Hello! Welcome back to your course on Engineering Economics.
In our last lesson, we learned how to convert a project's entire stream of cash flows into a single, equivalent uniform annual worth (AW). We saw that AW and Present Worth (PW) are two sides of the same coin, representing a project's value in different but equivalent terms.
Today, we'll leverage this new tool to solve one of the most common challenges in project comparison. Your learning outcome for this lesson is to compare alternatives with unequal useful lives using the annual worth method. Imagine choosing between a server that lasts 3 years and a more expensive one that lasts 5. How do you make a fair comparison? The AW method provides an elegant and powerful solution.
The Challenge of Unequal Lives
When you use Present Worth (PW) analysis, you are comparing the total value of projects over a defined study period. If two projects have the same lifespan, a direct comparison of their PW is fair and straightforward.
But what if the lifespans are different? Comparing the PW of a 3-year project to the PW of a 5-year project is like comparing the cost of a two-night hotel stay to a three-night stay—it's an apples-to-oranges comparison. The 5-year project provides two additional years of service, a benefit not captured if you only look at the initial project durations.
To make a valid comparison, a fundamental principle must be followed: All alternatives must be compared over the same time span.
There are two primary ways to do this:
- Least Common Multiple (LCM) Method with PW: You can find the least common multiple of the project lives and calculate the PW of the repeated projects over this extended period. For 3- and 5-year projects, you'd do a 15-year analysis. This is valid but can become incredibly cumbersome, especially for lives like 7 and 9 years (a 63-year analysis!).
- Annual Worth (AW) Method: This is the much simpler and more direct approach, and the one we will focus on today.
The Power of Annual Worth for Unequal Lives
The Annual Worth method elegantly solves the unequal lives problem by relying on a key assumption: the service provided by the asset is needed for the long term, and at the end of its life, the asset will be replaced by another one with similar costs and performance.
This is called the repeatability assumption.
When we calculate the AW of a project over its single life cycle, we are finding the equivalent annual cost for that cycle. Because we assume the cycle repeats, this annual cost is valid for any number of future cycles. The AW of a 3-year server is its equivalent annual cost whether you need it for 3, 6, 9, or 15 years.
This allows us to calculate the AW for each alternative over its own unique life and then directly compare the resulting annual values. The alternative with the most favorable AW (highest for revenue projects, lowest cost for service projects) is the best choice, regardless of the difference in their lifespans.
1.011 Textbook Resources, Comparing costs and benefits
This MIT textbook resource briefly lists several ways to handle unequal lives. Notice that it explicitly recommends using the AW method and clarifies the underlying assumption.
Please read the list under the heading 'If Project Lives Are Different'. Pay special attention to the bullet point that recommends using the AW method. The text in parentheses explains the core 'repeatability' assumption we just discussed.
Applying the AW Method: A Practical Example
Let's apply this to a scenario you might encounter: choosing between two server infrastructure options.
Your company needs a new database server. The MARR is 10%.
-
Alternative A (Standard Server):
- Initial Cost: $20,000
- Annual Maintenance: $3,000
- Useful Life: 3 years
- Salvage Value: $2,000
-
Alternative B (High-Performance Server):
- Initial Cost: $30,000
- Annual Maintenance: $1,500
- Useful Life: 5 years
- Salvage Value: $5,000
We simply calculate the Annual Worth (or Equivalent Uniform Annual Cost, EUAC) for each option over its own life.
