Module 5 Lecture 3

ENGINEERING ECONOMY

Ma. Cecilia C. Carlos

Rene D. Estember

Fundamentals of Engineering Economy

• Introduction

• Money-Time Relationships and Equivalence

Introduction:

• Engineering Economy is the systematic evaluation of the economic merits of proposed solutions to engineering problems.

• Economically acceptable solutions to engineering problems must:– Demonstrate positive balance of long-term benefits over

long-term costs– Promote the well-being and survival of the organization– Embody creative and innovative technology– Permit identification and scrutiny of estimated outcomes– Translate profitability to the “bottom line”

Money-Time Equivalence

• Capital refers to wealth in the form of money or property that can be used to produce more wealth. Equity capital – investment made by individuals in a business in the

hope of receiving a profit. Debt or Borrowed capital – money obtained from lenders for

investment. Interest and profit are payments for the risk of the investor

• Money has time value. A peso today does not have the same worth two years from today because of the interest it can earn.

• Simple interest – total interest is linearly proportional to the principal, interest rate and interest periods (I=Prt)

• Compound interest – interest based on the remaining principal plus any accumulated interest charges up to the beginning of that period.

Concept of Equivalence

• The concept of equivalence is used to compare alternatives by reducing them to an equivalent basis that is dependent on interest rate, amount of money involved, timing of receipts/expenses and manner in which the interest on invested capital is paid and the initial investment recovered

• NotationsI = effective interest rate per interest periodN = number of compounding periodsP = present sum of money; equivalent value of cashflows at a

reference point in time called the presentF = future sum of money; equivalent value of cashflows at a reference

point in time called the futureA = end-of-period cashflows in a uniform series continuing for a

specified number of periods, starting at the end of the first period and continuing through the last period.

• Time scale – horizontal line with progression moving from left to right

Time Scale

F

A or inflows

0 1 2 3 ……. N

P

Formulae

• Single CashflowF = P(1+i)N P = F(1+i)-N

• Annuity (Uniform Series)

F = A[(1+i)N – 1] ; F = A(F/A, i%, N)

i

P = A (1+i)N – 1 ; P = A (P/A, i%, N)

i(1+i)N

• Uniform Gradient N-1

F = G/i [(1+i)k] – NG/i

k=0

A = G [1/i – N/{(1 + i)N – 1}]

P = G/i [(1+i)N – 1- Ni]/ i (1+i)N

Applications of Money-Time Relationships

Present Worth Method (PW) – all cash flows are discounted to the present time at a desired interest rate

Future Worth Method (FW) – equivalent worth of all cash flows at the end of a planning horizon at a desired interest rate.

Annual Worth Method (AW) – equal annual amounts for a given period that is equivalent to the cash flows at a desired interest rate; it is the annual equivalent revenues minus expenses less the annual equivalent capital amount recovery

Internal Rate of Return Method (IRR) – solves for the interest rate that equates the equivalent worth of an alternative’s cash inflows to the equivalent worth of cash outflows.

Payback Method – number of years required for cash inflows to equal cash outflows.

Sample Problems 

1.     An investment of P150,000 will yield the following cash returns:

 

Year 1 P60,000 Year 3 80,000

Year 2 70,000 Year 4 85,000

Year 5 90,000

 The investment is on an equipment with no scrap value at the end of five years. Desired rate is 10%. Determine the following:

a. Accounting rate of return

b. Payback period

c. Net or excess present value

. d. Time adjusted rate of return

2.     A proposed product modification to avoid production difficulties will require an immediate expenditure of P140,000 to modify certain dies. What annual savings must be realized to recover this expenditure in five years with interest at 15% per year?

Sample Problems

3.       It is estimated that a certain piece of equipment can save $6,000 per year in labor and material costs. The equipment has an expected life of five years and no salvage value. If the company must earn a 15% annual return on such investments, how much could be justified now for the purchase of this piece of equipment?

4.      A debt of P20,000 was paid for as follows: P1,500 at the end of 3 months, P3,000 at the end of 6 months, P4,500 at the end of 9 months, P4,500 at the end of 15 months, P3,000 at the end of 18 months, and a final payment F at the end of 21 months. If the rate of interest was 15% compounded quarterly, find the final payment F.

 5.     Equal end-of-year payments of $263.80 each are being made on a $1,000 loan at 10% effective interest per year. How many payments are required to repay the entire loan? Immediately after the third payment, what lump-sum amount would completely pay off the loan?