Solutions to Practice Problems – Chapter 6

1 a)      Ignore interest charges.  Use the depreciation that will be used for tax returns.  The working capital employed at the beginning of the last year will be recovered at the end of the year.

                                    Revenue                        15,000

                                    Manufacturing cost       -  8,000

                                    Depreciation                 -     500

                                    Pre-tax income                 6,500

                                    Tax (@ 40%)                -  2,600

                                    “N.I.”                                3,900

                                    Dep                             +     500

                                    Recovery of W.C.        +  1,500                                                                            

                                    Net cash flows                 5,900

b) Accelerated depreciation is higher in the early years of a project and lower in the later years.

2.         At the end of 4 years, the assets will be fully depreciated, and the entire salvage value will be taxable.  This yields the following cashflows.

OCF for project :                         0           1                     2                     3                     4

Revenue                                    -           400                 660                 672                 693    

Variable Cost                             -          (220)              (375)               (432)               (462)

Fixed Cost                                 -            (55)                (65)                 (75)                 (85)

Depreciation                              -          (166.65)         (222.25)            (74.05)            (37.05)

EBIT                                          -            (41.65)             (2.25)             90.95             108.95

Taxes                                        -             14.16                0.76             (30.92)            (37.04)

Net Income                                -            (27.49)             (1.49)          60.03                  71.91

Depreciation                              -           166.65           222.25            74.05                 37.05

Increase in W.C.                      (5)            (3)                  (0.6)              (0.4)                    9

After-tax Salvage Value                                                                                                 19.8

Operating Cash Flow              (5)        136.16             220.16          133.68               137.76

Change in Operating cashflow for existing business:

Increase in EBIT                        -          (10)                   (8)                 (8)                      (8)

Increase in Taxes                      -            (3.4)                (2.7)              (2.7)                   (2.7)

Increase in OCF                        -            (6.6)                 (5.3)             (5.3)                   (5.3)

Capital Expenditure:             (500)            -                        -                  -                           -

Net Cash Flow:                    -505        129.56             214.86              128.38         132.46

Note: after-tax salvage value = 30,000 - .34*(30,000) = 19,800

3.         Since this company uses its tax depreciation in the income statement for tax purposes, we don’t need to change depreciation.  Just ignore interest and take working capital into account.  At time 4, you provide the additional working capital needed for the last year ($100); at time 5 you recover the total working capital of $1,900.

                                                                            Year 4             Year 5

                                    Revenue                          32,000             35,000

                                    Manufacturing cost          20,000             21,500

                                    Depreciation                      2,400               1,600

                                    Pre-tax income                  9,600              11,900

                                    Tax (@ 40%)                     3,840                4,760 

                                    N.I.                                     5,760                7,140

                                    Dep                               +  2,400                1,600

                                    W.C.                              -      100            +  1,900

                                    Net cash flows                   8,060              10,640

4.         At time zero, the net cashflow will be –100,000 (capital investment) – 2,00 (initial working capital) = -102,000.  The remaining cashflows are:

                                                  Year 1           Year 2             Year 3             Year 4

            Revenue                      23,000            24,500             26,250             28,300

            Manufacturing cost      11,000            11,550             12,200             13,000

            Depreciation                  4,400              3,200               2,400               1,600

           Pre-tax income               7,600               9,750             11,650             13,700

            Tax (@ 40%)                 3,040              3,900               4,660                5,480 

            N.I.                                 4,560              5,850               6,990                8,220

            Depreciation                  4,400              3,200               2,400                1,600

            W.C.                            -    100            -    150             -    150             + 2,400

            Net cashflows                8,860              8,900               9,240               12,220

5.         The project is an 8-year project (although the plant and equipment are depreciated over 10 years).  This simply means that when the project is terminated at the end of 8 years, the plant and equipment will not be fully depreciated.  The capital gain on the sale will have to be computed using the book value at time 8:

            Accumulated depreciation for 8 years = 960

            Book value at time 8 = 1200 – 960 = 240

            Capital gain = 400 – 240 = 160

            Tax @ 35% = 56

The net cashflows that would be obtained from renting the warehouse if it wasn’t used for the project must be charged to the project too, since it is an opportunity cost.

The cashflow at time zero will consist of the capital investment of $1.2m and the initial working capital.

6)            Machine A

NPV = 3600/1.15 + 3600/1.15² + … 3600/1.15⁶ - 8,000 = 13,624.14 - 8,000 = 5,624.14

EAC is given by              EAC/1.15 + EAC/1.15² + … EAC/1.15⁶  = 5,624.14

Solving, EAC_A = 1,486.10

=> Using machine A is like generating a cash inflow of 1,486.10 each year.

            Machine B

NPV = 5700/1.15 +  … 5700/1.15⁸ - 13,000 = 25,577.73 – 13,000 = 12,577.73

EAC is given by              EAC/1.15 + EAC/1.15² + … EAC/1.15⁸  = 12,577.73

Solving, EAC_B = 2,802.95

=> Using machine B is like generating a cash inflow of 2,802.95 each year.

