1. Analytical Tools Objectives: After reading this chapter, you will be able to Solve linear and quadratic equations, system of linear equations Use geometric series in financial calculations...

1. Analytical Tools


Objectives:
After reading this chapter, you will be able to

1. Solve linear and quadratic equations, system of linear equations


2. Use geometric series in financial calculations


3. Understand the basic concepts of statistics


4. Use WolframAlpha, Excel, or Maple to solve mathematical problems


5. Understand the concept of optimization

Before we actually start studying finance and the financial management as a discipline, it is worthwhile to review some of the fundamental concepts in mathematics first. This will help us appreciate the usefulness of analytical techniques as powerful tools in financial decision-making. We shall briefly review elementary algebra, basic concepts in statistics, and finally learn Excel or Maple as a handy way to cut through the mathematical details.
Our approach toward learning finance is to translate a word problem into a mathematical equation with some unknown quantity, solve the equation, and get the answer. This will help us determine an exact answer, rather than just an approximation. This will lead to a better decision.
1.1 WolframAlpha

The mathematical software, Mathematica, has the ability to solve a wide range of mathematical problems. Mathematica has a website at WolframAlpha, which is free to use. You should explore it and use for performing many of the mathematical operations. If you click on the button WRA , you will go to the website for WolframAlpha. The next sections will show a variety of problems that you can do at WolframAlpha.
1.2 Linear Equations

To review the basic concepts of algebra, we look at the simplest equations first, the linear equations. These equations do not have any squares, square roots, or trigonometric or other complicated mathematical functions.
Example

1. Suppose John buys 300 shares of AT&T stock at $26 a share and pays a commission of $10. When he sells the stock, he will have to pay $10 in commission again. Find the selling price of the stock, so that after paying all transaction costs, John’s profit is $200.

Let us define profit p as the difference between the final payoff
F, after commissions, and the initial investment
I0, including commissions. We can write it as a linear equation as follows
p =
F
-
I0

We require a profit of $200, thus, p = 200. Suppose the final selling price of the stock per share is
x, the number we want to calculate. Selling 300 shares at
x
dollars each, and paying a commission of $10, gives the final payoff as,
F
= 300x
- 10. The initial investment in the stock, including commission, is
I0 = 300(26) + 10 = $7810. Make these substitutions in the above equation to obtain
200 = 300x
- 10 - 7810
Moving things around, we get 200 + 10 + 7810 = 300x
Or, 8020 = 300x

Or,
x
= 8020/300 = 26.73333333 » $26.73 ?
This means that the stock price should rise to $26.73 to get the desired profit. Note that the answer has a dollar sign and it is truncated to a reasonable degree of accuracy, namely, to the nearest penny.
Consider another problem involving dollars, doughnuts, and coffee.

1. Jane works in a coffee shop. During the first half-hour, she sold 12 cups of coffee and 6 doughnuts, and collected $33 in sales. In the next hour, she served 17 cups of coffee and sold 8 doughnuts, for which she received $46. Find the price of a cup of coffee and that of a doughnut.

This is an example where we have to find the value of
two
unknown quantities. The general rule is that you need two equations to find two unknowns. We have to develop two equations by looking at the sales in the first half-hour and in the second hour. Suppose the price of a cup of coffee is
x
dollars, and that of a doughnut
y
dollars.
First half-hour, 12 cups and 6 doughnuts for $33, gives 12
x
+ 6
y
= 33 Second hour, 17 cups and 8 doughnuts for $46, gives 17
x
+ 8
y
= 46
Now we have to solve the above equations for
x
and
y.
First, try to eliminate one of the variables, say
y. You can do this by multiplying the first equation by 8 and the second one by 6, and then subtracting the second equation from the first. This gives
8*12
x
+ 8*6
y
= 8*33
6*17
x
+ 6*8
y
= 6*46 Subtracting second from first, (8*12 – 6*17)
x
= 8*33 – 6*46 Simplifying it, – 6
x
= – 12, or
x
= 2 ?
Substituting this value of
x
in the first equation, we have 12*2 + 6
y
= 33

Or, 6y
= 33 – 24 = 9
This gives
y
= 9/6 = 3/2 ?
The answer is that a cup of coffee sells for $2 and a doughnut for $1.50. To verify the result, use the following instruction at WolframAlpha. WRA 12*x+6*y=33,17*x+8*y=46

1.3 Non-linear Equations

Non-linear equations contain higher powers of the unknown variable, or the variable itself may show up in the power of a number. For instance, a quadratic equation is a non- linear equation. The general form of a quadratic equation is

ax2 +
bx
+
c
= 0 (1.1) The roots of this equation are
-
b
±
b2 - 4ac


x
= 2a
(1.2)

To verify equation (1.2), use the following instruction at WolframAlpha.
WRA
a*x^2+b*x+c=0

Consider the following examples of non-linear equations.
Examples

1. Solve for
   x: 1.113x
   = 2.678

First, we recall the basic property of logarithm functions, namely,
ln(a
x) =
x
ln
a

Taking the logarithm on both sides of the given equation, we obtain

x
ln(1.113) = ln(2.678)

ln(2.678)
0.9850702 Or,
x
= ln(1.113) = 0.1070591 = 9.201 ?
To verify the result, use the following instruction at WolframAlpha.
WRA
1.113^x=2.678


1.3.
Solve for
x, (2 +
x)2.11 = 16.55
This gives 2 +
x
= (16.55)1/2.11
Or,
x
= (16.55)1/2.11 – 2 = 1.781 ?
The keystrokes needed to perform the calculation on a TI-30X calculator are as follows: 16.55 ? ? 1 ? 2.11 ? ? 2 ?
To verify the result, use the following instruction at WolframAlpha.
WRA (2+x)^2.11=16.55

1. Find the roots of 5x2 + 6x
   - 11 = 0 Here
   a
   = 5,
   b
   = 6, and
   c
   = – 11. This gives us

–6 ± 36 – 4(5)(–11)

x
= 10 =
–6 ± 256
10 =
–6 ± 16
10 = -
11
5 or 1 ?

To verify the result, use the following instruction at WolframAlpha.
WRA
5*x^2+6*x-11=0

1.4 Geometric Series

In certain problems in financial management, we have to deal with a series of cash flows. When we look at the present value, or the future value, of these cash flows, the resulting series is a geometric series. Thus, geometric series will play an important role in managing money. Let us consider a series of numbers represented by the following sequence

a
,
ax
,
ax2 ,
ax3 , ... ,
ax
n-1
The sequence has the property that each number is multiplied by
x
to generate the next number in the list. There are altogether
n
terms in this series, the first one has no
x, the second one has an
x, and the third one has
x2. By this reasoning, we know that the
nth term must have
x
n-1 in it. This type of series is called a
geometric series. Our concern is to find the sum of such a series having
n
terms with the general form

S
=
a
+
ax
+
ax2 +
ax3 + ... +
ax
n-1 (1.3) To evaluate the sum, proceed as follows. Multiply each term by
x
and write the terms on
the right side of the equation one-step to the right of their original position. We can set up the original and the new series as follows: