IENG112
Notes #11
Engineering Economy
1. Time Diagram of Investments: Shows the net cash in-flow/out-flow.
 Salvage value: The value of the equipment at the end of its use.
 Costs in time diagram: 1. Investment, 2. Yearly installment, 3. Operation, 4.
Maintenance.
2. Net Present Value: What is the value of a sum S if it is obtained n years from now on and the
inflation rate is i?
𝑆
(1 + 𝑖)𝑛
The underlying equation is
𝑁𝑃𝑉(1 + 𝑖)𝑛 = 𝑆.
The factor (1 + 𝑖)𝑡 is called single-payment compound amount factor.
3. Examples for Interest Factors
Pages 341 and 342. In all examples the interest rate is 12 percent.
3.1. The value of an investment after n years
I = the invested amount
F = future value
𝐹 = 𝐼(1 + 𝑖)𝑛 = 1000 × (1 + 𝑖)5 = 1000 × 1.762 = 1762.
(1) 150 × 1.1210 = 465.90.
3.2. What is the sum which will have value F after n years? (2)
𝐼(1 + 𝑖)𝑛 = 𝐹
𝐼=
𝐹
(1+𝑖)𝑛
=
10,000
(1+𝑖)15
= 1827.
3.3. What is the total value at the end of n years if at the end of each year I is invested?
(3)
The value of the investment of the first year:
𝐼 × (1 + 𝑖)𝑛−1
The value of the investment of the second year: 𝐼 × (1 + 𝑖)𝑛−2
...
The value of the investment of the last year:
𝐼
Total value
𝑡
𝐼 × (1 + 𝑖)𝑛−1 + 𝐼 × (1 + 𝑖)𝑛−1 + ⋯ + 𝐼 = 𝐼 ∑𝑛−1
𝑡=0 (1 + 𝑖) = 𝐼
(1+𝑖)𝑛 −1
𝑖
1.128 − 1
100
= 1230.
0.12
3.4. What is the yearly installment of loan if it is paid back in n years? (4)
L = loan
y = yearly installment
The still unpaid loan at the end of the first year:
𝐿(1 + 𝑖) − 𝑦.
The still unpaid loan at the end of the second year:
(𝐿(1 + 𝑖) − 𝑦)(1 + 𝑖) − 𝑦 = 𝐿(1 + 𝑖)2 − 𝑦((1 + 𝑖) + 1).
...
The still unpaid loan at the end of the last year:
𝑡
𝐿(1 + 𝑖)𝑛 − 𝑦((1 + 𝑖)𝑛−1 + … + (1! 𝑖) + 1) = 𝐿(1 + 𝑖)𝑛 − 𝑦 ∑𝑛−1
𝑡=0 (1 + 𝑖)
(1 + 𝑖)𝑛 − 1
= 𝐿(1 + 𝑖)𝑛 − 𝑦
= 0.
𝑖
1.1210 − 1
10
1500 × 1.12 = 𝑦
0.12
y = 265.5.
3.5. How much should be saved at the end of each year of a period of ne years if the
future value is F? (5)
The saved value at the end of the first year: y.
The saved value at the end of the first year: y(1+i)+y = y(2+i).
...
The saved value at the end of the last year:
(1 + 𝑖)𝑛 − 1
𝐹 = 𝑦(1 + 𝑖)𝑛−1 + (1 + 𝑖)𝑛−2 + ⋯ + (1 + 𝑖) + 1 = 𝑦
.
𝑖
(1.12)9 − 1
2000 = 𝑦
0.12
𝑦 = 135.4
3.6. How much is the loan if the yearly installment is y in a period of n years? (6)
Exactly the same equations as in 3.4, however the unknown is L and not y.
y=100
n=5
𝐿= 𝑦
(1 + 𝑖)𝑛 − 1
(1.125 − 1
=
100
= 360.48
𝑖(1 + 𝑖)𝑛
0.12 × 1.125
4. Comparison of investment options
n=5
Alternative 1. No investment, yearly nominal cost is 10,000 USD.
Nominal value is the value which is not adjusted according to the inflation.
Alternative 2. 30,000 USD investment and yearly nominal cost 2000 USD. No salvage
value.
Alternative 3. 40,000 USD investment and yearly nominal cost 1000 USD. The salvage
value at the end of year 5 is 8000 USD.
Net Present Values of the alternatives.
Alternative 1.
∑40 1.12𝑡
1
1
1
1
1
10000 (
+
+
+
+
)
=
10000
1.121 1.122 1.123 1.124 1.125
1.125
1.125 − 1
= 10000
= 36,048.
0.12 × 1.125
Alternative 2.
1.125 − 1
30000 + 2000
= 37,209.6.
0.12 × 1.125
Alternative 3.
40000 + 1000
1.125 − 1
1
− 8000
= 38,498.2.
5
0.12 × 1.12
1.125