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Example 1: Service projects with equal project lives Finite service requirement: The project lives are short term and in this example it is 5 years. Equal project lives: The project lives are the same (5 years in this example). Service projects: All the cash flow, except for the salvage value at the end of the project, is negative cash flow. Figure 3.10 below illustrates cash flow diagrams for the two projects with equal project lives. Project A MARR=14% Salvage value=10 000 0 1 2 3 4 2 000 2000 2000 2 000 2 000 40 000 NPV=-40 000-2 000(PIA, 14,5)+10 000(P/F 14,5) - 40 000-2 000(3.4331)+10 000(0.5194) =- 41 672.2 Project B MARR=14% Salvage value=5 000 0 2 3 4 5 3 000 3 000 3 000 3 000 3 000 50 000 NPV=-50 000-3 000(P/A, 14,5)+5 000(P/F.14,5) --50 000-3 000(3.4331)+5 000(0.5194) =- 57 702.3 Figure 3.10: Cash flow diagrams for the two projects with equal project lives Project A more economical project