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 "

*." If the cash flows start at the end of the current period, rather than immediately, it is call an "*

**immediate annuity***."*

**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 t_{}1

12 _{A}_{1}_{}_{r}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 1r PV of the second cash flow = A

1r2 ª

PV of the last cash flow = A
1r12
Then PV of this annuity is given by t_{}1

12 A

1rt .

**** **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]:= **ClearA, r; A** **100; r0.01; Print"** ****
**t1**
**12** _{A}**1rt**
**** **",** ****
**t1**
**12** _{A}**1rt**
****
t1
12 _{A}
1rt
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**

**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 2106 Type B 10 years 4106

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

t_{}1

5 ca

1rt 210
6_{∫}_{ (1)}

Let *cb* be annualized capital cost of type B. It satisfies the following equation.
t_{}1

10 cb

1rt 410

6 _{∫}_{ (2) }

**ü** ** Finding value of annualized capital cost of Type A**

We are going to solve equation (1).

In[93]:= **Clearca, r, ansa; r0.1; ansaNSolve**
**t1**

**5** _{ca}

**1rt**

**2106, ca**

Out[93]= ca527 595.

In[94]:= **caca. ansa1; Print"ca** **** **", ca**

ca 527 595.

**ü** ** Finding value of annualized capital cost of Type B**

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

In[95]:= **Clearansb, cb; ansbNSolve**

**t1**
**10** _{cb}**1rt**
**4106, cb;**
**cbcb. ansb1; Print"cb** **** **", cb**
cb 650 982.

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

In[97]:= **Print"cacb** **", cacb**

cacb 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

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

1x 0.

In[98]:= **** **office building** **Clearx, ansx, IRRx;**
**ansxNSolve20**

**24**
**1x**

**, x;**

**IRRxx. ansx1; Print" IRRx", IRRx**

IRRx0.2

IRR for the office building is 20% .
NPV of the parking lot is equal to t_{}_{1} 10

1yt 10. We solve an equation; t1

10

In[101]:= **** **parking lot** **Cleary, IRRy, ansy;**
**ansyNSolve**
**t1**
**** _{10}**1yt**
**100, y;**

**IRRyy. ansy1; 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** **ClearNPVx, r; r0.15;**
**NPVx**

**24**
**1r**

**20** **** **106;**

**Print"NPV of the office building is ", NPVx**

NPV of the office building is 869 565.

In[107]:= **** **Parking lot** **ClearNPVy**
**NPVyN **

**t1**

**** _{10}**1rt**

**10** **103;**

**Print"** **NPV of the parking lot is ", NPVy , ",** **NPVxNPVy** **** **", NPVxNPVy**

NPV of the parking lot is 56 666.7, NPVxNPVy 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]:= **Clearr, NPV1, NPV2; r0.21;**
**NPV1**

**24**
**1r**

**20** **** **106; Print"NPV of office building when r0.21 :** **NPV1", NPV1**

NPV of office building when r0.21 : NPV1165 289.

In[112]:= **NPV2** **N **

**t1**

**** _{10}**1rt**

**10** **103;**
**Print"NPV of parking lot** **", NPV2**

NPV of parking lot 37 619.

Comparison: NPV1NPV2202 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]:= **Clearr; NSolve**
**24**
**1r**
**20** **** **106**
**** ****
**t1**
**** _{10}**1rt**
**10** **103** **0, r**
Out[115]= r0.197568,r0.00253204

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** **Clearf1, r, g1**
**f1r_:**

**24**
**1r**

**20** **** **106;**

In[118]:= **g1** **Plotf1r,r, 0, 0.22, ImageSize200, 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** **Clearf2, g2**
**f2r_:** ****

**t1**

**** _{10}**1rt**

In[121]:= **g2Plotf2r,r, 0, 0.22, ImageSize200, 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]:= **Showg1, 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.