COE ME Module 1 ECO - Lecture notes 1 PDF

Preview

Summary

President Ramon Magsaysay State UniversityCOLLEGE OF ENGINEERINGDepartment of Mechanical EngineeringBES 02: Engineering Economics2 nd Semester of A. 2020-MODULE NO. 01I. IntroductionThe need for engineering economy is primarily motivated by the work that engineers doin performing analyses, synthesiz...

Description

President Ramon Magsaysay State University

COLLEGE OF ENGINEERING Department of Mechanical Engineering BES 02: Engineering Economics 2nd Semester of A.Y. 2020-2021 MODULE NO. 01 I.

Introduction

The need for engineering economy is primarily motivated by the work that engineers do in performing analyses, synthesizing, and coming to a conclusion as they work on projects of all sizes. In other words, engineering economy is at the heart of making decisions. These decisions involve the fundamental elements of cash flows of money, time, and interest rates. This module introduces the basic concepts and terminology necessary for an engineer to combine these three essential elements in organized, mathematically correct ways to solve problems that will lead to better decisions. II.

Rationale

Understanding and applying economic principles to engineering have never been more important. Engineering is more than a problem-solving activity focusing on the development of products, systems, and processes to satisfy a need or demand. Beyond function and performance, solutions must also be viable economically. Design decisions affect limited resources such as time, material, labor, capital, and natural resources, not only initially (during conceptual design) but also through the remaining phases of the life cycle. A great solution can die a certain death if it is not profitable. III.

Intended Learning Outcomes

The students should be able to understand and apply fundamental concepts and use the terminology of engineering economics. IV.

Activity

For additional references you may visit the following sites: • Homemade Math (youtube.com) • ilectureonline (yotube.com) V.

Discussion

These are the following topics: ▪ Introduction ▪ Money-Time Relationships and Equivalence VI.

Exercise

(You may refer to succeeding pages for sample problems).

Course Module No. 01 BES 02 – Engineering Economics Page 2 College of Engineering

VII. Assessment (You may refer to succeeding pages for practice problems) Deadline/ submission will be determined later. VIII. Reflection What is the most important concept in engineering economy? Explain. IX.

Resources and Additional Resources

▪ ▪

Sta Maria, Hipolito., Engineering Economy Third Edition. Philippines: National Bookstore Sullivan, William G. (2015). Engineering Economy 16th Edition. New Jersey: Pearson Publishing

Additional Resources: Blank, Tarquin (2012). Engineering Economy 7th Edition. New York: McGraw-Hill

FEMA 2 – Fundamental Engineering Mathematics II Course Module No. 01 Department of Computer Engineering Page 3 INTRODUCTION The mathematics of business and finance is based on the time value of money. The analysis and evaluation of the factors that will affect the economic success of any engineering project is the very essence of Engineering Economy . The best used of borrowed capital is the fundamental concept behind all investments. ❖ INTEREST Interest is a fee paid for the use of money. Interest can be simple or compound. In simple interest, the interest is based on the principal only ; while the compound interest is calculated on the principal plus the interest per interest period . I.

SIMPLE INTEREST

𝑰 = 𝑷𝒊𝒏 𝑭=𝑷+𝑰 𝑭 = 𝑷(𝟏 + 𝒊) II.

where: I = interest

𝒅 𝟑𝟔𝟎

where:

𝒏=

𝟑𝟔𝟎

where:

𝑵𝑹 𝒎

where:

; d = no. of days invested (borrowed)

𝒅

for ordinary year;

𝒏=

𝒅

𝟑𝟔𝟔

for leap year

(𝟏 + 𝒊)𝒏 is the single payment compound amount factor (𝟏 + 𝒊)−𝒏 is the single payment present worth factor

i = is the rate of interest per interest period r = is the nominal rate of interest m = is the number of compounding periods per year

modes of compounding Annually Semi-annually Quarterly Bi-monthly Monthly Semi-monthly Daily

m 1 2 4 6 12 24 360

EFFECTIVE RATE (ER)

𝑬𝑹 = (𝟏 + 𝒊) 𝒎 − 𝟏 VII.

