## Highway & Rail Transit Tunnel Maintenance & Rehabilitation Manual

**2005 Edition**

## Appendix A: Life-Cycle Cost Methodology

To properly plan for future repairs or scheduled maintenance in a tunnel, it is beneficial to perform a life-cycle cost analysis of the different options involved for each anticipated major repair to ensure the greatest cost efficiency over the life of the tunnel. This process involves evaluating the alternatives over a given duration or economic life to determine specific costs involved for each option and then equating them through a series of mathematical formulas that enable the costs of each option to be compared at a common point in time. The life-cycle costs of a given alternative include all associated costs over the expected life of the option. In general these costs may include:

**Initial costs**- Engineering or design costs, price of equipment, construction costs, etc.**Operating/energy costs**- Annualized amount to operate (e.g. cost of electricity to run mechanical equipment).**Maintenance costs**- Annualized costs to maintain equipment or repair minor defects.**Rehabilitation costs**- Future expense for known procedure at specified time (e.g. certain type of light bulbs may need to be replaced every five years).**User costs**- Costs associated with impact on the functioning of tunnel (e.g. tunnel may need shut down for repair; therefore, impact to traffic can be shown by applying an annualized cost to each hour tunnel is closed).**Salvage value**- Sale value of equipment at end of expected life (e.g. mechanical fan or railroad tie may be of some value to others even after it has served its purpose in the tunnel).

There are two main methods for performing a life-cycle cost analysis, namely, the present worth and the annualized methods.

### 1. Present Worth Method

As the name implies, this method attempts to bring all of the present and future costs of a given option to present day values. This process should be completed for each major repair/rehabilitation and subsequently the options could be compared. Determining the present worth of a future expense is done by taking into account inflation of the dollar and therefore discounting the amount by a predetermined rate over the period between the future expense and the present time. The present worth of the future expense is also the amount that could be invested today with reinvested interest over the duration to equal the amount of the future expense. An example of a future expense would be the rehabilitation costs mentioned above. The general form of the equation for determining the present worth of a future expense is:

P = F 1 ( 1 + i ) ^{n}where

*P*= Present worth*F*= Future one-time expense*n*= Number of years*i*= Discount rateFuture expenses can also be uniform, in that the same expense occurs at the end of each year. An example of this would be the annualized maintenance costs described previously. The general form of the equation for determining the present worth of an end-of-year expense is:

P = A [ ( 1 + i) ^{n}- 1 ][ i ( 1 + i ) ^{n}]where

*P*= Present worth*A*= End-of-year payments*n*= Number of years*i*= Discount rate### 2. Annualized Method

The annualized method is used to transform present and future costs into a uniform annual expense. This annual expense can be compared to the annual expenses of the other repair/rehabilitation alternatives to determine which one is most cost effective. Converting all future expenses into a present value as before and then using the equation below to convert that value into an annual expense will provide a uniform annual cost.

A = P [ i ( 1 + i ) ^{n}][ (1 + i ) ^{n}- 1 ]where

*A*= End-of-year payments*P*= Present worth*n*= Number of years*i*= Discount rateThe above two methods can also be performed without using the actual equations given. Standard economic tables have been developed that give factors that are based on the discount rate and the economic life under consideration. These factors are also unique to the desired result. The procedure for using standard economic tables is as follows:

- Determine the discount rate
*(i)*and economic life*(n)*to be used for the analysis. It is important to choose an economic life that is equal for the given alternatives if the present worth method is to be used. Otherwise, the annualized method must be used. - Develop a cash flow diagram for each option, which shows all relevant costs described above on a timeline of years in the economic life.
- Take individual costs, whether uniform or one-time, and insert them in the proper formula given below along with the factor from the appropriate economic table.
*(P/F,i%,n)*- or present worth*(P)*given future expense*(F)*at discount rate*(i)*for number of years*(n)*.*(P/A,i%,n)*- or present worth*(P)*given end-of-year payments*(A)*at discount rate*(i)*for number of years*(n)*.*(A/P,i%,n)*- or end-of-year payments*(A)*given present worth*(P)*at discount rate*(i)*for number of years*(n)*.

Caution must be used in determining the appropriate discount rate. Because of the power of compounded interest, a difference in discount rate can actually change the final outcome of the analysis if the repair/rehabilitation options being considered have different arrangements of uniform and onetime costs. According to Peter Kleskovic who wrote

*A Discussion of Discount Rates for Economic Analysis of Pavements*, a draft report for FHWA:"The discount rate can affect the outcome of a life-cycle cost analysis in that certain alternatives may be favored by higher or lower discount rates. High discount rates favor alternatives that stretch out costs over a period of time, since the future costs are discounted in relation to the initial cost. A low discount rate favors high initial cost alternatives since future costs are added in at almost face value. In the case of a discount rate equal to 0, all costs are treated equally regardless of when they occur. Where alternative strategies have similar maintenance, rehabilitation, and operating costs, the discount rate will have a minor effect on the analysis and initial costs will have a larger effect."

