Chapter 5 Mathematics of Finance and Optimization.pdf

Mathematics
in the Modern
World

Chapter 5:
Mathematics of
Finance and
Optimization

According to Biehler (2008), interest is the “rent”
that a borrower pays a lender to use the lender’s
money. It is paid in addition to the original
amount borrowed. There are various types of
interest rates.
Most commonly used types of interest
• Simple interest
• compound interests

1. Fixed – a specific, fixed interest tied to a loan. This type of
interest is easy to compute, easy to understand and stable
both for the lender and the borrow.
2. Variable – an interest that can fluctuate depending on the
movement of base interest rates. As this interest changes
with time, it may be beneficial to the borrower (if the rate
decreases) or unfavorable (if the rate increases).
3. Annual percentage rate (APR) – the amount of your total
interest expressed annually on the total cost of the loan. This
is usually used by credit card companies when borrowers do
not pay their balance in full.
Definition of terms

4. Fixed – a specific, fixed interest tied to a loan. This type of
interest is easy to compute, easy to understand and stable
both for the lender and the borrow.
5. Variable – an interest that can fluctuate depending on the
movement of base interest rates. As this interest changes
with time, it may be beneficial to the borrower (if the rate
decreases) or unfavorable (if the rate increases).
6. Annual percentage rate (APR) – the amount of your total
interest expressed annually on the total cost of the loan. This
is usually used by credit card companies when borrowers
do not pay their balance in full.

Simple interest is calculated on a yearly basis (annually) and
depends on the interest rate (Karris, 2009). The rate is often given
per annum (p.a.), that is, per year. To calculate interest, the amount
borrowed is multiplied by the interest rate and the amount of time.
The simple interest formula is given by
I = PRT
where
I = the amount of simple interest
P = the principal (amount of money at the start)
R = the interest rate
T = term

Correspondingly, the total amount A to be
received at the end of the term is
A = P (1+RT)
Example 1
Marjorie borrowed ₱5000.00 at 10% simple
interest rate for 6 months. How much must she
pay at the end of the period?

Example 1
Marjorie borrowed ₱5000.00 at 10% simple interest rate for 6 months. How
much must she pay at the end of the period?
Solution:
The interest that Marjorie needs to pay for her
loan is
I = (5000)(0.1)(0.5) = ₽250.
Consequently, at the end of 6 months, she will have
to repay ₱5250.00.
Here, T is equal to 0.5 since 6 months is ½ of a year.

Example 2
Jenny needs to borrow some money for 4 months.
Her neighbor offers to lend her the money at a
simple interest rate of 12%. After 4 months, Jenny
needs to repay ₱10,660.00 to her neighbor. How
much was the money she borrowed?
Solution:
A = P(1 + RT) so that 𝑃 =
10660
1+(0.12)(
1
3
)
= ₱10,250.00

Compound interest includes interest earned on
interest.
In general, we let
A = the total amount to be earned at the end of the term
P = the amount of money borrowed (the principal)
R = the interest rate
n = term (in years).
Then the compound interest formula is
𝐴 = 𝑃 1 + 𝑅 𝑛

If the interest is compounded q times per year,
then
𝐴 = 𝑃〖(1 +
𝑅
𝑞
)〗𝑛𝑞
Example 3
Marjorie borrowed ₱ 5000.00 at 10% compound
interest rate for 6 months. How much must she
pay at the end of the period?
𝐴 = 5000(1 +
0.10
2
)2 = ₱ 5,512.50

Example 4. Determine how much ₱ 5000 will
grow in 10 years at 6% annual compound interest.
Solution:
𝐴 = 5000(1 + 0.06)10
= ₱ 8,954.24

An annuity is a series of payments made at equal
time periods with interest. It should be
emphasized that in annuities, there is a succession
of deposits or a succession of payments made at
equal time periods. Examples of annuities include
payment of social security premium, loans and
mortgages. The term of an annuity is the time
from the start of the first payment period to the
end of the last payment period.

A sum of money to which an annuity’s payments
and interest accumulate in the end is called the
annuity’s future value. A sum of money paid at the
beginning of an annuity to which the annuity’s
payments are accepted as equivalent, is called the
annuity’s present value. Corresponding to future
value, we ask how much money will be accrued,
including interest, if we make regular deposits into
a bank. In present value, we ask how much money
we need to have in the bank now, taking into
account interest, in order to make a sequence of
regular payments in the future.

Example 1
Suppose Jen invests ₱1,000.00 at the end of every
year for 5 years. If the interest rate is 8%, how much
is the worth of the investment after 5 years?
Solution:
Consider the first payment, ₱1,000.00. It remains in the
bank for 4 years, Then its future value is 1000 x (1+0.08)4. The
next deposit is in the bank for 3 years and is worth 1000 x
(1+0.08)3. The next deposit will stay in the bank for 2 years and
is equal to 1000 x (1+0.08)2 . Hence the other 2 deposits will be
worth 1000 x (1+0.08) and 1000.

Note that the last payment does not earn
interest anymore since it is made at the end
of the term.
Correspondingly, the total worth of the
investment at the end of 5 years is
1000 x (1+0.08)4+ 1000 x (1+0.08)3 + 1000 x
(1+0.08)2+1000 x (1+0.08)+1000.

