Transportation Asset Management
Asset management is the operation, maintenance, upgrade, and disposal of assets in a cost-effective manner. Transportation asset management is the deployment of sound financial and economic principles and processes for effective operation, maintenance, upgrade, and expansion of transportation infrastructure facilities throughout their life cycles.
The strategy and process for asset management often answer questions as listed below.
■ What is the current state or condition of an asset?
■ What is the desired service and performance condition?
■ What are investment approaches available to maintain the desired service and performance condition?
■ What is the best investment strategy?
Basic Engineering Economic Principle
The fundamental concept of engineering economics is that money generates money. $1,000 today is not the same as $1,000 one year later. This concept can be illustrated by depositing $1,000 in a savings account now with a 3% annual interest. One year from now, the current $1,000 is worth $1,003-
Interest Rate
An interest rate is an amount earned or charged, expressed as a percentage of the principal, for the use of the principle for a period of time. For lenders, the higher the interest rate is, the bigger the return their investments will get. For borrowers, the higher the interest rate is, the more expensive their loans are.
There are two types of interest, simple and compound. In the case of the simple interest, the only earning or cost of using a principle is the original principle. For example, $100 is invested in a 5% simple interest account
(e.g., a certificate of deposit account). At the end of the first year, $5.00 interest is earned. For the second year, it is still the original $100, which makes the 5% simple interest.
where
F is the future value P is the present value+ i_{s} is the simple interest rate n is the length of time
For the above example, the total amount at the end of two years is F= 100+ 100X
5%x2= 100+10 = $110.
As opposed to simple interest, compound interest earns interest on both the original principle and the accrued interest. With the above example, if the account is a savings account with a compound 5% interest rate, at the end of the first year, $5 interest is earned based on the $100 principle. At the end of the second year, both the original $100 plus the $5 interest earned in year one will earn interest. So, at the end of the second year, interest earned is $5.25 vs. the otherwise $5-00 in the simple interest case.
where
F is the future value P is the present value i is the compound interest rate n is the length of time
For the above example, the total amount at the end of the two-year time is
F= 100(1 +5%)^{2} = $110.25
A compound interest rate is often specified on an annual basis, known as the
effective annual percentage rate (APR).
If an interest rate is compounded more than once in a year, it can be converted to an effective APR. For example, a 2% monthly compound interest is equivalent to an APR of 26.82%.
TYPES OF INTERESTED RATE
- • Simple interest
- • Compound interest
- • Annual percentage rate (APR)
- • Conversion of others to APR where
APR is the effective annual percentage rate i= interest rate for one compounding period w=the number of compounding periods in a year.
With the above example, i=2% with a compounding period of one month, m= 12 compounding periods (1 year= 12 months), so the APR=(1 +2%)^{12} —1 =26.82%.
Time Value of Money
Present Value: The concept of present value is to determine the current worth of a future amount of money. It is often used to compare different investment scenarios.
Scenario studies such as determining which investment option is worth the most or least (e.g., $100 today, $110 one year from now, $1,100 ten years from now, and $10 per month for next year) are often carried out with the present value procedure.
Present Worth (P): equivalent amount at present when the time is zero (?= 0). P/(F, i, ri) — convert a future value F to a present value P knowing F, i, and n.
where
A is the future value P is the present value i is the compound interest rate n is the length of time
For example, $1,000 (F) promised at the end of three years (ri) with an APR of 5%
- (/) is worth P = —--т = $863-84 now.
- (1 + 5%)^{3}
PI(A, i, ri) — convert equal annual value A to present value P knowing the equal annual A, i, and n.
where
P is the present value i is the compound interest rate
n is the length of time
A is the annual amount to be saved or borrowed For example, a payment of $100 every year (A) for 15 years («) with an APR of 3%
- 100Г(1 + 3%)'^{5}-11
- (/) is equivalent to P =---гг—- = $1,194.86 now.
