Last Updated on January 25, 2026 by Maged kamel
A solved problem for EAW/ Equivalent Annual Worth.
From our last post, we have considered a machine that costs $20,000. The salvage value is $4000, with revenue of $5000 per year for the next five years, while the operating and maintenance costs are $500 per year for the next five years.
We have converted the Initial investment and the salvage to annual Cost or CR and added the Cost of operation and maintenance. The sum is $5120.76, and the Revenue value is $5000. We have concluded that this investment will lead to a loss.
What is the relation between EAB, EAC, and EAW?
The next slide contains three expressions. We will introduce definitions for EAB, EAC, and EAW. The first item, EAB (sometimes written EUAB), stands for equivalent uniform annual benefits.
EUAB is the revenue you earn from operating a machine or the income you earn due to cash inflows, and this is represented by a positive sign(+).
The second expression is EAC, sometimes called EUAC, which represents the equivalent uniform Annual Cost. The uniform term expresses that these costs are equal in value, and the term Annual means that the cost is equivalent per year.
The second expression is EAC, sometimes called EUAC, which represents the equivalent—uniform Annual Cost. The uniform term expresses that these costs are equal in value, and the term Annual means that the Cost is per year and equivalent.
EAC, we obtain from two sources; the first is the Capital recovery we have expressed.
 The second source is the Cost of operation and maintenance. Both sources are represented by downward arrows with negative (-) signs.
The Third expression, EAW (EUAW) or the equivalent uniform annual worth, EAB-EAC (EUAB-EUAC), will yield a positive profit value.
If the difference is positive, it indicates a good investment. If the difference is negative, this indicates that EUAC is larger than EUAB, concluding that costs exceed benefits and the investment is not profitable.
A solved problem for EAW or the Equivalent Annual Worth.
We will have a solved problem for EAW (Equivalent Annual Worth). An asset has an initial cost of $100,000 and an estimated salvage value of $40,000 at the end of its 6-year service life.
I draw a time scale from t=0 to t=6 years. Estimated O&M costs are $50,000 in year one, increasing by $6,000 per year after that.
We have an item to consider: the uniform gradient. The starting value is $50,000, and a G Gradient of $ 6,000 is added at the end of each year from year 1 to year 6.
The initial value is $50,000. At the end of year 2, we have $56000; at the end of year 3, $62000; then $68000, $ 74000, and finally $80,000.
The increase is the Gradient value G. The G value is $6000. The assets are expected to generate annual benefits of $110,000. This $110,000 is represented by upward arrows on the time scale over the next six years, from t=1 to t=6.
Is this a desirable investment if MARR is 20%? How can we judge or decide on this investment? This is done by using the Annual Equivalent Cost.
It is important to convert the costs in the solved problem, whether initial or final, to an annual cost. That is why we will use the conversion table relations. Using the table for the relation between A/F& A/P and G/A.
How to estimate EAC?
On the next slide of the solved problem, the EAC consists of two items: capital recovery and operations and maintenance. I will estimate one by one. The Capital recovery CR can be calculated from the initial cost of -100,000 and the salvage value of 40,000.
For the first estimate, we have – 100,000*(A/P, MARR=20%,n=6 years). The slide image shows both the cash-in and cash-out diagrams with the years from t=0 to t=6.
The second term is+40,000*(A/F,20%,6). We will use the conversion equations if we do not have the table. A/P=(i)*(i+i)^n/(1+i)^n-1, for i=20%, we can get the value of A/p as equals (20*1/100)*(1+0.20)^6/(1+0.20)^(6-1)=0.300071.
For A/F, it is (A/P)*(P/F)= From the Previous relation of A/P and then multiplied by (1+i)^n. The final value of A/F=i/(1+i)^n-1. A/F=0.20/(1+0.20)^6-1=0.1007. I have completed the capital recovery estimate.
On the next slide, for EAC, CR = -100,000 (0.30071) + 40,000 (0.1071). EAC=-26,042. For the estimate of EAC for operation and maintenance. We have an annual gradient cost; the initial value is $50,000, and the G value is 6000.
I
The annual gradient cost can be estimated as =-50,000, will be left unchanged drawn as downward arrows,+(- G(A/G,i=20%,6). A/G=(1/i)-((n/(1+i)^n)-1)). i=20%, then A/G=(1/0.20)-((6/(1.20)^6-1))=1.9788.
In the next slide, the total value of EAC for operation and maintenance, EUAC, is the sum of -87.715. The last step is to calculate EUAB, which is easy since the data are given as $110,000, distributed evenly over the next six years.
The last equation is EUAW or the equivalent annual worth =EUAB -EUAC=110,000-87,915=22,085. This investment yields a profit because the difference is positive, and the investment is acceptable.
The PDF used to illustrate this post can be downloaded from the following document.
The following post will introduce the different types of assets.
For a useful external resource, Engineering Economy A good referance.