4. 0 1 2 3 4 Initial Outlay OCF 1 OCF 2 OCF 3 OCF 4 + Terminal CF NCF 0 NCF 1 NCF 2 NCF 3 NCF 4 Set up without numbers a time line for the project CFs.

10. Net Investment Outlay at t = 0 (000s) Equipment Freight + Inst. Change in NWC Net CF 0 ($200) (40) (20) ($260)  NWC = $25,000 - $5,000 = $20,000.

11. Basis = Cost + Shipping + Installation $240,000 Depreciation Basics

12. Year 1 2 3 4 % 0.33 0.45 0.15 0.07 Depr. $ 79 108 36 17 x Basis = Annual Depreciation Expense (000s) $240

13. Net revenue Depreciation Before-tax income Taxes (40%) Net income Depreciation Net operating CF $125 (79 ) $ 46 (18 ) $ 28 79 $107 Year 1 Year 1 Operating Cash Flows (000s)

14. Net revenue Depreciation Before-tax income Taxes (40%) Net income Depreciation Net operating CF $125 (79 ) $ 46 (18 ) $ 28 79 $107 $125 (17 ) $108 (43 ) $ 65 17 $ 82 Year 4 Year 1 Year 4 Operating Cash Flows (000s)

15. Net Terminal Cash Flow at t = 4 (000s) Salvage value Tax on SV Recovery on NWC Net terminal CF $25 (10) 20 $35

16. What if you terminate a project before the asset is fully depreciated? Cash flow from sale = Sale proceeds - taxes paid. Taxes are based on difference between sales price and tax basis, where: Basis = Original basis - Accum. deprec.

18. Project Net CFs on a Time Line Enter CFs in CFLO register and I = 10. NPV = $81,573. IRR = 23.8%. *In thousands. 0 1 2 3 4 (260)* 107 118 89 117

19. What is the project’s MIRR? (000s) ( 260 ) MIRR = ? 0 1 2 3 4 (260)* 107 118 89 117.0 97.9 142.8 142.4 500.1

20. 1. Enter positive CFs in CFLO: I = 10; Solve for NPV = $341.60. 2. Use TVM keys: PV = 341.60, N = 4 I = 10; PMT = 0; Solve for FV = 500.10. (TV of inflows) 3. Use TVM keys: N = 4; FV = 500.10; PV = -260; PMT= 0; Solve for I = 17.8. MIRR = 17.8%. Calculator Solution

21. What is the project’s payback? (000s) Cumulative: Payback = 2 + 35/89 = 2.4 years. 0 1 2 3 4 (260)* (260) 107 (153) 118 (35) 89 54 117 171

22. If this were a replacement rather than a new project, would the analysis change? Yes. The old equipment would be sold and the incremental CFs would be the changes from the old to the new situation.

30. S and L are mutually exclusive and will be repeated. k = 10%. Which is better? (000s) 0 1 2 3 4 Project S: (100) Project L: (100) 60 33.5 60 33.5 33.5 33.5

31. S L CF 0 -100,000 -100,000 CF 1 60,000 33,500 N j 2 4 I 10 10 NPV 4,132 6,190 NPV L > NPV S . But is L better? Can’t say yet. Need to perform common life analysis.

33. Project S with Replication: NPV = $7,547. Replacement Chain Approach (000s) 0 1 2 3 4 Project S: (100) ( 100 ) 60 60 60 ( 100 ) (40 ) 60 60 60 60

34. Compare to Project L NPV = $6,190. Or, use NPVs: 0 1 2 3 4 4,132 3,415 7,547 4,132 10%

38. If the cost to repeat S in two years rises to $105,000, which is best? (000s) NPV S = $3,415 < NPV L = $6,190. Now choose L. 0 1 2 3 4 Project S: (100) 60 60 ( 105 ) (45 ) 60 60

40. Year 0 1 2 3 CF ($5,000) 2,100 2,000 1,750 Abandonment Value $5,000 3,100 2,000 0 Consider another project with a 3-year life. If abandoned prior to Year 3, the machinery will have positive abandonment value.

41. 1.75 1. No abandonment 2. Abandon 2 years 3. Abandon 1 year (5) (5) (5) 2.1 2.1 5.2 2 4 0 1 2 3 CFs Under Each Alternative (000s)

42. NPV (no) = -$123. NPV (2) = $215. NPV (1) = -$273. Assuming a 10% cost of capital, what is the project’s optimal life?