- 4 3%(l + 3%)^{15}
Future Value: Future value is to convert all past or present values into future values where scenarios can be compared and ranked. The above comparison example covering $100 today, $110 one year from now, $1,100 ten years from now, $10 per month for next year can also be analyzed with the future worth procedure. Future Worth (F): equivalent future amount at a time of year n (t=n).
F/(P, i, n) — convert current value P to future value F knowing P, i, and n.
For example, a deposit of $100 now in an APR of 3% savings account for 15 years is worth 100(1 +3%)^{15}=$155.80 in the future.
FI (A, i, n) — convert equal annual value A to future value /"’knowing A, i, and n.
For example, a deposit of $100 every year into an APR of 3% savings account for
ioo[(i + 3%)^{15}-i]
- 15 years is worth-= $1,860 at the end of the 15th year.
- 3%
Annual Worth or Cost: The concept of annual worth or cost is to convert present or future worthiness into an annualized term.
A/(P, i, ri) — convert present value P to equal annual return A knowing P, i, and n.
For example, with a 3% APR, an investment worth $1,000 now will enable an annual withdraw of $117-18 for each of the next ten years.
Another example is the estimation of monthly payment (equal payment) for a five-year $15,000 loan with an APR of 6%.
First, convert the 6% APR to a compound monthly average interest rate (i) by using the formula APR = (1 + /)" — 1. The period for the interest is one month.
There are 12months (periods) in a year. So, 6% = (1+/)^{12} — 1. /= 0.4866%. Total number of monthly payments (n) is 5 years X 12 payment/year=60 payment.
The equal monthly payment (A) is:
AI(F, i, n) — convert future value P to equal annual return A knowing P, i, and n.
Equivalence
Tlie ability to compare and contrast the values of different investment alternatives is critical for decision-making. Creating a common benchmark value offers such possibilities. The annual return or annual cost of an alternative is often used as a benchmark value for such comparison.
For alternative analysis about investment and return, three scenarios listed below are often encountered.
a. A single investment right now delivers “n” years of acceptable service.
For example, a full-depth roadway resurfacing project costing $6.2 million now can provide 24 years of acceptable ride quality service.
b. A single investment right now plus a uniform annual investment delivers “k” years of acceptable service.
For example, a full-depth roadway resurfacing project costing $6.1 million now plus an annual maintenance investment of $45,000 can provide 32 years of acceptable ride quality services.
c. A single investment right now delivers “m” years of service + another single investment at the end of “m" years gaining additional “x” years of service.
For example, a full-depth roadway resurfacing project costing $5-1 million now can provide 20 years of acceptable service. At the end of the 20 years, another $2.2 million (present value) of major rehabilitation work can extend the adequate ride quality service for another 14 years.
To determine which alternative offers the best return will require the values of “и,” “k,” “m” and “x” in addition to an investment interest rate.
Tlie commonly adopted method for such analysis relies on the life cycle “annual cost” concept. With the above example, the life cycle for Alternative a is
years. It is “m” years for Alternative b. And it is “(m+х)” years for Alternative c (Table 10.2).
Table 10.2 Illustration of Alternative Analysis Based on Annualized Life Cycle Cost
Parameters |
Alternative a |
Alternative b |
Alternative c |
Service life (life cycle - years) |
24 |
32 |
(20+ 14) = 34 |
Interest rate (APR) |
3.6% |
3.6% |
3.6% |
Value ($) |
$6.2M (present) |
$45,000 (present annual (A) maintenance) + $6.1 M (present) |
$5.1 M (present) + $2.2M (present) |
Formula |
|||
Application |
|||
Annual cost |
$390,031 |
$369,114 |
$427,125 |
The above analysis shows that investment Alternative 2 is the most economically efficient way and has the lowest cost among all alternatives.
The objective of transportation asset management is to minimize the life cycle cost for managing and maintaining transportation assets such as bridges, tunnels, roads, and other road features. By adopting sound engineering in concert with robust engineering economic analysis, a long-lasting and efficiently invested highway system can be built, operated, and maintained.