On solving the above expression, the value of i_r is found to be 13.55% per year which is greater than MARR (10%). Now using the using Microsoft Excel spreadsheet and entering year-wise cash inflows and cash out flows, the value of rate of return is found to be 13.53% (using the function ‘IRR’). However this minor difference in the value of i_r obtained from both the methods can be minimized by finding out the net present worth at narrow range of interest rate values and carrying out linear interpolation between these values (trial and error method) to find out the more precise value close to the actual rate of return.

The net present worth of the construction equipment at MARR i.e. 10% is given by;

PW = Rs.137344 at MARR (10%)
The net present worth of the construction equipment at MARR is greater than zero and the rate of return is greater than MARR. Thus the purchase of the construction equipment is economically justified. It may be noted here that when the equivalent worth of an investment is greater than zero at interest rate equal to MARR, then the rate of return of the investment is greater than MARR.

The rate of return ‘i_r’ can also be determined by equating the net annual worth to zero. For the above construction equipment, the net equivalent annual worth at different values of i_r are calculated as follows;