Question 175 pbIHLs A project requires $160,000 as first cost (investment). Thre

Question

Question 175 pbIHLs A project requires $160,000 as first cost (investment). Three scenarios were identified, and the following table presents the values of the project key parameters associated with these scenarios: Key project parameters Scenario 1 Scenario 2Scenario 3 Net annual savings, $ 20,000 25,000 30,000 Depreciation rate, % Service life, years 4 10 Probabilities of the first two scenarios are 0.1 and 0.5 respectively. Calculate the expected value of the project's annual worth if market interest rate is 5.5%

Explanation / Answer

Depreciation, being non-cash expense, is excluded from cash flow analysis using annual worth.

Probability of scenario 3 = 1 - (0.1 + 0.5) = 1 - 0.6 = 0.4

Using non-probability weighted cash flows,

AW, Scenario 1 ($) = - 160,000 x A/P(5.5%, 5) + 20,000 = - 160,000 x 0.2342** + 20,000

= - 37,472 + 20,000 = - 17,472

AW, Scenario 2 ($) = - 160,000 x A/P(5.5%, 7) + 25,000 = - 160,000 x 0.176** + 25,000

= - 28,160 + 25,000 = - 3,160

AW, Scenario 3 ($) = - 160,000 x A/P(5.5%, 10) + 30,000 = - 160,000 x 0.1327** + 30,000

= - 21,232 + 30,000 = 8,768

Therefore,

Expected AW ($) = 0.1 x (- 17,472) + 0.5 x (- 3,160) + 0.4 x 8,768 = - 1,747.2 - 1,580 + 3,507.2

= 180

**A/P(r%, N) = r / [1 - (1 + r)-N]

A/P(5.5%, 5) = 0.055 / [1 - (1.055)-5] = 0.055 / (1 - 0.7651) = 0.055 / 0.2349 = 0.2342

A/P(5.5%, 7) = 0.055 / [1 - (1.055)-7] = 0.055 / (1 - 0.6874) = 0.055 / 0.3126 = 0.1760

A/P(5.5%, 10) = 0.055 / [1 - (1.055)-10] = 0.055 / (1 - 0.5854) = 0.055 / 0.4146 = 0.1327

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