Chapter 6, Problem 36P

To determine

Calculate the equivalent annual LCC.

Explanation of Solution

Alternative A: First cost of development (CD) is $250,000. Annual cost of development (AD) is $150,000 from year 1 to 4. First cost of programming (CP) is $45,000. Annual cost of programming (AP) is $35,000 from year 1 to 2. Operation cost (O) is $50,000 from year 1 to 10. Supporting cost (S) is $30,000 year 1 to 5. Time period (n) is 4. Time period 1 (n1) is 2. Time period 2 (n2) is 10. Time period 3 (n3) is 5. Interest rate (i) is 8%.

Alternative B: First cost of development (CD) is $10,000. First cost of programming (CP) is $45,000. Annual cost of programming (AP) is $30,000 from year 1 to 3. Operation cost (O) is $80,000 from year 1 to 10. Supporting cost (S) is $40,000 year 1 to 10. Time period (n) is 3. Time period 1 (n1) is 10. Interest rate (i) is 8%.

Alternative C: Operation cost (O) is $175,000 from year 1 to 10. Time period (n) is 10. Interest rate (i) is 8%.

The equivalent annual LCC for project A (AA) can be calculated as follows:

$\begin{array}{c}\text{AA}=-\left(\begin{array}{c}\text{CD+CP}+\text{AD}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)+\text{AP}\left(\frac{{\left(1+\text{i}\right)}^{\text{n1}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n1}}}\right)\\ +\text{O}\left(\frac{{\left(1+\text{i}\right)}^{\text{n2}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n2}}}\right)+\text{S}\left(\frac{{\left(1+\text{i}\right)}^{\text{n3}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n3}}}\right)\end{array}\right)\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n2}}}{{\left(1+\text{i}\right)}^{\text{n2}}-1}\right)\\ =-\left(\begin{array}{c}\text{250,000+45,000}+\text{150,000}\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{4}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{4}}}\right)\\ +\text{35,000}\left(\frac{{\left(1+\text{0}\text{.08}\right)}^2-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{2}}}\right)\\ +\text{50,000}\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{10}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{10}}}\right)+\text{30,000}\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{5}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{5}}}\right)\end{array}\right)\left(\frac{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{10}}}{{\left(1+\text{0}\text{.08}\right)}^{\text{10}}-1}\right)\end{array}$

$\begin{array}{c}=-\left(\begin{array}{c}\text{295,000}+\text{150,000}\left(\frac{1.360489-1}{\text{0}\text{.08}\left(1.360489\right)}\right)\\ +\text{35,000}\left(\frac{1.1664-1}{\text{0}\text{.08}\left(1.1664\right)}\right)\\ +\text{50,000}\left(\frac{2.158925-1}{\text{0}\text{.08}\left(2.158925\right)}\right)+\text{30,000}\left(\frac{1.469328-1}{\text{0}\text{.08}\left(1.469328\right)}\right)\end{array}\right)\left(\frac{\text{0}\text{.08}\left(2.158925\right)}{2.158925-1}\right)\\ =-\left(\begin{array}{c}\text{295,000}+\text{150,000}\left(\frac{0.360489}{0.108839}\right)+\text{35,000}\left(\frac{0.1664}{0.093312}\right)\\ +\text{50,000}\left(\frac{1.158925}{0.172714}\right)+\text{30,000}\left(\frac{0.469328}{0.117546}\right)\end{array}\right)\left(\frac{0.172714}{1.158925}\right)\\ =-\left(\begin{array}{c}\text{295,000}+\text{150,000}\left(3.312131\right)+\text{35,000}\left(1.783265\right)\\ +\text{50,000}\left(6.710081\right)+\text{30,000}\left(3.992718\right)\end{array}\right)\left(0.149029\right)\\ =-\left(\begin{array}{c}\text{295,000}+496,819.65+62,414.275\\ +335,504.05+119,781.54\end{array}\right)\left(0.149029\right)\\ =-\left(1,309,519.515\right)\left(0.149029\right)\\ =-195,156.38\end{array}$

The equivalent annual value of the project cash flow is -$195,156.38.