𝒏=

NOMINAL RATE OF INTEREST (NR)

𝒊=

VI.

where:

COMPOUND INTEREST

𝑭 = 𝑷(𝟏 + 𝒊)𝒏 𝑷 = 𝑭(𝟏 + 𝒊)−𝒏 V.

i = interest rate

EXACT SIMPLE INTEREST

𝑰 = 𝑷𝒊𝒏 IV.

F = future worth

ORDINARY SIMPLE INTEREST

𝑰 = 𝑷𝒊𝒏 III.

P = principal

; for two or more nominal rates to be equal, their effective rates must be equal.

CONTINUOUS COMPOUNDING

𝑭 = 𝑷𝒆𝒓𝒏 𝑷 = 𝑭𝒆−𝒓𝒏

where: F = future worth

P = present worth

r = nominal interest rate

n = no. of interest period

Course Module No. 01 BES 02 – Engineering Economics Page 4 College of Engineering

VIII. DISCOUNT Discount refers to an interest charge that is collected at the beginning of term of a loan. ➢

Discount (D) = Future worth - Present worth; 𝑫

➢

Rate of discount (d) ; 𝒅

➢

Relation between rate of discount (d) and rate of interest (i);

= 𝟏 − (𝟏 + 𝒊) −𝟏

=𝑭−𝑷 or

𝒅=

𝒅𝒊𝒔𝒄𝒐𝒖𝒏𝒕

𝒊=

𝒑𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝒅

𝟏−𝒅

IX. ANNUITIES Annuity is the series of equal payment occurring at equal interval of time. This includes anything involving fixed weekly, monthly, or yearly payments. ➢

Ordinary Annuity is one where the payments are made at the end of each period. o

Present worth (P) of annuity; 𝟏−(𝟏+𝒊)−𝒏

𝑷 = 𝑨[

o

𝒊

]

Future worth (F) of annuity; (𝟏+𝒊)𝒏 −𝟏

𝑭 = 𝑨[

➢

𝒊

]

Annuity Due is one where the payments are made at the beginning of each period .

FEMA 2 – Fundamental Engineering Mathematics II Course Module No. 01 Department of Computer Engineering Page 5 ➢

Deferred Annuity is one where the first payment is made several periods after the beginning of the annuity.

𝟏 − (𝟏 + 𝒊)−𝒏 ] (𝟏 + 𝒊) −𝒎 𝑷 = 𝑨[ 𝒊

➢

Perpetuity is an annuity in which the payment continues indefinitely. 𝟏−(𝟏+𝒊)−𝒏

𝑷 = 𝑨[

]

𝒊

as n → ∞; 𝟏−𝟎

(𝟏 + 𝒊)−∞→ 0 then; 𝑷 = 𝑨 [

𝑷=

X.

𝑨

𝒊

] or

𝒊

UNIFORM ARITHMETIC GRADIENT 𝑮

(𝟏+𝒊)𝒏 −𝟏

𝑷𝑮 = [ 𝒊 ] [

𝑮

𝒊

− 𝒏] [

(𝟏+𝒊)𝒏 −𝟏 − 𝒊

𝑭𝑮 = [ ] [ 𝒊

𝟏

(𝟏+𝒊)𝒏

𝒏]

]

Course Module No. 01 BES 02 – Engineering Economics Page 6 College of Engineering

𝟏−(𝟏+𝒊)−𝒏

𝑷𝑨 = 𝑨 [

𝒊

]

This is the Present worth of A

(𝟏+𝒊)𝒏 −𝟏

𝑭𝑨 = 𝑨 [

𝒊

]

This is the Future worth of A

The total accumulated amount F:

𝑭 = 𝑭𝑨 + 𝑭𝑮 ;

𝑮

(𝟏+𝒊)𝒏 −𝟏

𝑭 = 𝑨[

(𝟏+𝒊)𝒏 −𝟏

] + [ ][ 𝒊

𝒊

𝒊

− 𝒏]

The Present Worth P can be solved:

𝑷 = 𝑭(𝟏 + 𝒊)−𝒏 or 𝑷 = 𝑷𝑨 + 𝑷𝑮 ; 𝟏−(𝟏+𝒊)−𝒏

𝑷 = 𝑨[

𝒊

𝑮

(𝟏+𝒊)𝒏 −𝟏

] + [ ][ 𝒊

𝒊

− 𝒏] [

𝟏

(𝟏+𝒊)𝒏

]

FEMA 2 – Fundamental Engineering Mathematics II Course Module No. 01 Department of Computer Engineering Page 7 XI.

