SUBJECT: ENGINEERING ECONOMICS TOPIC: AMORTIZATION PLEASE ANSWER THE QUESTION ON THE FIRST PICTURE AND MAKE IT A BASIS ON THE SECOND EQUATION. THANK YOU 4. A loan of P8,000 has an interest of 10%compounded annually is to be paid byequal annual payments over the next 3years. Find the annual payment andconstruct an amortization schedule. m=4A debt of P10000 with interest at the rate of 15% compounded quarterly is to be amortized by 7equal quarterly payments at the end of each months, the first payment is to be made after Iyear. Find the quarterly payment and construct an amortization schedule. i=0-154PERIODPo=P+Iv10 0001254P-(A-1)12V345678910TOTALOUTSTANDINGPRINCIPAL ATBEGINNING OFPERIODA=?A= P= 10764.0625to 764.06 +41000037510 000 +3750 375 x 0:15=10375389.062510 375 +389. 0625 10 764-06 x 0-15+-403.6511167.71484167-71-1424-79743.008159743.008-1478.138264.874956+= 10 0006731.3117675140.2399583489.502957PX1776.863317H[INTERESTTO OBOX0-15403.6523438Pxi418.7893066Pxi365.3628056Pxi252.4241913192.7589984309.93281091843.496130.856360966.63237441-(1)-"0-15/4APAYMENT1843.4961- (1+0)-71843.496PERIOD=101843.496✓1843.496✓1843.4961843.49630 0002904.471692 12904.472•] (151) m4045)-7] (14 015) ³3A-IPRINCIPAL REPAID AT ENDOF PERIOD1843.496-418.7891424.7066931843-496 365-301478.133194A-I1533.5631891591.0718091650.7370021712.6396391776.86362611167.71515= $1843.496

Answered: 4. A loan of P8,000 has an interest of…

4. A loan of P8,000 has an interest of 10% compounded annually is to be paid by equal annual payments over the next 3 years. Find the annual payment and construct an amortization schedule.

ENGR.ECONOMIC ANALYSIS

14th Edition

ISBN:9780190931919

Author:NEWNAN

Publisher:NEWNAN

Chapter1: Making Economics Decisions

Section: Chapter Questions

Problem 1QTC

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Question

SUBJECT: ENGINEERING ECONOMICS
TOPIC: AMORTIZATION

PLEASE ANSWER THE QUESTION ON THE FIRST PICTURE AND MAKE IT A BASIS ON THE SECOND EQUATION. THANK YOU

4. A loan of P8,000 has an interest of 10%
compounded annually is to be paid by
equal annual payments over the next 3
years. Find the annual payment and
construct an amortization schedule.

m=4
A debt of P10000 with interest at the rate of 15% compounded quarterly is to be amortized by 7
equal quarterly payments at the end of each months, the first payment is to be made after I
year. Find the quarterly payment and construct an amortization schedule. i=0-15
4
PERIOD
Po=P+Iv
10 000
1254
P-(A-1)
1
2
V
3
4
5
6
7
8
9
10
TOTAL
OUTSTANDING
PRINCIPAL AT
BEGINNING OF
PERIOD
A=?
A= P
= 10764.0625
to 764.06 +4
10000
375
10 000 +3750 375 x 0:15
=10375
389.0625
10 375 +389. 0625 10 764-06 x 0-15
+-403.65
11167.71484
167-71-1424-7
9743.00815
9743.008-1478.13
8264.874956
+= 10 000
6731.311767
5140.239958
3489.502957
PX
1776.863317
H
[
INTEREST
TO OBOX
0-15
403.6523438
Pxi
418.7893066
Pxi
365.3628056
Pxi
252.4241913
192.7589984
309.93281091843.496
130.8563609
66.6323744
1-(1)-"
0-15/4
A
PAYMENT
1843.496
1- (1+0)-7
1843.496
PERIOD=10
1843.496
✓1843.496
✓1843.496
1843.496
30 000
2904.471692 12904.472
•] (151) m
4045)-7] (14 015) ³3
A-I
PRINCIPAL REPAID AT END
OF PERIOD
1843.496-418.789
1424.706693
1843-496 365-30
1478.133194
A-I
1533.563189
1591.071809
1650.737002
1712.639639
1776.863626
11167.71515
= $1843.496

Application of economic techniques to the evaluation of design and engineering alternatives. The role of engineering economics is to assess the appropriateness of a given project, estimate its value, and justify it from an engineering standpoint.

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