ü 6.3 The Net Present Value Investment Rule ü 4.6 Annuities

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Cash flows of the same amount is called "annuity". Paying 10,000 yen per month for gift certificate of the department store constitutes an annuity. If cash flows start immediately, it is called an "immediate annuity." If the cash flows start at the end of the current period, rather than immediately, it is call an "ordinary annuity."

Let r be interest rate per month. Suppose you keep paying amount $A each month for a year. Starting today, you pay twelve times. Apply monthly compounding. This is an immediate annuity. FV is given by FV t1

12 A1rt

Next suppose that you will receive amount $A per month 12 times for a year. The first payment is a month from now. What is PV of this ordinary annuity? Apply monthly compounding. PV of the annuity is expressed as a sum of geometric sequence.

PV of the first cash flow = A 1r PV of the second cash flow = A

1r2 ª

PV of the last cash flow = A 1r12 Then PV of this annuity is given by t1

12 A

1rt .

Example

Suppose that you will receive amount $100 per month 12 times for a year. Interest rate per month is 1 percent.

Then PV of this annuity is $1,125.51 as shown below. In terms of PV, having $1,125.51 today is equivalent to receiving $100 each month for a year.

In[92]:= ClearA, r; A 100; r0.01; Print" t1 12 A 1rt ", t1 12 A 1rt  t1 12 A 1rt  1125.51 ¤ Mathematica

•Print[ expression you want to show]

For example command if you input Print[ variable name ] then you will see value of variable name. •Print[ "expression as you see " , expression to be calculated]

You have comma here. You use comma to differentiate the end of one expression from the next one.

Chapter 6. How to Analyze Investment Projects

ü

6.3 The Net Present Value Investment Rule

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ü

6.5 Cost of Capital

Cost of Capital is the risk-adjusted discount rate to use in computing a project's NPV. It depends on riskiness of the project. It can vary from project to project.

ü 6.8 Projects with Different Lives

When the lives of projects are different, we can compare their annualized capital costs.

Annualized capital cost: an annual cash payment that has a present value equal to the initial outlay. In other words, annual-ized capital cost is a constant payment per year over the investment period which is calculated so that total of their present values is equal to the cost.

Decision rule based on annualized capital cost: We choose the project with smaller annualized capital cost. ü Example: Two types of machines with different lives

Suppose that we are considering to install a machine to filter water from the hot spring. Two types of machines are available. Which machine type to choose? Type A works 5 years. Type B works 10 years. Type B lasts longer and costs more. Suppose that interest rate to borrow is 10%.

machine type life year price yen Type A 5 years 2106 Type B 10 years 4106

Let ca be annual cash payment for type A machine. Then annualized capital cost is value of ca which solves the following equation.

t1

5 ca

1rt 210 6 (1)

Let cb be annualized capital cost of type B. It satisfies the following equation. t1

10 cb

1rt 410

6 (2)

ü Finding value of annualized capital cost of Type A

We are going to solve equation (1).

In[93]:= Clearca, r, ansa; r0.1; ansaNSolve t1

5 ca

1rt

2106, ca

Out[93]= ca527 595.

In[94]:= caca. ansa1; Print"ca ", ca

ca  527 595.

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ü Finding value of annualized capital cost of Type B

Annualized cost of capital of type B is solution to equation (2).

In[95]:= Clearansb, cb; ansbNSolve

t1 10 cb 1rt 4106, cb; cbcb. ansb1; Print"cb ", cb cb  650 982.

Annualized cost of capital of type B is $650,982 .

In[97]:= Print"cacb ", cacb

cacb 123 387.

Conclusion: Type A has lower annualized capital cost. We choose type A.

Mathematically, finding value of annualized capital cost is the same the following: Interest rate is given. Value of annuaity is given. You find value of value of constant payment.

ü 6.9 Ranking Mutually Exclusive Projects

If projects under consideration require exclusive use of unique asset, we call them mutually exclusive projects. "Internal Rate of Return" may not be a good measure for ranking such mutually exclusive projects. You should use NPV method. ü Example: Plans with different scales for the same parcel of land

Suppose you have a peace of land. You have two choices. building office building and making parking lot. 1. office building: initial outlay $20×106. You can sell it for $24×106 in one year.