Calculation for Alternative A (n=3, i=10%):
- Factors: (A/P, 10%, 3) = 0.40211, (A/F, 10%, 3) = 0.30211
- AW_A =
-[P(A/P) - S(A/F)] - Annual Cost - AW_A =
-[20000(0.40211) - 2000(0.30211)] - 3000 - AW_A =
-[8042.20 - 604.22] - 3000 - AW_A =
-7437.98 - 3000= -$10,437.98
Calculation for Alternative B (n=5, i=10%):
- Factors: (A/P, 10%, 5) = 0.26380, (A/F, 10%, 5) = 0.16380
- AW_B =
-[P(A/P) - S(A/F)] - Annual Cost - AW_B =
-[30000(0.26380) - 5000(0.16380)] - 1500 - AW_B =
-[7914 - 819] - 1500 - AW_B =
-7095 - 1500= -$8,595.00
Decision:
The equivalent annual cost of Alternative A is $10,438, while for Alternative B it is $8,595. Since we want to minimize cost, Alternative B is the better economic choice.
The AW method allowed us to make a direct, fair comparison without the complexity of a 15-year PW analysis.
Test your understanding!
Your team is choosing between two CRM software licensing models. Your company's MARR is 15%.
- Model 1 (Long-Term): An upfront fee of $50,000 for a 4-year license. There are no annual fees.
- Model 2 (Short-Term): An upfront fee of $35,000 for a 2-year license. There are no annual fees.
Which model has the lower equivalent annual cost?
(Factors: (A/P, 15%, 4) = 0.35027; (A/P, 15%, 2) = 0.61512)
Show answer
For cost-only alternatives with no salvage value, the AW is simply the annualized initial cost.
- AW (Model 1):
-$50,000 * (A/P, 15%, 4) = -$50,000 * 0.35027 = -$17,513.50 - AW (Model 2):
-$35,000 * (A/P, 15%, 2) = -$35,000 * 0.61512 = -$21,529.20
The equivalent annual cost of Model 1 is significantly lower. Even though its upfront cost is higher, its longer duration makes it the more economical choice on an annual basis.
Proving the Power of AW
To solidify why AW is the superior method here, let's examine an analysis that compares the three approaches: incorrect PW, correct (but complex) PW, and correct (and simple) AW.
Engineering Risk Benefit Analysis
To see a detailed comparison of these methods, let's look at a case study from MIT's 'Engineering Risk Benefit Analysis'. This will show why simply comparing Present Worth over different lifespans is wrong, how complex the correct PW method is, and why Annual Worth is the preferred approach.
Please review slides 39 through 46. Start with slide 40, which sets up a problem comparing three alternatives (A, B, C) with unequal lives of 6, 3, and 4 years. Look at slide 41. This shows the straightforward Annual Equivalent (AE, or AW) calculation for each, concluding that C, with an annual cost of $7,000, is the best. Now, examine slide 46. This demonstrates the common mistake of comparing the Present Worths over their individual lives. This incorrect method suggests B is the best choice, leading to the wrong decision. Finally, look at slide 45. This shows the correct but much more complicated PW analysis over a 12-year common lifespan (the LCM of 3, 4, and 6). This confirms that C is indeed the best option, but notice how much more calculation is required. This three-step comparison powerfully illustrates that the simple AW/AE calculation gives the right answer with far less effort.
This example from the MIT materials is a definitive demonstration. The Annual Worth method is not just a shortcut; it is the most efficient and direct path to the correct answer when comparing alternatives with unequal lives, provided the repeatability assumption holds.
Conclusion
You have now learned the primary application that makes the Annual Worth method so essential in engineering economics. While PW analysis is excellent for many situations, AW shines when project lifespans differ.
- Key Takeaways:
- To compare alternatives fairly, you must evaluate them over an equal time horizon.
- Simply comparing the PW of projects over their different individual lives is a common but serious error that can lead to poor decisions.
- The Annual Worth method provides a direct and simple way to compare alternatives with unequal lives by converting each to an equivalent annual value.
- This method assumes that each alternative can be repeated with the same cash flows in subsequent cycles.
Preview of the Next Lesson
In our next lesson, we will put this new skill to immediate use in a highly relevant context for any tech leader today. Your goal will be to evaluate cloud computing costs vs. on-premise infrastructure using annual worth. This will be a practical application that combines what you've just learned about unequal lives with the build-vs-lease decisions we've previously discussed.