-
8/12/2019 Eng Econ Ass 1
1/4
Katrina Joana Marie G. MagsayoBS EcE-5B
Engr. Leonel PabilonaEngineering EconomyQuiz no. 2
1. The sum of P20,000 was borrowed on January 1, 1974. The terms
for repaying the loan were asfollows: The borrower will make 5
annual payments, the first payment being made on January 1975,and
each payment will be P1000 more than the preceding one. If the
interest rate of the loan is 8%,what was the amount of the first
payment? Verify the solution by constructing an amortization
schedule of the loan.
1 2 3 4 5 F =
20,000 =
A =
A = 3,162.66
Period Outstanding Principal at
beginning of period
Interest due at 8%
per period
Total payment at
end of period
Principal repaid at
end of period
1 20,000 1,600 3,162.66 1,562.66
2 18,437.34 1,474.99 4,162.66 2,687.67
3 15,749.67 1,259.97 5,162.66 3,902.68
4 11,846.98 947.76 6,162.66 5,214.89
5 6,632.09 530.56 7,162.66 6,632.09
TOTAL 72,666.08 5,813.28 25,878.64 20,000
20, 000
A
A + 1,000
A +2,000
A + 3,000
A + 4,000
-
8/12/2019 Eng Econ Ass 1
2/4
2. Pinoy has just offered to make you wealthy! For every peso
that you save in an insured, continuouslycompounded, bank account
during the next 10 years, he will give you a peso to match it.
Because
your modest income permits you to save P3,000 per year for each
of the next 10 years, your unclewill be willing to give you P30,000
at the end of the 10
thyear. If you desire a total of P75,000 ten
year from now, what annual interest rate would you have to earn
on your insured bank account tomake your goal possible?
Solution:
- - -
F = A
+ 30, 000
75, 000 = 3, 000
+ 30, 000
=
15 =
By interpolation:
Assume i = 1% :
= 10.4622
Assume i = 10%:
= 15.9374
=
x = 0.08459 or 8.46%
3, 000
30, 000
-
8/12/2019 Eng Econ Ass 1
3/4
3. An individual makes six annual deposits of P2,000 in a
savings account that pays interest at a rate of4% compounded
annually. Two years after making the last deposit, the interest
rate changes to 7%
compounded annually. Twelve years after the last deposit, the
accumulated money is withdrawnfrom the account. How much is
withdrawn?
Solution:
F = A
= 2,000
= 13, 265.95
In two years after the last deposit
= 13, 265.95 = 14, 348.45
12 years after the last deposit
= 14, 348.45 (1 + 0.07)10
= 28, 225.57 is the accumulated money withdrawn
4. A certain fluidized-bed combustion vessel has an investment
cost P100,000 a life of 10 years andnegligible market (resale)
value. Annual cost of material, maintenance, and electric power for
the
vessel are expected to total P8,000. A major relining of the
combustion vessel will occur during thefifth year at a cost of
P20,000, during this year, the vessel will not be in service. If
the interest rate is
15% per year, what is the lump sum equivalent cost of this
project at the present time?
Solution:
P = 100,000 + 8,000
+ [ ]
= 146, 116.27
5. The heat loss through the exterior walls of a certain poultry
processing plant is estimated to cost theowner P3,000 next year. A
salesman from Superfiber Insulation, Inc., the plant manager, has
toldyou, that he can reduce heat loss by 80% with the installation
of P18,000 worth of Superfiber now. If
the cost of heat loss rises by P200 per year (gradient) after
the next year and the owner plans to keep
the building for 15 more years, what would you recommend if the
interest rate is 10% per year?
Given:G = P200 n = 15
i = 0.10
-
8/12/2019 Eng Econ Ass 1
4/4
Solution:
Total heat loss without insulation:
=
+ A
= + A = P8, 030.40 + P22, 818.24
= P 30, 848.64
With installed Superfiber:
= P18,000 + (1-0.80)(P30,848.64)
= P24, 169.73
Profit when installing Superfiber:= P30, 848.64P24, 169.73= P6,
678.91
Therefore, I recommend installing Superfiber.
6. Find the value of the unknown quantity Z in the following
diagram such that the equivalent cashoutflow equals equivalent
inflow when r= 20% compounded continuously:
Solution:P4 = A (P/A, r%, N)
= 500
= 1,427. 65
Present value of year 3
P3 = F (P/F, r%, N)
= 1,427.54 (= 1,168.77
Therefore,
Z = 1,168.77
= 1, 421.65
