Engineering Economy
Engineering Economy
8th Edition
ISBN: 9780073523439
Author: Leland T Blank Professor Emeritus, Anthony Tarquin
Publisher: McGraw-Hill Education
Question
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Chapter 6, Problem 36P
To determine

Calculate the equivalent annual LCC.

Expert Solution & Answer
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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:

AA=(CD+CP+AD((1+i)n1i(1+i)n)+AP((1+i)n11i(1+i)n1)+O((1+i)n21i(1+i)n2)+S((1+i)n31i(1+i)n3))(i(1+i)n2(1+i)n21)=(250,000+45,000+150,000((1+0.08)410.08(1+0.08)4)+35,000((1+0.08)210.08(1+0.08)2)+50,000((1+0.08)1010.08(1+0.08)10)+30,000((1+0.08)510.08(1+0.08)5))(0.08(1+0.08)10(1+0.08)101)

=(295,000+150,000(1.36048910.08(1.360489))+35,000(1.166410.08(1.1664))+50,000(2.15892510.08(2.158925))+30,000(1.46932810.08(1.469328)))(0.08(2.158925)2.1589251)=(295,000+150,000(0.3604890.108839)+35,000(0.16640.093312)+50,000(1.1589250.172714)+30,000(0.4693280.117546))(0.1727141.158925)=(295,000+150,000(3.312131)+35,000(1.783265)+50,000(6.710081)+30,000(3.992718))(0.149029)=(295,000+496,819.65+62,414.275+335,504.05+119,781.54)(0.149029)=(1,309,519.515)(0.149029)=195,156.38

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:

AB=(CD+CP+AP((1+i)n1i(1+i)n)+O((1+i)n11i(1+i)n1)+S((1+i)n11i(1+i)n1))(i(1+i)n1(1+i)n11)

AB=(10,000+45,000+30,000((1+0.08)310.08(1+0.08)3)+80,000((1+0.08)1010.08(1+0.08)10)+40,000((1+i)n11i(1+i)n1))(0.08(1+0.08)10(1+0.08)101)=(10,000+45,000+30,000(1.25971210.08(1.259712))+80,000(2.15892510.08(2.158925))+40,000(2.15892510.08(2.158925)))(0.08(2.158925)2.1589251)=(10,000+45,000+30,000(0.2597120.100777)+80,000(1.1589250.172714)+40,000(1.1589250.172714))(0.1727141.158925)=(10,000+45,000+30,000(2.577096)+80,000(6.710081)+40,000(6.710081))(0.149029)=(10,000+45,000+77,312.88+536,806.48+268,403.24)(0.149029)=(937,522.6)(0.149029)=139,718.06

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:

AC=O((1+i)n1i(1+i)n)(i(1+i)n(1+i)n1)=175,000((1+0.08)1010.08(1+0.08)10)(0.08(1+0.08)10(1+0.08)101)=175,000(2.15892510.08(2.158925))(0.08(2.158925)2.1589251)=175,000(1.1589250.172714)(0.1727141.158925)=175,000(1)=175,000

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.

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