In general, if we let P be the annual rent, n be
the term and i is the interest rate per time interval
is
𝑆 =
𝑃( 1+𝑖 𝑛−1)
𝑖
.
The present value of an annuity, on the other
hand, is computed using the formula
𝑎
𝑛|𝑖=
1−(1+𝑖)−𝑛
𝑖
.

Example 2
What is the amount for an investment of ₱5000
where payments are made quarterly over 5 years at
a nominal interest rate of 6% compounded
quarterly?

5.3 Credit Cards and other Loans

Credit cards are known to be very useful to both
businesses and individual consumers. However, it is
also no secret that letting things get out of control
with credit cards can be disastrous to one’s financial
well-being. Hence, it is necessary for us to be familiar
with how credit cards work.

Credit cards are a convenient and flexible way of paying for
things with borrowed money. Swiping your credit card when
shopping for your favorite brand of shoes or when eating at a
fancy restaurant may not feel like you are taking out a loan,
but that is exactly what you are doing every time you pay
with a credit card. The bank (or other financial institution)
that issued your card pays the department store or the
restaurant on your behalf, and then you pay them back later.
You are borrowing money. People do use credit cards to buy
things that they do not have the money to pay for up front.
There are, though, many other reasons to use a credit card to
pay for something beyond needing to borrow the money to
pay.

Not everyone wants a credit card and not
everyone gets a credit card, too. Debit cards
present an alternative for those who want to
enjoy almost the same convenience that credit
cards offer, except that it has nothing to do with
borrowing money. Purchases made with a debit
card are paid for out of a checking, savings, or
similar account immediately. Since the money
comes directly from your account, you are not
borrowing when you make a payment with one
of these cards. Hence debit cards are easier to
possess.

While debit cards work just like credit cards at the point of
use, they are really quite a different thing. Since no loans are
involved, there is no interest to consider with debit cards
(which makes them a lot less interesting mathematically).
Also, a consumer should be aware that despite their many
resemblances to credit cards, the regulations and consumer
protection laws that apply to credit cards may not apply, or
may apply differently, to debit cards. It is also important to
remember that, while credit cards allow some flexibility as to
when you actually pay for your purchases, with a debit card
the money is withdrawn from your account right away. You
need to have the money in your account at the moment that
you use the card

Types of Credit Card Fraud
In the following discussion, we present
common credit card frauds that can be observed in
the Philippines and what we can do to protect
ourselves from fraud. We understand that the
more educated become educated against these
scams, the more measures we can take to protect
oneself.

Lost or stolen cards  Report loss to the issuing
bank immediately
 Ask bank to block card to
minimize any damage
Account takeover
 Fraudster gets access to
personal information of a
cardholder reports t
issuing bank that the card
was lost or stolen for the
fraudster to secure a new
credit card under the same
account.
 Report to the issuing bank
immediately
minimize any damage

Counterfeit cards
 When a card is “cloned” from
another and then used to make
purchases
immediately
minimize any damage
Collusive merchant
 When merchant employees work
with fraudsters to defraud banks
and customers
immediately
 Ask bank to block card to minimize
any damage
Card-not-present (CNP) fraud
 when your card information (such
as account number and expiry
date) become known to
fraudsters, they can make online
transactions which do not require
your card to be physically
present.
 Scrutinize your bank statements
for these types of charges
immediately

In summary, as cardholders, we should be very
vigilant in all our actions and transactions. If you
connect your phone or laptop to public networks
such as in hotels or malls, be very careful with
entering personal information. Also, keep your
cards safe all the time and do not share personal
information to strangers.

Housing Loans
Before you decide which housing loan to
avail of, it is best to first identify what type of
house you would want to have. The most
commonly seen types in the Philippines are
presented in the following table.

Bungalow -- a single-storey house
with all rooms located in the
same floor (the ground floor)
2-storey – a house with 2 floors
Single-family -- a house that
stands on its own lot and offers
space and privacy to the
homeowner
Multi-family -- a house that is
attached to each other at the
sides. This includes row houses,
apartments, condominium units
and town houses.
Detached – a house with spaces
around it. This usually requires
installation of fence to secure
the whole house.
Attached – a house that is
attached to one boundary of the
lot where the house stands. That
side is most likely to have a
firewall that separates it from the
other house next to it.

It can be seen that some houses in the
Philippines fall under a combination of these
types. Deciding what type of house to build or
buy is necessary before you venture on getting a
housing loan.
There are two types of housing loans in the
Philippines – namely, conventional and
flexible.

a. Conventional -- in this type of housing loan, one
is expected to make fixed payment over a
particular duration. This type of loan is ideal for
someone who is looking for a predictable payment
scheme that he can plan ahead for.
b.Flexible – this type of loan is often available in
banks. Here one can control the interest rate
depending on how much he or she will deposit.

Bonds
A bond is a written contract by a debtor to pay the
final redemption value on an indicated redemption
date, or maturity date, and to pay a certain sum
periodically. The typical bond mentions a borrowed
principal called the face value or par value of the bond,
and describes the periodic sum as the payment of
interest on the principal at a specified nominal rate
called the bond rate. In many cases, a small dated
coupon is attached to the bond corresponding to each
payment of periodic sum.