=>            Choose machine B

7 a)      Machine A

NPV = 15,000/1.15 + … 15,000/1.15⁴ - 36,000 = 42,824.68 - 36,000 = 6,824.68

EAC is given by              EAC/1.15 +  … EAC/1.15⁴  = 6,824.68

Solving, EAC_A = 2,390.45

=> Using machine A is like generating a cash inflow of 2,390.45 each year.

            Machine B

NPV = 16,000/1.15 +  … 16,000/1.15⁶ - 52,000 = 60,551.72 - 52,000 = 8,551.72

EAC is given by              EAC/1.15 + … EAC/1.15⁶  = 8,551.72

Solving, EAC_B = 2,259.68

=> Using machine B is like generating a cash inflow of 2,259.68 each year.

=>            Choose machine A

b)  Trying to use IRR, we would proceed as follows:

Using a financial calculator, IRR_A = 24.10%, and IRR_B = 20.93%

The incremental cashflows for B - A are:

                   0             1             2            3             4              5              6

            -16,000      1,000      1,000      1,000      1,000     16,000     16,000

These are conventional cashflows; IRR turns out to be 17.41% which is greater than the required return of 15%.  This seems to say that the incremental project B-A is worth taking => choose project B.

The reason why we get the wrong answer is that we are comparing non-comparable alternatives.  Project B is looking better because with B we are including the benefit of 6 year’s production whereas with A we are including the benefit of only 4 year’s production. Just like the NPV method, the IRR method can only be applied to cash flows over a common production horizon.

8.         First compute Equivalent Annual Cashflow for the new machine:

NPV = 4600/1.12 +  … 4600/1.12⁸ - 20,000 = 22,851.14 - 20,000 = 2,851.14

EAC is given by              EAC/1.12 + … EAC/1.12⁸  = 2,851.14

Solving, EAC = 573.94

The cashflows for the different replacement alternatives are then:

            Time       C₀       C₁       C₂     C₃

               0         600     574     574    574  . . .

               1         600     500     574    574  . . .

               2         600     500     400    574  . . .

Cashflows beyond time 2 are 573.94 each year in each case, and can be ignored.

It is clear that the old machine should be replaced right away (at time 0).  It generates only $500 the first year, whereas once you put in a new machine it is like getting $574 forever.

9.         First compute Equivalent Annual Cashflow for the new machine:

NPV = 5000/1.08 +  … 5000/1.08⁶ - 9,000 = 23,114.40 - 9,000 = 14,114.40

EAC is given by              EAC/1.08 + … EAC/1.08⁶  = 14,114.40

Solving, EAC = 3053.16

Assume the salvage values given are after-tax values (since no tax rate is given).  If you replace at time 0, you will get the cashflow of 3,000 today plus the salvage value of 1,500, so the cashflows will be 4500 today and then 3053 forever.  If you replace at time 1, you will get 3,000 at time 0 followed by 2,000 + 1,000 at time 1 and then 3053 forever. If you replace at time 2, you will get 3,000 at time 0, 2,000 at time 1 followed by 1200 + 500 at time 2 and then 3053 forever.   If you replace at time 3, you will get 3,000 at time 0, 2,000 at time 1, 1200 at time 2, 500 at time 3 and then 3053 forever. 

So the cashflows are:

            Time                    C₀          C₁        C₂         C₃        C₄        C₅

               0                   4500      3053    3053     3053     3053    3053  . . .

               1                   3000      3000    3053     3053     3053    3053  . . .

               2                   3000      2000    1700     3053     3053    3053  . . .

               3                   3000      2000    1200       500     3053    3053  . . .

Cashflows beyond time 3 are 3053.16 each year in all four cases, and can be ignored.

Clearly, replacing at time 2 is better than replacing at time 3; replacing at time 1 is better than replacing at time 2, and replacing at time 0 is better than replacing at time 1.  So the machine should be replaced right away.

10.         First compute Equivalent Annual Cost for the new circuits.

PV of costs = 1200 + 200/1.15³ +  320/1.15⁴ +  545/1.15⁵ = 1785.45

=>                    EAC/1.15 + EAC/1.15² + … EAC/1.15⁵  = 1785.45

Solving, EAC =  532.62

Since the existing circuits will last only for 3 years, they must be replaced at time 0, 1, 2 or 3.  Using the equivalent annual cashflows for the new circuits instead of the original cashflows, the equivalent cashflows from different replacement options are as follows:

            Time      C₀         C₁        C₂         C₃         C₄        C₅

               0          0          533      533       533       533      533  . . .

               1          0          350      533       533       533      533  . . .

               2          0          350      600       533       533      533  . . .

               3          0          350      600       460       533      533  . . .

Cashflows beyond time 5 are 533 each year in all four cases (as new circuits are replaced by new circuits).  We can ignore cashflows beyond time 3, since they are equal in all four cases.