GEOMETRIC GRADIENT

𝑷=[

o

CASE I - If z ≠ 1 ( r ≠ 1 ) :

o

CASE II - If ( r = i ) : then z = 1

The Future accumulated amount F:

where:

𝑮

] [ 𝟏−𝒛 ] 𝟏+𝒊

𝑷=

𝟏−𝒛𝒏

𝑮(𝒏)

where

𝒛=

𝟏+𝒊

𝑭 = 𝑷(𝟏 + 𝒊)𝒏

i = interest rate r = common ration/ rate of increase/ decrease in disbursement P = equivalent present worth F = future worth

𝟏+𝒓 𝟏+𝒊

Course Module No. 01 BES 02 – Engineering Economics Page 8 College of Engineering

SAMPLE PROBLEMS 1.

If P5000.00 shall accumulate for 10 years at 8% compounded quarterly. Find the compounded interest at the end of 10 years. 𝑭 = 𝑷(𝟏 + 𝒊)𝒏 = 𝟓𝟎𝟎𝟎 (𝟏 +

𝟎. 𝟎𝟖 𝟏𝟎∗𝟒 ) = 𝟏𝟏 𝟎𝟒𝟎 𝟒

𝑭=𝑷+𝑰

𝟏𝟏 𝟎𝟒𝟎 = 𝟓𝟎𝟎𝟎 + 𝑰 𝑰 = 𝑷 𝟔 𝟎𝟒𝟎. 𝟎𝟎

2.

How much must be deposited at 6% each year beginning on Jan. 1, year 1, in order to accumulate P5000.00 on the date of the last deposit, Jan 1, year 6? 𝑭 = 𝑨[

( 𝟏 + 𝒊) 𝒏 − 𝟏 ] 𝒊

(𝟏 + 𝟎. 𝟎𝟔)𝟔 − 𝟏 ] 𝟓𝟎𝟎𝟎 = 𝑨 [ 𝟎. 𝟎𝟔 𝑨 = 𝑷 𝟕𝟏𝟕. 𝟎𝟎

3.

An instructor plans to retire in exactly one year and want an account that will pay him P25,000.00 a year for the next 15 years. Assuming a 6% annual effective interest rate, what is the amount he would need to deposit now? (The fund will be depleted after 15 years) 𝑷 = 𝑨[

4.

𝟏 − (𝟏 + 𝟎. 𝟎𝟔)−𝟏𝟓 𝟏 − (𝟏 + 𝒊)−𝒏 ] = 𝟐𝟓 𝟎𝟎𝟎 [ ]= 𝒊 𝟎. 𝟎𝟔

𝑷 𝟐𝟒𝟐 𝟖𝟎𝟔. 𝟎𝟎

P1,500.00 was deposited in a bank account 20 years ago. Today it is worth P3,000.00. Interest is paid semi-annually. Determine the interest rate paid on this account? 𝑭 = 𝑷(𝟏 + 𝒊)𝒏

𝒊 𝟐∗𝟐𝟎 𝟑 𝟎𝟎𝟎 = 𝟏 𝟓𝟎𝟎 (𝟏 + ) 𝟐 𝒊 = 𝟑. 𝟒𝟗%

5.

If the nominal interest rate is 3%, how much is P5000.00 worth in 10 years in a continuously compounded account? 𝑭 = 𝑷𝒆𝒓𝒏 = 𝟓 𝟎𝟎𝟎 𝒆 𝟎.𝟎𝟑∗𝟏𝟎 =

𝑷 𝟔 𝟕𝟓𝟎. 𝟎𝟎

FEMA 2 – Fundamental Engineering Mathematics II Course Module No. 01 Department of Computer Engineering Page 9 6.

John borrowed P50,000.00 from the bank at 25% compounded semi-annually. What is the corresponding effective rate of interest? 𝟎. 𝟐𝟓 𝟐 𝑬𝑹𝑰 = [𝟏 + ] −𝟏= 𝟐𝟔. 𝟓𝟔% 𝟐

7.

Find the present worth of a future payment of P300,000 to be made in 5 years with an interest rate of 8% per annum. 𝑷 = 𝑭(𝟏 + 𝒊)−𝒏 = 𝟑𝟎𝟎 𝟎𝟎𝟎 (𝟏 + 𝟎. 𝟎𝟖)−𝟓 =

8.