The equivalent annual LCC for project B (AB) can be calculated as follows:

$\text{AB}=-\left(\begin{array}{c}\text{CD+CP}+\text{AP}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)\\ +\text{O}\left(\frac{{\left(1+\text{i}\right)}^{\text{n1}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n1}}}\right)+\text{S}\left(\frac{{\left(1+\text{i}\right)}^{\text{n1}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n1}}}\right)\end{array}\right)\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n1}}}{{\left(1+\text{i}\right)}^{\text{n1}}-1}\right)$

$\begin{array}{c}\text{AB}=-\left(\begin{array}{c}\text{10,000+45,000}+\text{30,000}\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{3}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{3}}}\right)\\ +\text{80,000}\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{10}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{10}}}\right)+\text{40,000}\left(\frac{{\left(1+\text{i}\right)}^{\text{n1}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n1}}}\right)\end{array}\right)\left(\frac{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{10}}}{{\left(1+\text{0}\text{.08}\right)}^{\text{10}}-1}\right)\\ =-\left(\begin{array}{c}\text{10,000+45,000}+\text{30,000}\left(\frac{1.259712-1}{\text{0}\text{.08}\left(1.259712\right)}\right)\\ +\text{80,000}\left(\frac{2.158925-1}{\text{0}\text{.08}\left(2.158925\right)}\right)+\text{40,000}\left(\frac{2.158925-1}{\text{0}\text{.08}\left(2.158925\right)}\right)\end{array}\right)\left(\frac{\text{0}\text{.08}\left(2.158925\right)}{2.158925-1}\right)\\ =-\left(\begin{array}{c}\text{10,000+45,000}+\text{30,000}\left(\frac{0.259712}{0.100777}\right)\\ +\text{80,000}\left(\frac{1.158925}{0.172714}\right)+\text{40,000}\left(\frac{1.158925}{0.172714}\right)\end{array}\right)\left(\frac{0.172714}{1.158925}\right)\\ =-\left(\begin{array}{c}\text{10,000+45,000}+\text{30,000}\left(2.577096\right)\\ +\text{80,000}\left(6.710081\right)+\text{40,000}\left(6.710081\right)\end{array}\right)\left(0.149029\right)\\ =-\left(\begin{array}{c}\text{10,000+45,000}+77,312.88\\ +536,806.48+268,403.24\end{array}\right)\left(0.149029\right)\\ =-\left(937,522.6\right)\left(0.149029\right)\\ =-139,718.06\end{array}$

The equivalent annual value of the project cash flow is -$139,718.06.

The equivalent annual LCC for project C (AC) can be calculated as follows:

$\begin{array}{c}\text{AC}=-\text{O}\left(\frac{{\left(1+\text{i}\right)}^{\text{n}}-1}{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}\right)\left(\frac{\text{i}{\left(1+\text{i}\right)}^{\text{n}}}{{\left(1+\text{i}\right)}^{\text{n}}-1}\right)\\ =-175,000\left(\frac{{\left(1+\text{0}\text{.08}\right)}^{\text{10}}-1}{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{10}}}\right)\left(\frac{\text{0}\text{.08}{\left(1+\text{0}\text{.08}\right)}^{\text{10}}}{{\left(1+\text{0}\text{.08}\right)}^{\text{10}}-1}\right)\\ =-175,000\left(\frac{2.158925-1}{\text{0}\text{.08}\left(2.158925\right)}\right)\left(\frac{\text{0}\text{.08}\left(2.158925\right)}{2.158925-1}\right)\\ =-175,000\left(\frac{1.158925}{0.172714}\right)\left(\frac{0.172714}{1.158925}\right)\\ =-175,000\left(1\right)\\ =-175,000\end{array}$

The equivalent annual value of the project cash flow is -$175,000. Since the equivalent annual value of project B is greater than the other two projects, select project B.