The most simple and commonly used method of determining the present value of an ordinary annuity is to multiply the incremental payout by the appropriate rate found on the present value of an ordinary annuity table.

+ Present Value of an Ordinary Annuity

Table 2 - Present Value of an Ordinary Annuity of $1 at Compound Interest
Period	5%	6%	7%	8%	9%	10%	11%	12%
1	0.952	0.943	0.935	0.926	0.917	0.909	0.901	0.893
2	1.859	1.833	1.808	1.783	1.759	1.736	1.713	1.690
3	2.723	2.673	2.624	2.577	2.531	2.487	2.444	2.402
4	3.546	3.465	3.387	3.312	3.240	3.170	3.102	3.037
5	4.329	4.212	4.100	3.993	3.890	3.791	3.696	3.605
6	5.076	4.917	4.767	4.623	4.486	4.355	4.231	4.111
7	5.786	5.582	5.389	5.206	5.033	4.868	4.712	4.564
8	6.463	6.210	5.971	5.747	5.535	5.335	5.146	4.968
9	7.108	6.802	6.515	6.247	5.995	5.759	5.537	5.328
10	7.722	7.360	7.024	6.710	6.418	6.145	5.889	5.650
11	8.306	7.887	7.499	7.139	6.805	6.495	6.207	5.938
12	8.863	8.384	7.943	7.536	7.161	6.814	6.492	6.194
13	9.394	8.853	8.358	7.904	7.487	7.103	6.750	6.424
14	9.899	9.295	8.745	8.244	7.786	7.367	6.982	6.628
15	10.380	9.712	9.108	8.559	8.061	7.606	7.191	6.811
16	10.838	10.106	9.447	8.851	8.313	7.824	7.379	6.974
17	11.274	10.477	9.763	9.122	8.544	8.022	7.549	7.120
18	11.690	10.828	10.059	9.372	8.756	8.201	7.702	7.250
19	12.085	11.158	10.336	9.604	8.950	8.365	7.839	7.366
20	12.462	11.470	10.594	9.818	9.129	8.514	7.963	7.469

Using the previous table, enter the correct factor for three periods at 5%:

Periodic payment	x	Factor	=	Present value
$6,000	x	2.723	=	$16,338

The controller at Ross has determined that the company could save $8,000 per year in engineering costs by purchasing a new machine. The new machine would last 10 years and provide the aforementioned annual monetary benefit throughout its entire life. Assuming the interest rate at which Ross purchases this type of machinery is 8%, what is the maximum amount the company should pay for the machine? $fill in the blank