Time Value of Money – Part II
Economic basis to
evaluate engineering
projects
Time Management

The expected time to deliver this
module is 50 minutes. 20 minutes
are reserved for team practices and
exercises and 30 minutes for
lecture.
Learning Objective
After this class the students should be
able to:
 Describe the concept of equivalence as
applied to cash flow diagrams;
 Understand the difference between
Compound and Simple interest
 Define Present Value
 Define Future Value
Warm-up (Reflection)
If the monthly inflation rate expected
for this year is 0.5%. Which will be
annual inflation rate expected?
Each team is invited to answer this question and list
the reasons that justify their respective answers.(5
minutes)
Warm-up (Refletion)
If $100 is deposited in a saving
account that pays 5% annual
interest, what amount has
accumulated by the end of the
eight year?
 Each team is invited to answer this question and list the
reasons that justify their respective answers.(5 minutes)
Simple vs. Compound interest


engineering economy use
Compound interest.
Terminology:




i: interest rate per year
N: number of years
P: initial deposit, or Present Value
F: future value after N years
Simple interest

With simple interest, the interest
calculated for years 2, 3,... is based on
the initial deposit. There is no interest
computed on the accrued interest.
F1 = P+Pxi
0
1
P

F = P(1+ Ni)
F2 = P+Pxi+ Pxi
=P(1+2xi)
2
FN = P(1 + Nxi)
…
N
5
Compound Interest

The standard assumption is that interest
is computed on the current balance which
includes accrued interest that has not yet
been paid.
F1 = P+Pxi
= P(1 + i)
0
1
P

F = P(1 + i)N
F2 = P(1 + i)+ P(1 + i)xi
= P(1 + i)2
2
FN = P(1 + i)N
…
N
5
Simple and Compound interest
Figure 1
Simple and Compound interest
Compound Interest
Simple Interest
Year Interest Future Value Interest Future Value
0
5.00
100.00
5.00
100.00
1
5.25
105.00
5.00
105.00
2
5.51
110.25
5.00
110.00
3
5.79
115.76
5.00
115.00
4
6.08
121.55
5.00
120.00
5
6.38
127.63
5.00
125.00
6
6.70
134.01
5.00
130.00
7
7.04
140.71
5.00
135.00
8
7.39
147.75
5.00
140.00
9
7.76
155.13
5.00
145.00
10
8.14
162.89
5.00
150.00
11
8.55
171.03
5.00
155.00
12
8.98
179.59
5.00
160.00
13
9.43
188.56
5.00
165.00
14
9.90
197.99
5.00
170.00
15
10.39
207.89
5.00
175.00
16
10.91
218.29
5.00
180.00
17
11.46
229.20
5.00
185.00
18
12.03
240.66
5.00
190.00
19
12.63
252.70
5.00
195.00
20
265.33
200.00
Table 2
Simple and Compound interest


The graph (figure 1 e table 2) illustrates
the difference between simple and
compound interest. This difference
increases as the interest rate and/or the
number of years increases
I if simple interest were used, it would
result in a linear increase in the amount
in the bank. While, if Compound interest
is used it would result in a exponential
increase.
Purchasing an Equipment

Ernie's Earthmoving is considering the
purchase of a piece of heavy equipment.
What is the cash flow diagram if the
following cash flows are anticipated?
(each team has 5 minutes)
First Cost
O&M Cost
Overhaul Cost
Salvage value
$120K
$30K per year
$35K per year
$40K after 5 years
Purchasing an equipment cash flow diagram
$40.00
0
1
$30K
$120K
Figure 4
2
3
$30K
4
$30K
$30K +
$35K
5
Leasing Equipment

Rather than purchasing the heavy
equipment, Ernie's Earthmoving is
planning on leasing it. The lease
payments will be $25K per year.
What is the cash flow diagram?
(each team has 5 minutes)
Leasing equipment cash flow diagram
Depositor
0
$25K
1
$30K
2
3
4
$30K
$30K
$30K +
$25K
$25K +
$30K +
$35K
Figure 5
5
Equivalence for four loans
Equivalence is adjusting for the time
value of money. Equivalence
means that different cash flows at
different times are equal in
economic value at a given interest
rate.
$100 at year 0 and the $127.63 at
year 5 were equivalent at 5 % (see
figure 1) .

Figure 6 illustrates four ways to
repay an initial loan of $1000 at
10% interest over 5 years. The
business loan, bond, and consumer
loan patterns are labeled according
to their typical use. The constant
principal payment pattern is rarely
used, because the early payments
are more burdensome than in the
other patterns.
Cash flow for four loans
1. Business Loan
$1000
0
1
2
3
4
2. Bond
$1610.51
27.63
$1000
0
Figure 6
5
1
2
3
4
$100
$100
$100
$100
5
$1000
+$100
3. Consumer Loan
$1000
0
1
2
3
4
5
$263
$263
$263
$263
$263
3
4
5
$240
$220
4. Consumer Principal Payment
$1000
0
1
2
$260
$280
$300
Equivalent Present Value.
The compound interest formula
F = P(I + i)N,
can be applied to each loan payment. The
payment occurs later so it is the F, Future
Value
 This P is calculated for each payment by
dividing F by (I + i).
 This P is the amount of initial borrowing
that could be repaid by that F.



The final column, Present worth, or PV,
is the amount of initial borrowing that
could be repaid by each payment. Each
present worth amount is equivalent at
10% to a payment.
Each equivalent amount is called a
present worth, because each is at the
initial point of the loan, or time 0. If the
present worths of all of the payments
are summed, the total is $ 1000, which
was the initial loan amount.
Payments to repay $1000 at 10%
Present worth for consumer loan
Year
Payment
1
2
3
4
5
$263.80
263.797
263.797
263.797
263.797
Year
(I + i)
1.1
1.21
1.331
1.4641
1.6105
Present
Worth
$239.82
218.01
198.19
180.18
163.8
Total $ 1000.00
Exercises (10 minutes)




Draw the cash flow diagram if
Strong Metalworking borrows
$100,000, to be repaid in 5 years.
The interest rate is 11%, and no
payments are made before the end
of year 5. Compute the present
worth of the principal and interest
payments at an interest rate of
a. 7%. (Answer: -$120AK)
b. 11%.
c. 15%.
Exercises (10 minutes)




Draw the cash flow diagram if
Strong Metalworking issues a
$100,000 bond that pays 11%
annual interest. The bond is repaid
in 5 years. Compute the present
worth of the principal and interest
payments at an interest rate of
a. 7%. (Answer: -$116.4K)
b. 11%.
c. 15%.
Exercises (10 minutes)




Draw the cash flow diagram if Strong
Metalworking repays a $100,000 loan
with 5 annual payments of $27,057.
Construct an interest table to show that
this repays the loan at an interest rate of
11%. Compute the present worth of the
principal and interest payments in at an
interest rate of
a. 7%. (Answer: -$110.94K)
b. 11%.
c. 15%.
Reference

“Engineering Economic: Appling
Theory to Practice”
Ted G. Eschenbach
Oxford University Press
2002
Part I