ISU CCEE
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
Time Value of MoneyTime Value of MoneyTime Value of MoneyTime Value of Money
ISU CCEE
$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”)
Pre-Course EEA Pre-Course EEA AssessmentAssessmentPre-Course EEA Pre-Course EEA AssessmentAssessment
ISU CCEE
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?
Your time value of Your time value of money?money?Your time value of Your time value of money?money?
ISU CCEE
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 interestI = $1000 x .08 x 2 = $160
Simple InterestSimple InterestSimple InterestSimple Interest
ISU CCEE
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
Simple InterestSimple InterestFuture Value, F, of a Loan, PFuture Value, F, of a Loan, P Simple InterestSimple InterestFuture Value, F, of a Loan, PFuture Value, F, of a Loan, P
ISU CCEE
Compound InterestCompound Interest Compound InterestCompound InterestInterest on original sum + on interest
Example $1000 @ 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(1+0.1)(1+0.1) = $1210or $1000(1+0.1)2
F3 = $1000(1+0.1)3;
Fn = $1000(1+0.1)n
General: Fn = P(1+i)n
ISU CCEE
Interest computed on the original sum and any unpaid interest
Compound InterestCompound InterestCompound InterestCompound 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
ISU CCEE
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 = $1469.33F = P (F/P, i, n) = $1000 (F/P, .08, 5) = $1469.00 (see EEA p. 576)
Compound InterestCompound InterestCompound InterestCompound Interest
Note: Simple interestyield would be $1400
ISU CCEE
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
Compound InterestCompound InterestCompound InterestCompound Interest
ISU CCEE
Future value, F, for P = $1000 at 8%
Simple vs. Compound Simple vs. Compound InterestInterestSimple vs. Compound Simple vs. Compound InterestInterest
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
ISU CCEE
Specification of Interest Specification of Interest Rate, iRate, iSpecification of Interest Specification of Interest Rate, iRate, i1) “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
ISU CCEE
In-class ExampleIn-class ExampleIn-class ExampleIn-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
ISU CCEE
Four Ways to Repay a Four Ways to Repay a DebtDebtFour Ways to Repay a Four Ways to Repay a DebtDebt
Plan
Principal repaid …
Interest paid …
Trend on interest earned
1in equal annual
installments
on unpaid balance
declines
2at end of
loan
on unpaid balance constant
3in equal annual
installments
declines at increasing
rate
4at end of
loan
at end of loan
(compounded)
increases at increasing
rate
ISU CCEE
Loan Repayment Plan 1 Loan Repayment Plan 1 ($5000, 5 ($5000, 5 years, 8%)years, 8%)Loan Repayment Plan 1 Loan Repayment Plan 1 ($5000, 5 ($5000, 5 years, 8%)years, 8%)
Year
Owed at
begin-ning
Annual Interes
t
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 $6200Bank loans with yearly payments: Principal
repaid in equal installments
ISU CCEE
Loan Repayment Plan 2 Loan Repayment Plan 2 ($5000, 5 ($5000, 5 years, 8%)years, 8%)Loan Repayment Plan 2 Loan Repayment Plan 2 ($5000, 5 ($5000, 5 years, 8%)years, 8%)
Year
Owed at
begin-ning
Annual Interes
t
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 $7000Interest only loans:
used in bonds and international loans
ISU CCEE
Loan Repayment Plan 3 Loan Repayment Plan 3 ($5000, 5 ($5000, 5 years, 8%)years, 8%)Loan Repayment Plan 3 Loan Repayment Plan 3 ($5000, 5 ($5000, 5 years, 8%)years, 8%)
Year
Owed at
begin-ning
Annual Interes
t
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 $6260Equal annual installments:
Auto/home loans (but usually monthly payments)
ISU CCEE
Loan Repayment Plan 4 Loan Repayment Plan 4 ($5000, 5 ($5000, 5 years, 8%)years, 8%)Loan Repayment Plan 4 Loan Repayment Plan 4 ($5000, 5 ($5000, 5 years, 8%)years, 8%)
Year
Owed at
begin-ning
Annual Interes
t
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 $7347Interest and principal repaid at end of loan:
Certificates of deposit (CD’s) and IRA’s
ISU CCEE
Loan repayment plans 1-4Loan repayment plans 1-4Loan repayment plans 1-4Loan 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”
ISU CCEE
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)
EquivalenceEquivalenceEquivalenceEquivalence
ISU CCEE
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
EquivalenceEquivalenceEquivalenceEquivalence
ISU CCEE
You borrow $8000 to help pay for You borrow $8000 to help pay for senior year at ISU – the bank offers senior year at ISU – the bank offers you two repayment plans for paying you two repayment plans for paying off the loan in four years:off the loan in four years:
You borrow $8000 to help pay for You borrow $8000 to help pay for senior year at ISU – the bank offers senior year at ISU – the bank offers you two repayment plans for paying you two repayment plans for paying off the loan in four years:off the loan in four years:
YearPlan 1
paymentPlan 2
payment
1 $0 $0
2 $0 $3300
3 $0 $3300
4 $10,400 $3300
Total $10,400 $9900
Which would you choose?
