The method is about the progress of getting back the worth of future benefit or cash flows of every alternative are going to be lessen to time zero by presume a charge per unit which is i in the present of kick-off point in time. Afterwards, the MARR is acting as an interest rate to all cash inflows and outflows at discounted to the present point in time. Furthermore, the investments by the investor of a projects get the minimum profit if the project is at positive present worth. If P( i%)>0 the project is consider as profitable but if P( i%)<0 then the project is consider as loss-making and if the net present worth is =0 then it means precisely i% return. It is important to know which is the highest rate of interest and the within limit of cash flows occurs but if the lower the PW is the better. For example, the PW of Rm 1,000 10 years from now is Rm613.90 when i = 5% per year. However, if i =10%, that same Rm 1,000 is only worth Rm 385.50 now.

Let’s see another example:

If continuous compounding had been specified for a nominal interest rate (r) of 20% per year, the PW would have been calculated by using the interest factors. 

PW (r = 20%)  = -RM25,000 + $8,000 (P/A, r = 20%, 5) +$5,000 (P/F, r =20%,5)

=-RM25,000 + RM8,000 (2.8551) + RM5000 (0.3679)

=-RM319.60

Indeed, the project is not acceptable because it is negative.

In this method, it is to maximize the future luxury for the investors. It is based on the parallel worth of all inflows and outflows at the lattermost of the process period if interest rate is similar to MARR. Basically if Future Worth (i= MARR) is equal to 0 then the project is acceptable. This method is compatible to apply for the big project.

For example:

RM 110,000, energy sports complex was projected to save RM 30,000 per year in electrical power and be worth RM 8,000 at the end of the 6-year study period. By using the FW method to determine whether the project is still acceptable if the system has zero market value after six years. The MARR is 15% per year.

In this example, we need to find the future equal of the $110,000 investment and the $30,000, annual savings at an interest rate of 15% per year.

FW (15%) = -RM 110,000 (F/P, 15%, 6) + RM 30,000 (F/A, 15%,6)

=-RM 110,000 (2.3131) + RM 30,000 (8.7537)

=RM 8,170

This project shown that it is acceptable and profitable even though it has no market value at the lattermost of the time.

The objective of Annual Worth is to make an identical annual series of ringgit amounts at the beginning of the time that is same to cash inflows and outflows at an interest rate (=MARR). This is method where revenues ( R ) – expenses (E) – capital recovery (CR) amount.

AW (i%)= R – E – CR (i%). AW of a project is parallel to its PW and FW, if AW (i=MARR) > , then the project is economically acceptable.

-          AW = PW (A/P,i%,N)

-          AW = FW (A/F,i%,N)

 

The CR is formulated as , CR (i%)= I (A/P, i%N) – S(A/F,i%, N). 

Consider a construction material that costs RM10,000 last five years, and its salvage (market) value now is RM2,000. If MARR is 10% per year, calculate CR and AW of the materials. So, 

AW (10%) = R – E – CR (10%) 

= RM10,000 – RM2,000 – RM2,310.4 = RM5,689.6

 

So in conclusion, the single evaluating project is a project selection problem. This problem are being help with the method to use for the best solution of a project. Indeed the 3 method that were suggested either the project is returning the profit or making a loss.