A man borrowed P2000 from a bank and promises to pay the amount for one year. He received only the amount of P1920.00 after the bank collected an advance interest of P80.00. What was the rate of interest that the bank collected in advance? 𝒅= 𝒊=

9.

𝑷 𝟐𝟎𝟒 𝟏𝟕𝟒. 𝟎𝟎

𝒊𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝟖𝟎 = 𝟒% = 𝒐𝒓𝒊𝒈𝒊𝒏𝒂𝒍 𝒑𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝟐 𝟎𝟎𝟎

𝒊𝒏𝒕𝒆𝒓𝒆𝒔𝒕 𝟖𝟎 = = 𝒂𝒄𝒕𝒖𝒂𝒍 𝒑𝒓𝒊𝒏𝒄𝒊𝒑𝒂𝒍 𝟏 𝟗𝟐𝟎

𝟒. 𝟏𝟕%

An engine is expected to have a maintenance cost of P1000 the first year. It is believed that the maintenance cost will increase P500 per year. The interest rate is 6% compounded annually. Over a 10-year period, what will be the approximate effective annual maintenance cost? 𝟏 − (𝟏 + 𝒊)−𝒏 𝟏 𝑮 (𝟏 + 𝒊)𝒏 − 𝟏 𝑷 = 𝑨[ − 𝒏] [ ] +[ ][ ] = 𝑷 𝟐𝟐 𝟏𝟔𝟏. 𝟎𝟎 𝒊 𝒊 𝒊 (𝟏 + 𝒊) 𝒏 𝑷 = 𝑨[

𝟏 − (𝟏 + 𝒊)−𝒏 ] 𝒊

𝟏 − (𝟏 + 𝟎. 𝟎𝟔)−𝟏𝟎 ] 𝟐𝟐 𝟏𝟔𝟏 = 𝑨 [ 𝟎. 𝟎𝟔 𝑨 = 𝑷 𝟑 𝟎𝟏𝟏. 𝟎𝟎

Course Module No. 01 BES 02 – Engineering Economics Page 10 College of Engineering

PRACTICE PROBLEMS 1.

What is the annual rate of interest if P265 is earned in four months on an investment of P15, 000?

2.

A loan of P2, 000 is made for a period of 13 months, from January 1 to January 31 the following year, at a simple interest of 20%. What is the future amount is due at the end of the loan period?

3.

If you borrow money from your friend with simple interest of 12%, find the present worth of P20, 000, which is due at the end of nine months.

4.

Determine the exact simple interest on P5, 000 for the period from Jan.15 to Nov.28, 1992, if the rate of interest is 22%.

5.

A man wishes his son to receive P200, 000 ten years from now. What amount should he invest if it will earn interest of 10% compounded annually during the first 5 years and 12% compounded quarterly during the next 5 years?

6.

By the condition of a will the sum of P25, 000 is left to be held in trust by her guardian until it amounts to P45, 000. When will the girl receive the money if the fund is invested at 8% compounded quarterly?

7.

At a certain interest rate compounded semiannually P5, 000 will amount to P20, 000 after 10 years. What is the amount at the end of 15 years?

8.

Jones Corporation borrowed P9, 000 from Brown Corporation on Jan. 1, 1978 and P12, 000 on Jan. 1, 1980. Jones Corporation made a partial payment of P7, 000 on Jan. 1, 1981. It was agreed that the balance of the loan would be amortizes by two payments one of Jan. 1, 1982 and the other on Jan. 1, 1983, the second being 50%larger than the first. If the interest rate is 12%. What is the amount of each payment?

9.

A woman borrowed P3, 000 to be paid after 1.5 years with interest at 12% compounded semiannually and P5, 000 to be paid after 3 years at 12% compounded monthly. What single payment must she pay after 3.5 years at an interest rate of 16% compounded quarterly to settle the two obligations?