2. parking lot: initial outlay $10,000. You can expect $10,000 per year forever. We like to compare these two alternatives and choose by using IRR and NPV methods. ü Comparison by Using IRR

Net present value of the project of the office building is NPV  20 24

1x . Internal rate of return is an interest rate which make NPV equal to zero. We solve an equation; 20 24

1x  0.

In[98]:=  office building Clearx, ansx, IRRx; ansxNSolve20

24 1x

, x;

IRRxx. ansx1; Print" IRRx", IRRx

IRRx0.2

IRR for the office building is 20% . NPV of the parking lot is equal to t1 10

1yt 10. We solve an equation; t1

 10

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In[101]:=  parking lot Cleary, IRRy, ansy; ansyNSolve t1 10 1yt 100, y;

IRRyy. ansy1; Print"IRRy ", IRRy

IRRy  1.

IRR of Parking lot is 100%. So parking lot has higher IRR. ü Comparison by Using NPV

We want to choose the one which has higher NPV. Suppose that cost of capital is 15%.

In[104]:=  office building ClearNPVx, r; r0.15; NPVx

24 1r

20 106;

Print"NPV of the office building is ", NPVx

NPV of the office building is 869 565.

In[107]:=  Parking lot ClearNPVy NPVyN 

t1

10 1rt

10 103;

Print" NPV of the parking lot is ", NPVy , ", NPVxNPVy ", NPVxNPVy

NPV of the parking lot is 56 666.7, NPVxNPVy  812 899.

As shown above, if cost of capital is 15%, then the office building has higher NPV. You should choose office building.

ü Result can be reversed

However, if cost of capital is higher than 20%, the result is reversed. The result depends on cost of capital. For example, suppose cost of capital is 21%.

ü Cost of Capital = 21%

In[110]:= Clearr, NPV1, NPV2; r0.21; NPV1

24 1r

20 106; Print"NPV of office building when r0.21 : NPV1", NPV1

NPV of office building when r0.21 : NPV1165 289.

In[112]:= NPV2 N 

t1

10 1rt

10 103; Print"NPV of parking lot ", NPV2

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NPV of parking lot 37 619.

Comparison: NPV1NPV2202 908.0

If the cost of capital is 21%, then the parking lot is the project to choose. ü Switch-over Point

When is the result reversed? Where is "switch-over point"? It is 19.7568% as shown below.

In[115]:= Clearr; NSolve 24 1r 20 106 t1 10 1rt 10 103 0, r Out[115]= r0.197568,r0.00253204

How can such a reversal happen? One of the reasons is different lives of the projects. Consider NPV as a function of cost of capital. It is denoted as r in our example. If cost of capital is below 0.2532%, then parking lot has higher NPV again. ü NPV as a function of cost of capital

Let's draw graphs and see how NPV's changeas cost of capital changes.

In[116]:=  NPV of Office Building Clearf1, r, g1 f1r_:

24 1r

20 106;

In[118]:= g1 Plotf1r,r, 0, 0.22, ImageSize200, PlotLabel"NPV of office building"

Out[118]= 0.05 0.10 0.15 0.20 1μ106 2μ106 3μ106 4μ106 NPV of office building

In[119]:= NPV of Parking Lot Clearf2, g2 f2r_:

t1

10 1rt

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In[121]:= g2Plotf2r,r, 0, 0.22, ImageSize200, PlotLabel"NPV of parking lot" Out[121]= 0.05 0.10 0.15 0.20 100 000 200 000 300 000 400 000 500 000 NPV of parking lot

In[122]:= Showg1, g2, PlotLabel"comparison of NPV's"

Out[122]= 0.05 0.10 0.15 0.20 1μ106 2μ106 3μ106 4μ106 comparison of NPV's

ü Homework No. 2, Due next class

Q1. p.146, Problem 36. Hint: interest rate per month = APR

12 . Consider initially there were 13 loans. Sammy paid back 12 of

them in a year.

Q2. Suppose cost of capital is 18%. At what scale would the NPV of the parking lot be equal to the office building? Hint: Quick Check 6-8.

Q3. p190, Problem 1 . Also calculate IRR.

Q4. p.193, Problem 18.

Figure

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