The above procedures will allow the most economical repair/rehabilitation alternative to be identified, but as can be expected, the least costly is not always the best. Therefore further comparison can sometimes be utilized to take into account the "human factors" of the alternatives. In his book entitled

*Value Engineering: Practical Applications ... for Design, Construction, Maintenance & Operations*, Alphonse Dell'Isola, P.E., has developed a procedure for weighted evaluation of human factors such as comfort, appearance, performance, and safety along with the economic costs. It is suggested that his procedure or something similar be used if the effects of the human factors are of concern during the economic life of the alternatives.- Determine the discount rate
### 3. Example

The following example uses completely arbitrary costs to properly show the benefits of a life-cycle cost analysis.

Consider a transit tunnel in which the track support system is in need of replacement. Currently the system is ballasted track and can either be replaced with direct fixation slab track or a new, ballasted track system. Costs given to the different options are shown below for every 150 m (500 ft) of track.

#### a) Direct Fixation Slab Track:

Initial Construction Costs $500,000 Joint/Crack Sealing (years 10, 20, 30 and 40) $20,000 Annual Maintenance $1,000 Salvage ($100,000) Estimated Life 50 years #### b) Ballasted Track:

Initial Construction Costs $250,000 Replacement Ties (years 12 and 24) $200,000 Annual Maintenance $20,000 Salvage ($50,000) Estimated Life 35 years #### c) Factors from Standard Economic Table (assume 7 percent discount rate)

n (P/F) (P/A) (A/P) 10 0.5083 7.0236 0.1424 12 0.4440 7.9427 0.1259 20 0.2584 10.5940 0.0944 24 0.1971 11.4693 0.0872 30 0.1314 12.4090 0.0806 35 0.0937 12.9477 0.0772 40 0.0668 13.3317 0.0750 50 0.0339 13.8007 0.0725 #### d) Alternate 1 - Direct Fixation Slab Track

##### (1) Present Worth Method

P = $500,000 + $1,000(P/A,7%,50) + $20,000(P/F,7%,10) + $20,000(P/F,7%,20) + $20,000(P/F,7%,30) + $20,000(P/F,7%,40) - $100,000(P/F,7%,50) P = $500,000 + $1,000(13.8007) + $20,000(0.5083) + $20,000(0.2584) + $20,000(0.1314) + $20,000(0.0668) - $100,000(0.0339) P = $529,709 ##### (2) Annualized Method

A = $500,000(A/P,7%,50) + $1,000 + $20,000(P/F,7%,10)(A/P,7%,50) + $20,000(P/F,7%,20)(A/P,7%,50) + $20,000(P/F,7%,30)(A/P,7%,50) + $20,000(P/F,7%,40)(A/P,7%,50) - $100,000(P/F,7%,50)(A/P,7%,50) A = $500,000(0.0725) + $1,000 + $20,000(0.5083)(0.0725) + $20,000(0.2584)(0.0725) + $20,000(0.1314)(0.0725) + $20,000(0.0668)(0.0725) - $100,000(0.0339)(0.0725) A = $38,403/year

#### e) Alternate 2 - Ballasted Track

##### (1) Present Worth Method

P = $250,000 + $20,000(P/A,7%,35) + $200,000(P/F,7%,12) + $200,000(P/F,7%,24) - $50,000(P/F,7%,35) P = $250,000 + $20,000(12.9477) + $200,000(0.444) + $200,000(0.1971) - $50,000(0.0937) P = $631,689 ##### (2) Annualized Method

A = $250,000(A/P,7%,35) + $20,000 + $200,000(P/F,7%,12)(A/P,7%,35) + $200,000(P/F,7%,24)(A/P,7%,35) - $50,000(P/F,7%,35)(A/P,7%,35) A = $250,000(0.0772) + $20,000 + $200,000(0.444)(0.0772) + $200,000(0.1971)(0.0772) - $50,000(0.0937)(0.0772) A = $48,837/year Since this example used two different time periods the present worth results are not useful in comparing costs, but the annualized method is since the outcome is a cost per year. From this example it can be seen that even though Alternative 2 had a lower initial cost, it turned out to be more expensive over the long term due to greater costs to repair and maintain that alternative. This process can be used in evaluating many different aspects of tunnel maintenance and repairs from structural aspects like the example above to fan model selection for the mechanical ventilation system to which light bulb manufacturer is better over the long term. There are multiple reference materials that could be of assistance if a more detailed analysis is desired.