Tutorial 5 (Chapter 6-The application of project evaluation methods)

4.

Define the term ‘mutually exclusive projects’ and provide a simple example. Outline and justify the basic net present value rule applicable to them. How should this rule be modified when such projects have unequal lives? Two projects are said to be mutually exclusive if the selection of one project will automatically preclude the other project from being selected. This can be generalised to an n-project case. If a company owns land on which it can build either a factory or a warehouse, these two projects are said to be mutually exclusive because if a decision is made to build the factory, it is then impossible for the company to build the warehouse. The NPV rule for mutually exclusive projects with equal lives is to choose the project that gives the largest positive NPV. By choosing the project with the largest positive NPV, the company’s market value is maximised.

Solutions to problems

1.

Compare machines over a life of 10 years. Assume zero scrap value in each case. Continue to operate old machine: Net cash flow = $10 000 – $7 000 = $3 000 per year 1  1  (1.10)10 NPV = $3 000   0.10   = $18 434 Purchase new machine now: Cost = $16 000 Disposal value of old machine = $1 000 Net cash flow = $11 000 – $5 000 = $6000 NPV
1  1  (1.10)10 = –$16 000 + $1 000 + $6 000   0.10   = $21 867      

     

As the NPV of purchasing the new machine exceeds the NPV of continuing to operate the old machine, the new machine should be purchased now. 2. The aim is to minimise the NPV of cash outflows.

(a) Assuming a 10% required rate of return: Machine A 1  1  (1.10)10 NPV = –$50 000 – $20 000   0.10   = –$172 891 Machine B
1  1  (1.10)10 NPV = –$85 000 – $15 000   0.10   = –$177 169 Decision: Choose A

     

     

(b) Assuming a 7% required rate of return: Machine A 1  1  (1.07 ) 10 NPV = –$50 000 – $20 000  0.07    = –$190 472 Machine B 1  1  (1.07 ) 10 NPV = –$85 000 – $15 000  0.07    = –$190 354 Decision: Choose B
1   1  (1.10)12   = Savings: $20 000   0.10      less Expenditure: Initial outlay Overhaul after 2 years $3 000/(1.10)2 Overhaul after 4 years $3 000/(1.10)4 Overhaul after 6 years $3 000/(1.10)6 Overhaul after 8 years $3 000/(1.10)8 Overhaul after 10 years $3 000/(1.10)10 Net present value

     

     

3.

$136 274

–$125 000 –$ 2 479 –$ 2 049 –$ 1 693 –$ 1 400 –$ 1 157 $ 2 496

Decision: As the net present value is positive, the equipment should be purchased. 4. Cash flows ($) associated with the project Year Land Farm buildings Farm equipment Current assets Sales Expenses 0 –1 000 000 –200 000 –400 000 –250 000 1 2 ... 10 500 000 50 000 0 250 000 2 480 000 2 200 000

2 480 000 2 200 000

2 480 000 2 200 000 $ = (1 850 000) = 1 720 479 = $308 435 $178 914

Calculation of net present value: Initial outlay Present value of annual net cash flow ($2 480 000 – $2 200 000) Residual value and recovery of current assets $800 000/(1.10)10 Net present value Decision: The project is acceptable. 6.

If continue to operate old machine $ (a) Initial cost not relevant (b) Disposal value of old machine not relevant 1   1  (1.10) 3   (c) Operating costs (22 000)   0.10      = (54 711) (54 711)

If purchase machine $ (60 000) 15 000 1   1  (1.10) 3   (11 000)   0.10      = (27 355) (72 355)

Decision: Continue to operate the old machine which has a lower net present value cost. 7. (a) Replace in 5 years’ time: $
1   1  (1.1) 5   Net cash flow: 10  $100 000   0.10      plus Residual value at the end of 5 years: less Cash outflow: Net present value:

=

3 790 787

0 0 3 790 787

Replace now Present value of net cash flows: 1   1  (1.1) 5   5  $1 000 000  =  0.10      Proceeds from sale of piston aircraft: 10  $10 000 = Present value of residual value of turboprops