time value of money – part ii economic basis to evaluate engineering projects

Time Value of Money – Part II

Economic basis to evaluate engineering

projects

Time Management

The expected time to deliver this module is 50 minutes. 20 minutes are reserved for team practices and exercises and 30 minutes for lecture.

Learning Objective

After this class the students should be able to:

Describe the concept of equivalence as applied to cash flow diagrams;

Understand the difference between Compound and Simple interest

Define Present Value Define Future Value

Warm-up (Reflection)

If the monthly inflation rate expected for this year is 0.5%. Which will be annual inflation rate expected?

Each team is invited to answer this question and list the reasons that justify their respective answers.(5 minutes)

Warm-up (Refletion)

If $100 is deposited in a saving account that pays 5% annual interest, what amount has accumulated by the end of the eight year?

Each team is invited to answer this question and list the reasons that justify their respective answers.(5 minutes)

Simple vs. Compound interest

engineering economy use Compound interest.

Terminology: i: interest rate per yearN: number of years P: initial deposit, or Present ValueF: future value after N years

Simple interest

With simple interest, the interest calculated for years 2, 3,... is based on the initial deposit. There is no interest computed on the accrued interest.

F = P(1+ Ni)

… N5

P

0 1 2

F1 = P+Pxi

F2 = P+Pxi+ Pxi =P(1+2xi)

FN = P(1 + Nxi)

Compound Interest The standard assumption is that interest

is computed on the current balance which includes accrued interest that has not yet been paid.

F = P(1 + i)N

FN = P(1 + i)N

… N5

P

0 1 2

F1 = P+Pxi = P(1 + i)

F2 = P(1 + i)+ P(1 + i)xi = P(1 + i)2

Simple and Compound interest

Figure 1

Simple and Compound interest

Year Interest Future Value Interest Future Value0 5.00 100.00 5.00 100.001 5.25 105.00 5.00 105.002 5.51 110.25 5.00 110.003 5.79 115.76 5.00 115.004 6.08 121.55 5.00 120.005 6.38 127.63 5.00 125.006 6.70 134.01 5.00 130.007 7.04 140.71 5.00 135.008 7.39 147.75 5.00 140.009 7.76 155.13 5.00 145.00

10 8.14 162.89 5.00 150.0011 8.55 171.03 5.00 155.0012 8.98 179.59 5.00 160.0013 9.43 188.56 5.00 165.0014 9.90 197.99 5.00 170.0015 10.39 207.89 5.00 175.0016 10.91 218.29 5.00 180.0017 11.46 229.20 5.00 185.0018 12.03 240.66 5.00 190.0019 12.63 252.70 5.00 195.0020 265.33 200.00

Compound Interest Simple Interest

Table 2

Simple and Compound interest

The graph (figure 1 e table 2) illustrates the difference between simple and compound interest. This difference increases as the interest rate and/or the number of years increases

I if simple interest were used, it would result in a linear increase in the amount in the bank. While, if Compound interest is used it would result in a exponential increase.

Purchasing an Equipment

Ernie's Earthmoving is considering the purchase of a piece of heavy equipment. What is the cash flow diagram if the following cash flows are anticipated? (each team has 5 minutes)

First Cost $120KO&M Cost $30K per yearOverhaul Cost $35K per yearSalvage value $40K after 5 years

Purchasing an equipment cash flow diagram

$120K

0 1 2 3 4 5

$40.00

$30K $30K $30K

$30K + $35K

Figure 4

Leasing Equipment

Rather than purchasing the heavy equipment, Ernie's Earthmoving is planning on leasing it. The lease payments will be $25K per year. What is the cash flow diagram?(each team has 5 minutes)

Leasing equipment cash flow diagram

$25K

0 1 2 3 4 5

$30K

$30K + $25K

$30K

$25K + $30K + $35K

$30K

Depositor

Figure 5

Equivalence for four loans

Equivalence is adjusting for the time value of money. Equivalence means that different cash flows at different times are equal in economic value at a given interest rate.

$100 at year 0 and the $127.63 at year 5 were equivalent at 5 % (see figure 1) .

Figure 6 illustrates four ways to repay an initial loan of $1000 at 10% interest over 5 years. The business loan, bond, and consumer loan patterns are labeled according to their typical use. The constant principal payment pattern is rarely used, because the early payments are more burdensome than in the other patterns.

Cash flow for four loans

0 1 2 3 4 5

$1610.5127.63

$1000

1. Business Loan

0 1 2 3 4 5

$1000 +$100

$1000

2. Bond

$100 $100 $100 $100

0 1 2 3 4 5

$1000

3. Consumer Loan

$263 $263 $263 $263 $263

0 1 2 3 4 5

$1000

4. Consumer Principal Payment

$300

$280 $260

$240 $220

Figure 6

Equivalent Present Value.

The compound interest formula F = P(I + i)N, can be applied to each loan payment. The

payment occurs later so it is the F, Future Value

This P is calculated for each payment by dividing F by (I + i).

This P is the amount of initial borrowing that could be repaid by that F.

The final column, Present worth, or PV, is the amount of initial borrowing that could be repaid by each payment. Each present worth amount is equivalent at 10% to a payment.

Each equivalent amount is called a present worth, because each is at the initial point of the loan, or time 0. If the present worths of all of the payments are summed, the total is $ 1000, which was the initial loan amount.

Payments to repay $1000 at 10%

Year Payment (I + i)Year Present Worth

1 $263.80 1.1 $239.82 2 263.797 1.21 218.013 263.797 1.331 198.194 263.797 1.4641 180.185 263.797 1.6105 163.8

Total $ 1000.00

Present worth for consumer loan

Exercises (10 minutes) Draw the cash flow diagram if

Strong Metalworking borrows $100,000, to be repaid in 5 years. The interest rate is 11%, and no payments are made before the end of year 5. Compute the present worth of the principal and interest payments at an interest rate of

a. 7%. (Answer: ‑$120AK) b. 11%. c. 15%.

Exercises (10 minutes) Draw the cash flow diagram if

Strong Metalworking issues a $100,000 bond that pays 11% annual interest. The bond is repaid in 5 years. Compute the present worth of the principal and interest payments at an interest rate of

a. 7%. (Answer: ‑$116.4K) b. 11%. c. 15%.

Exercises (10 minutes) Draw the cash flow diagram if Strong

Metalworking repays a $100,000 loan with 5 annual payments of $27,057. Construct an interest table to show that this repays the loan at an interest rate of 11%. Compute the present worth of the principal and interest payments in at an interest rate of

a. 7%. (Answer: ‑$110.94K) b. 11%. c. 15%.

Reference

“Engineering Economic: Appling Theory to Practice”Ted G. EschenbachOxford University Press2002Part I