ne 364 engineering economy - cloudecampus.org · money − time value capital refers to wealth in...

NE 364

Engineering EconomyLecture 3

Money-Time Relationships and Equivalence

(Part 1: Single Payment)

1

Time

F

P

NE 364 Engineering Economy

The objective of this lecture is to

explain the time value of money.

2NE 364 Engineering Economy

Money − Time Value

Capital refers to wealth in the form of money or

property that can be used to produce more wealth.

Engineering economy studies involve the

commitment of capital for extended periods of time.

A dollar today is worth more than a dollar one or

more years from now.


Return to capital or Interest

Interests and profit are payments for the risk the investor takes in letting another use their capital.

Any project or venture must provide a sufficient return to be financially attractive to the suppliers of money or property.


Simple InterestWhen interest earned or charged is not accumulated.

I = P * N * i

P = principal amount lent or borrowed

N = number of interest periods (e.g., years, months,…)

i = interest rate per interest period

The total amount repaid at the end of N interest periods is

F = P + I.


Example on Simple Interest

If $5,000 were loaned for five years at a simple interest

rate of 7% per year, the interest earned would be

I = $5,000 * 5 * 0.07 = $1,750

So, the total amount repaid at the end of five years

would be

F = P + I = $5000 + $1,750 = $6,750

NE 364 Engineering Economy 6

Compound InterestWhenever the interest charge for any interest period is

based on the

remaining principal amount + any accumulated interest

charges

up to the beginning of that period, the interest is said to

be compound.


Compound Interest Calculations

for $1,000 at 10%

Period

(1)

Amount owed

at beginning of

period

(2)=(1)x10%

Interest

amount for

period

(3)=(1)+(2)

Amount owed

at end of

period

1 $1,000 $100 $1,100

2 $1,100 $110 $1,210

3 $1,210 $121 $1,331

8

Compound interest is commonly used in personal and professional financial

transactions.


Simple versus Compound Interest


Cash-flow Diagram

10

Horizontal Line is the Time Scale

Arrow Down=

Expenses

Arrow Up=

Receipts

The Cash-flow is dependent on the point of view


Example on Cash-flow Before evaluating the economic merits of a proposed

investment, the XYZ Corporation insists that its engineers develop a cash-flow diagram of the proposal.

An investment of $10,000 can be made

that will produce uniform annual revenue of $5,310 for five years and

then have a market (recovery) value of $2,000 at the end of year (EOY) five.

Annual expenses will be $3,000 at the end of each year for operating and maintaining the project.

Draw a cash-flow diagram for the five-year life of the project.

Use the corporation's viewpoint.


Example on Cash-flow

(Solution)


Single Payment


Cash-flow Calculation Rules

• Rule 1: Cash flows cannot be added or

subtracted unless they occur at the same time.

• Rule 2: To move a cash flow forward in time by

one time unit, multiply the magnitude of the cash

flow by (1 + i), where i is the interest rate.

• Rule 3: to move a cash flow backward in time by

one time unit, divide the magnitude of the cash

flow by (1 + i).


Arithmetic on Cash flow

0 1 2 3 4 5 6

Time (years)i%

P

Multiply by (1+i) Compounding

Divide by (1+i) Discounting



0 1 2 3 4 5 6

Time (years)

P*(1+i)



i%



0 1 2 3 4 5 6

Time (years)

P*(1+i)2



i%



0 1 2 3 4 5 6

Time (years)

P*(1+i)3



i%


Single Payment (cont.) If an amount of P dollars is invested at a point in time

with interest rate i%, the amount will grow to a future

amount :

after period one

P+Pi =P(1+i)

after period two

P(1+i)+P(1+i)i=P(1+i)[1+i]=P(1+i)2


Single Payment (cont.) After period 3

P(1+i)2

+ P(1+i)2

i= P(1+i)2

[1+i]= P(1+i)3

After period N:

P(1+i)N-1

+ P(1+i)N-1

i= P(1+i)N-1

[1+i]= P(1+i)N

Hence ,

F=P(1+i)N


We can apply compound interest formulas

to “move” cash flows along the cash flow

diagram.


Example on Compound Interest

Estimating the Future

You decide to invest $2,500 at the bank. The bank

offers 8% yearly interest rate.

How much will you have in six years?

F=$2,500 * (1+ 0.08)6 = $3,967.19



Estimating the Present

An investor (owner) has an option to purchase a tract of

land that will be worth $10,000 in six years.

If the value of the land increases at 8% each year, how

much should the investor be willing to pay now for this

property?

P = $10,000 * (1+ 0.08)− 6 = $6301.70


We can use the

discrete compounding tables

to compute F or P


Interest Tables




using Discrete Compounding Tables




F=$2,500 * (F/P, 8%, 6)


Interest Tables








F=$2,500 * (F/P, 8%, 6)

= $2,500 * 1.5869

=$3,967.25









property?

P = $10,000 * (P/F, 8%, 6)









property?

P = $10,000 * (P/F, 8%, 6)

= $10,000 * 0.6302

=$6302.00


Thank you


ne 364 engineering economy - cloudecampus.org · money − time value capital refers to wealth in...

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