10. Mr. J. de la Cruz borrowed money from a bank. He received from the bank P1, 342 and promises to repay P1, 500 at the

end of 9 months. Determine the simple interest rate and the corresponding discount rate or often referred to as the “Banker’s discount.” 11. A man deposits P50, 000 in a bank account at 6% compounded monthly for 5 years. If the inflation rate of 6.5% per year

continues for this period, will this effectively protect the purchasing power of the original principal? 12. What is the future worth of P600 deposited at the end of every month for 4 years if the interest rate is 12% compounded

quarterly? 13. What is the future worth of P600 deposited at the end of every month for 4 years if the interest is 12% compounded

quarterly? 14. Mr. Reyes borrows P600, 000 at 12% compounded annually, agreeing to repay the loan in 15 equal annual payments. How

much of the original principal is still unpaid after he has made the 8th payment? 15. M purchased a small lot in a subdivision, paying P200, 000 down and promising to pay P15, 000 every 3 months for the next

10 years. The seller figured interest at 12% compounded quarterly. (a) What was the cash price of the lot? (b) If M missed the first 12 payments, what must he pay at the time the 13th is due to bring him up to date? (c) After making 8 payments, M wished to discharge his remaining indebtedness by a single payment at the time when the 9th regular payment was due, what must he pay in addition to the regular payment then due? (d) If M missed the first 10 payments, what must he pay when the 11th payment is due to discharge his entire indebtedness? 16. A man approaches the ABC Loan Agency for P100, 000 to be paid in 24 monthly installments. The agency advertises an

interest rate of 1.5% per month. They proceed to calculate the amount of his monthly payment in the following manner. 17. A new office building was constructed 5 years ago by a consulting engineering firm. At that time the firm obtained the bank

loan for P 10,000,000 with a 20% annual interest rate, compounded quarterly. The terms of the loan called for equal quarterly payments for a 10-year period with the right of prepayment any time without penalty. Due to internal changes in the firm, it is now proposed to refinance the loan through an insurance company. The new loan is planned for a 20- year term with an interest rate of 24% per annum, compounded quarterly. The insurance company has a onetime service charge 5% of the balance. This new loan also calls for equal quarterly payments. a.) What is the balance due on the original mortgage

FEMA 2 – Fundamental Engineering Mathematics II Course Module No. 01 Department of Computer Engineering Page 11 (principal) if all payments have been made through a full five years? b.) What will be the difference between the equal quarterly payments in the existing arrangement and the revised proposal? 18. An asphalt road requires no upkeep until the end of 2 years when P60, 000 will be needed for repairs. After this P90, 000

will be needed for repairs at the end of each year for the next 5 years, then P120, 000 at the end of each year for the next 5 years. If money is worth 14% compounded annually, what was the equivalent uniform annual cost for the 12-year period? 19. A man wishes to provide a fund for his retirement such that from his 60th to 70th birthdays he will be able to withdraw equal

sums of P18, 000 for his yearly expenses. He invests equal amount for his 41st to 59th birthdays in a fund earning 10% compounded annually. How much should each of these amounts be? 20. Determine the present worth and the accumulated amount of an annuity consisting of 6 payments of P120, 000 each, the

payment is made at the beginning of each year. Money is worth 15% compounded annually. 21. Calculate the capitalized cost of a project that has an initial cost of P3, 000,000 and an additional cost of P100, 000 at the

end of every 10 yrs. The annual operating costs will be P100, 000 at the end of every year for the first 4 years and P160, 000 thereafter. In addition, there is expected to be recurring major rework cost of P300, 000 every 13 yrs. Assume i =15%. 22. The will of a wealthy philanthropist left P5, 000,000 to establish a perpetual charitable foundation. The foundation trustees

decided to spend P1, 200,000 to provide facilities immediately and to provide P100, 000 of capital replacement at the end of each 5-year period. If the invested funds earned 12% per annum, what would be the year end amount available in perpetuity from the endowment for charitable purposes? 23. The surface area of a certain plant requires painting is 8,000 sq. ft. Two kinds of paint are available whose brands are A and

B. Paint A cost P 1.40 per sq. ft. but needs renewal at the end of 4 yrs., while paint B cost P 1.80 per sq. ft. If money is worth 12% effective, how often should paint B be renewed so that it will be economical as point A? 24. A contract has been signed to lease a building at P20,000 per year with an annual increase of P1,500 for 8 years. Payments

are to be made at the end of each year, starting one year from now. The prevailing interest rate is 7%. What lump sum paid today would be equivalent to the 8-year lease payment plan?...