Hello! Welcome back to our study of engineering economics.
In our last lesson, we saw how the Annual Worth (AW) method allows us to fairly compare alternatives with unequal but finite lives, like deciding between on-premise servers and a cloud subscription. This naturally leads to a new question: what if a project or service is intended to last forever?
This is where today's topic, capitalized cost, comes in. Your learning outcome for this lesson is to calculate capitalized cost for projects or services with a perpetual lifespan. Capitalized cost (CC) is simply the present worth of a project with an infinite life. This concept is essential for evaluating long-term infrastructure, permanent endowments, or even core software platforms that a company intends to maintain indefinitely.
Understanding Perpetual Cash Flows
Imagine you have a sum of money, the capitalized cost CC, invested at an interest rate i. Each year, this investment generates interest equal to CC * i. If you withdraw only the interest amount each year, you can do so forever without depleting the original principal. This annual withdrawal amount is the equivalent uniform annual worth, A.
This gives us the relationship:
By rearranging this, we get the fundamental formula for capitalized cost:
This elegant formula states that the present worth of a perpetual series of uniform annual payments A is simply A divided by the interest rate i.
The image below provides a great visual breakdown of the cash flows involved in a perpetual project.

As the diagram shows, a perpetual project typically involves:
- An initial, non-recurring cost (like the
FCor an initial investment). - Recurring costs, which can be uniform annual costs (
A) or large periodic costs that repeat over a cycle (RC).
Our task is to combine all these into a single present value—the capitalized cost.
How to Calculate Capitalized Cost
To handle these different types of cash flows, we can use a straightforward, five-step procedure.
First, let's establish the procedure by reading a brief overview.
The following text from the textbook 'LEVEL TWO' introduces capitalized cost and outlines a clear, five-step method for its calculation. Focus on understanding the distinction it makes between 'recurring' and 'nonrecurring' costs.
Please read the section titled '5.5 CAPITALIZED COST CALCULATION AND ANALYSIS'. You can stop when you see the numbered list of the five steps. This will formally introduce the concept and the calculation framework.
As you just read, the procedure is:
- Draw a cash flow diagram showing all non-recurring costs and at least two cycles of all recurring costs.
- Find the present worth of all non-recurring amounts.
- Find the equivalent uniform annual worth (A) for one life cycle of all recurring amounts.
- Calculate the capitalized cost of the recurring portion by dividing the total annual worth from step 3 by the interest rate
i. - Add the present worth from step 2 and the capitalized cost from step 4 to get the total capitalized cost.
Applied Example: The Total Cost of a Software System
This procedure is best understood through an example. Given your background, let's analyze a scenario that's highly relevant to the tech industry: determining the total long-term cost of a new software system.
Now, let's apply the five-step procedure to a practical example. The following text analyzes the capitalized cost of a new software system with an indefinite life, accounting for initial costs, future one-time costs, changing maintenance fees, and periodic upgrades.
Please read Example 5.4 about the Marin County software system. Follow how it breaks down the different costs and applies the 5-step method. Pay close attention to how they handle the non-recurring costs (initial purchase + future cost) versus the recurring costs (maintenance + upgrades).
Let's summarize the logic from that example (using i = 5%):
- Step 1 (Cash Flow Diagram): The diagram in the text helps visualize the different cash flows and their timing.
- Step 2 (PW of Non-recurring Costs):
- Initial Cost: -$150,000 (already at present, year 0).
- Additional Cost in 10 years: -$50,000. Its PW is
-$50,000(P/F, 5%, 10). - The total PW of these one-time costs is
CC1 = -$180,695.
- Step 3 (AW of Recurring Costs):
- Upgrade Cost: -$15,000 every 13 years. This is converted to an annual cost using the A/F factor:
A1 = -$15,000(A/F, 5%, 13) = -$847. - Maintenance Cost: -$5,000 per year for years 1-4, then -$8,000 thereafter. The example cleverly handles this as a perpetual cost of -$5,000 per year, plus an additional perpetual cost of -$3,000 starting in year 5.
- Upgrade Cost: -$15,000 every 13 years. This is converted to an annual cost using the A/F factor:
- Step 4 (CC of Recurring Costs):
- The total annual worth of all recurring costs is calculated, and then divided by
ito find its capitalized cost.
- The total annual worth of all recurring costs is calculated, and then divided by
- Step 5 (Total CC):
- The PW of non-recurring costs is added to the CC of all recurring costs to get the final answer:
CCT = -$346,997.
- The PW of non-recurring costs is added to the CC of all recurring costs to get the final answer:
This means the county has committed to an expense that is equivalent to spending nearly $347,000 today. Alternatively, A = CCT * i = $346,997 * 0.05 = $17,350 per year, forever.
Comparing Finite vs. Infinite Life Alternatives
A powerful application of capitalized cost is comparing an alternative with a finite life to one with a perpetual life. This directly builds on our last lesson.
How does it work?
- For the infinite-life alternative, calculate its capitalized cost directly.
- For the finite-life alternative, you assume you will replace it with an identical replacement at the end of each life cycle, forever.
- First, calculate its Annual Worth (AW) over a single life cycle.
- Then, use the formula
CC = AW / ito find the capitalized cost of running that cycle in perpetuity.
This converts both alternatives to a common, infinite time horizon for a fair comparison.
Test your understanding!
A municipality needs a small bridge. It can choose between two options. The interest rate is 8%.
-
Option A (Steel Bridge): A permanent steel structure.
- Initial Cost: $900,000
- Annual Maintenance: $25,000
-
Option B (Wooden Bridge): A treated timber bridge.
- Initial Cost: $200,000
- Annual Maintenance: $40,000
- Life: 25 years
- Salvage Value: $0
Which bridge is the more economical choice based on capitalized cost?
(Factor: (A/P, 8%, 25) = 0.09368)
Show answer
1. Calculate the CC for Option A (Infinite Life):
The costs are an initial cost (P) and a perpetual annual cost (A).
The capitalized cost of the steel bridge is $1,212,500.
2. Calculate the CC for Option B (Finite Life):
First, find the Annual Worth (AW) for one 25-year cycle.
Now, find the capitalized cost of this perpetual annual cost.
The capitalized cost of the wooden bridge is $734,200.
Conclusion:
Comparing the two, CC_B ($734,200) is significantly lower than CC_A ($1,212,500). Therefore, repeatedly building the wooden bridge is the more economical choice.
Conclusion
Today, we've extended our analytical toolkit to handle projects with perpetual lifespans. This is a crucial concept for making decisions about long-term investments where the service or asset is needed indefinitely.
- Key Takeaways:
- Capitalized Cost (CC) is the present worth of a project or service that lasts forever.
- The fundamental relationship is
CC = A / i, whereAis the equivalent uniform annual worth of all recurring cash flows. - To calculate CC for a project, you find the present worth of all non-recurring costs and add it to the capitalized cost (
A/i) of all recurring costs. - This method allows for the comparison of alternatives with different lifespans, including finite vs. infinite, by converting both to an infinite time horizon.
Preview of the Next Lesson
We have now examined how to compare alternatives using both Annual Worth (for unequal, finite lives) and Capitalized Cost (for infinite lives). Our next lesson will address a related and very practical question: when is the optimal time to replace an existing asset? We will introduce the concepts of defender and challenger assets and begin our exploration of replacement analysis.