Clearly, replacing at 1 is cheaper than replacing at time 0 or 2.  But we cannot choose between replacing at 1 or 3 just by looking at the cashflows.  So compare PV₂ of the time 2 and 3 cashflows:

PV₂(if replace at 1) = 533 + 533/1.15 = 996.48

PV₂(if replace at 3) = 600 + 460/1.15 = 1000

=>  1 has the lower PV of costs, and the circuits should be replaced at time 1

The assumptions made here are the standard assumptions that must be made whenever we use EAC:

11.            Compute the E(R) from waiting or the PV at time 0 of the future values:

Time 0            Time 1            Time 2            Time 3            Time 4

$8.99              $10.75            $12.50            $13.75            $14.99

E(R)                                        19.58%          16.28%             10%               9.02%

PV₀(FV_t)                                         $  9.77            $10.33            $10.33            $10.24

You do want to wait for the first two years.  Whether you wait for the third year or not doesn’t matter.  The expected return equals the required return, so it’s a zero NPV investment.  (That’s reflected in PV₀ being equal at time 2 and time 3.)  A zero NPV investment has no effect on s/h wealth, so stockholders don’t care whether you wait for the third year or not.  Thus, the wine can be sold at either time 2 or time 3.  

12 a)      You can compute either the E(R) from waiting, or the PV of the future NPVs:

Time t                 0                  1                  2                 3                   4

NPV_t               2,400           2,850           3,225           3,515           3,725

E(R)                                 18.75%        13.16%          8.99%           5.97%

PV₀NPV_t                            2,567.57      2,617.48       2,570.14       2,453.77

Either way the project should be taken at time 2.  The expected return from waiting > required return for the first two years, so you wait and take it at the end of two years.  Or taking it at time 2 produces the future NPV with the highest PV today. 

b) The asking price for the project would have to be the PV today of the future NPV you get, when you take it at time 2:  2,617.48. 

13.       First we need to figure out what’s the best time to cut the trees.  Then the price we should ask for today is the PV today of the future NPV from the best choice.

Putting on our thinking cap before we start will reduce the number of tedious computations we need to make.  The amount we receive consists of the value of the timber + the value of the land.  As long as this combined value increases faster than the required return of 9%, we should wait.  Up to time 8, the value of the timber increases more than 9%, the value of the land increases by 4% => computations are necessary to  see whether the combined value increase by more than 9% or less than 9% each year.

But after time 8, the value of the timber increases by 8.16% each year (1.04*1.04 – 1).  Both timber and land values increase less than 9%, so the combined value increases less than 9%.  Thus, we only need to make computations for time 1 through 8:

Clearly, the trees should be cut at time 8.  The fair value of the property today is then $109,858.  The offer of $140,000 should be accepted forthwith!

14.    The total output per machine is 2,500 + 1,000 = 3,500 gallons

If you keep the old machines:

Annual operating cost = 2*3,500*1.50 = 10,500

PV of operating costs = 10,500/0.05 = 210,000

If you replace both machines:

Initial investment = 2*1,500 = 3,000

Annual operating cost = 2*3,500*1 = 7,000

PV of costs = 2*30,000 + 7,000/0.05 = 200,000

Replace just one machine:

The new machine is cheaper per unit ($1 instead of $1.50).  In the fall/winter, you’ll use only the new machine to get the 2000 gallons you need.  In spring/summer, you’ll use both machines, to get the 5000 gallons you need.

Output of new machine = 4,500; output of old machine = 2,500

Initial investment = 30,000

Annual operating cost = 4,500*1 + 2,500*1.5 = 8,250

PV of costs = 30,000 + 8,250/0.05 = 195,000

=> It’s best to replace just one machine.  Replacing the second machine just to produce 2,500 gallons is not cost effective.

15.     The total output per machine is 6,000 + 3,600 = 9,600 shovels

If you keep the old machines:

Annual operating cost = 2*9,600*4 = 76,800

PV of operating costs = 76,800/0.05 = 1,536,000

If you replace both machines:

Initial investment = 2*100,000 = 200,000

Annual operating cost = 2*9,600*3.25 = 62,400

PV of costs = 200,000 + 62,400/0.05 = 1,448,000

Replace just one machine:

The new machine is cheaper per unit ($3.25 instead of $4).  In the spring/summer you’ll use only the new machine at 100% capacity to produce 6000 shovels, and the old machine just to produce the remaining 1,200 shovels you need.  In fall/winter, you’ll use both machines, to produce 6000 shovels each.

Output of new machine = 12,000; output of old machine = 7,200

Initial investment = 100,000

Annual operating cost = 12,000*3.25 + 7,200*4 = 67,800

PV of costs = 100,000 + 67,800/0.05 = 1,456,000

=> This time it’s best to replace both machines.  Replacing the second machine to produce 7,200 is cost effective.

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