounded daily.
• How much could you withdraw from the account at
  the end of two years?
• F 3000 (F/P, .05/365, 730) 3,315.49
• How much could you withdraw if the compounding
  were monthly?
• F 3000 (F/P, .05/12, 24) 3,314.82

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In-class Example

• You are considering two designs for a wastewater
  treatment facility.
• Plan 1 Costs 1,700,000 to construct and will
  have to be replaced every 20 years.
• Plan 2 Costs 2,100,000 to construct and will
  have to be replaced every 30 years.
• Which is the better of the two designs?
• Assume 7 APR and neglect inflation and
  operating, maintenance and disposal costs.
• Sketch CFDs for each plan and compare.

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How to compare two schemes with different lives

• One plant lasts 20y and the other 30y
• Could we make a comparison over 20y?
• Could we make a comparison over 30y?
• What would be a decent period over which to
  compare?
• The smallest common denominator, i.e. in
  this case 60y.

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Technique of Equivalence Present Worth
Based on dollars only, which is the better plan?
2.254 M (if provision is made for final
replacement, 2.283M)
10
20
30
40
50
60
repeats
Plan 1
0
1.7 M
1.7 M
1.7 M
1.7 M
P 1.7 1.7 (1.07)-20 1.7
(1.07)-40 1.7 (1.07)-60

• This term will only apply when it is a
  requirement to replace at the end, not normally
  and it
• has been excluded from this analysis. You could
  get to 60 years without a final replacement.

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Technique of Equivalence Present Worth
Based on dollars only, which is the better plan?
2.376 M
P 2.1 2.1(1.07)-30 the 60y
replacement, which does not apply
10
20
30
40
50
60
repeats
0
Plan 2
Not required
2.1 M
2.1 M
Other option was 2.254 M