ISU CCEE
“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 13) Determine a single equivalent value at
the same selected time for Plan 24) Compare the two values
Comparing Economic Comparing Economic AlternativesAlternativesComparing Economic Comparing Economic AlternativesAlternatives
ISU CCEE
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
Technique of EquivalenceTechnique of EquivalenceTechnique of EquivalenceTechnique of Equivalence
ISU CCEE
Technique of EquivalenceTechnique of EquivalenceTechnique of EquivalenceTechnique of Equivalence
YearPlan 1
paymentPlan 2
payment
1 $0 $0
2 $0 $3300
3 $0 $3300
4 $10,400 $3300
Total $10,400 $9900
To compare payment plans, “move” all amounts to now (Present Worth)
or to 4 years hence (Future Worth)
ISU CCEE
Technique of Equivalence – Technique of Equivalence – Present WorthPresent WorthTechnique of Equivalence – Technique of Equivalence – Present WorthPresent WorthAssume our “time value of money” is 7% APR
00
11 22 33 44
P4 = 10,400(1 + .07)-4 = 7,934.11
Total PW = $7,934.11
Or P = $10,400 (P/F, .07, 4) = .7629 (10,400) = $7934.16
$10,4$10,40000
$10,4$10,40000
Present Worth:Payment Plan 1
P = F(1 + i)-n
PW?PW?
ISU CCEE
Technique of Equivalence – Technique of Equivalence – Present WorthPresent WorthTechnique of Equivalence – Technique of Equivalence – Present WorthPresent WorthAssume our “time
value of money” is 7% APR
0011 22 33 44
$330$33000
$330$33000P2 = 3300(1 + .07)-2 = 2882.35
P3 = 3300(1 + .07)-3 = 2693.78P4 = 3300(1 + .07)-4 = 2517.55Total PW = $8093.68
$330$33000
$330$33000
$330$33000
$330$33000
PW?PW?
Present Worth:Payment Plan 1
P = F(1 + i)-n
ISU CCEE
Technique of Equivalence – Technique of Equivalence – Present WorthPresent WorthTechnique of Equivalence – Technique of Equivalence – Present WorthPresent WorthWhich would you choose?
0011 22 33 44
$330$33000
$330$33000
$330$33000
$330$33000
$330$33000
$330$33000
$8093.68$8093.68
00
11 22 33 44
$10,4$10,40000
$10,4$10,40000
$7934.11$7934.11
Plan 2
Plan 1
ISU CCEE
Technique of Equivalence – Future Technique of Equivalence – Future WorthWorthTechnique of Equivalence – Future Technique of Equivalence – Future WorthWorthAssume our “time value of money” is 7% APR
00
11 22 33 44
F4 = 10,400(1 + .07)0 = 10,400
Total FW = $10,400.00$10,4$10,4
0000$10,4$10,4
0000
Future Worth:Payment Plan 1
F = P(1 + i)n
ISU CCEE
Technique of Equivalence – Technique of Equivalence – Future WorthFuture WorthTechnique of Equivalence – Technique of Equivalence – Future WorthFuture WorthAssume our “time value of money” is 7% APR
0011 22 33 44
$330$33000
$330$33000
F2 = 3300(1 + .07)2 = 3778.17F3 = 3300(1 + .07)1 = 3531.00F4 = 3300(1 + .07)0 = 3300.00Total FW = $10609.17
Again, larger than Plan 1
$330$33000
$330$33000
$330$33000
$330$33000
Future Worth:Payment Plan 2
P = F(1 + i)-n
FW?FW?
ISU CCEE
In-class ExampleIn-class ExampleIn-class ExampleIn-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
ISU CCEE
In-class ExampleIn-class ExampleIn-class ExampleIn-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 CFD’s for each plan and compare.
ISU CCEE
How to compare two How to compare two schemes with different livesschemes with different livesHow to compare two How to compare two schemes with different livesschemes 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.
ISU CCEE
Technique of Equivalence – Technique of Equivalence – Present WorthPresent WorthTechnique of Equivalence – Technique of Equivalence – Present WorthPresent WorthBased on dollars only, which is the better plan?
00
1010
$1.7 $1.7 MM
$1.7 $1.7 MM
$2.254 M (if provision is made for final replacement, $ 2.283M)$2.254 M (if provision is made for final replacement, $ 2.283M)
Plan 12020
3030
4040
5050
6060
$1.7 $1.7 MM
$1.7 $1.7 MM
$1.7 $1.7 MM
$1.7 $1.7 MM
repeats
repeats
P = $ 1.7 + $ 1.7 (1.07)-20 + $ 1.7 (1.07)-40 [ + $ 1.7 (1.07)-60]*
[$1.7 M]*[$1.7 M]*[$1.7 M]*[$1.7 M]*
•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.
ISU CCEE
Technique of Equivalence – Technique of Equivalence – Present WorthPresent WorthTechnique of Equivalence – Technique of Equivalence – Present WorthPresent WorthBased on dollars only, which is the better plan?
00
1010
$2.1 $2.1 MM
$2.1 $2.1 MM
$2.376 M$2.376 M
Plan 2
2020 3030 4040 5050 6060
$2.1 $2.1 MM
$2.1 $2.1 MM
repeats
repeats
Other option was $2.254 M
Not Not requiredrequired
Not Not requiredrequired
= P = 2.1 + 2.1(1.07)-30 + the 60y replacement, which does not apply