School Building Bond of 2016
City School District I
Parañaque, Philippines
December 23, 2016
The City School District I of Parañaque, Philippines acknowledges itself to owe and, for value
received, promises to pay the bearer SIX HUNDRED THOUSAND PESOS on December 23, 2019, with interest
on said sum after December 23, 2016, at the rate of 4.3% per annum, payable semi-annually, until the
principal is paid. Furthermore, all such sums are payable at the main office of the Metro National Bank of
the Philippines. Signed by authority of the School Board of said school district.
Hernan Luna ___
Chairman of the School Board
No. 6 ₱12, 900
On December 23, 2019, City School District I,
Parañaque, Philippines, will pay the bearer at the
Metro National Bank of the Philippines the sum of
Twelve Thousand and Nine Hundred Pesos for
interest due then on its School Building Bond of
2016 dated December 23, 2016.
___ Hernan Luna ___
No. 5 ₱12, 900
On June 23, 2019, City School District I,
interest due then on its School Building Bond of 2016
dated December 23, 2016.
___ Hernan Luna ___

No. 4 ₱12, 900
___ Hernan Luna ___
No. 3 ₱12, 900
___ Hernan Luna ___
No. 2 Bond 136 ₱12, 900
___ Hernan Luna ___
No. 1 Bond 136 ₱12, 900
___ Hernan Luna ___

Most bonds promise coupon payments semi-
annually and are redeemable at par. The bond
illustrated above has a face value of ₱600, 000.
The redemption value is ₱600,000, payable on
the redemption date December 23, 2019. The
bond rate is 4.3%. Each coupon payment is
2.15% of ₱400,000 or periodic payment is
₱12,900, payable semi-annually. This bond is
₱600,000, 4.3% bond.

To find the price of a bond on a coupon date to yield a
specified investment rate, when the coupon period is
equal to the investor’s interest period:
𝑃𝑟𝑖𝑐𝑒 = 𝑉 1 +
𝑗
𝑚
−𝑚𝑡
+ 𝐻
𝑟
𝑚
1 − 1 +
𝑗
𝑚
−𝑚𝑡
𝑗
𝑚
where:
V = redemption payment j = nominal rate
r = bond rate H = face value
m = period t = time in years

Example 1
₱50,000, 6% bond pays coupons semi-annually and
it will be redeemed on August 05, 2016. Find the
price if the bond is bought on August 05, 2012 to
yield 4.5% compounded semi-annually, (a) if the
bond is redeemable at par; (b) if the bond is
redeemable at 105%.
Given:
H = ₱50,000 t = 4 yrs
r = 0.06 m = 2
j = 0.045 V = (a) ₱50, 000 (b) = ₱52, 500
Required: P = (a) if the bond is redeemable at par
(b) if the bond is redeemable at 105%

Solution (a):
𝑃 = 𝑉 1 +
𝑗
𝑚
−𝑚𝑡
+ 𝐻
𝑟
𝑚
1 − 1 +
𝑗
𝑚
−𝑚𝑡
𝑗
𝑚
𝑃 = 50,000 1 +
0.045
2
−(2)(4)
+ 50,000
0.06
2
1 − 1 +
0.045
2
−(2)(4)
0.045
2
𝑃 = ₱41,846.91732 + 10,870.77691
𝑃 = ₱52,717.69423

Solution (b):
𝑃 = 𝑉 1 +
𝑗
𝑚
−𝑚𝑡
+ 𝐻
𝑟
𝑚
1 − 1 +
𝑗
𝑚
−𝑚𝑡
𝑗
𝑚
𝑃
= 52,500 1 +
0.045
2
−(2)(4)
+ 50,000
0.06
2
1 − 1 +
0.045
2
−(2)(4)
0.045
2
𝑃 = ₱43,939.26318 + 10,870.77691
𝑃 = ₱54,810.04009

Stocks
A stock is a written contract that gives the holder
ownership of a portion of a company and the right to receive
portion of the company’s income that might be distributed.
The owner of stocks is called a stockholder. Some of the
companies give portion of the company’s income to their
stockholders. This portion is called dividend. There are two
ways that a stockholder can gain: (a) through dividends (if
the stocks pay dividends) and (2) through capital gain or
capital appreciation – it is the difference between the price
you sold the stock to others and the price at which you
bought the stock.

Example 1
Sheena bought 25 shares of MISORTEL for ₱3,200
per share in 2013. ₱800 was paid as dividends in
2013, ₱850 was paid as dividends in 2014 , and
₱900 was paid as dividends for 2015. At the end of
2015, the closing price of MISORTEL is ₱3650.
Since Sheena has 25 shares of MISORTEL and its
price per share is ₱3200, then the value of her
MISORTEL stocks is (25)( ₱3,200)= ₱80,000.

Since the stock is paying dividends, then you have two
sources of gain:
1.Dividends: ₱800 in 2013, ₱850 in 2014 and ₱900 in
2015
2.Capital gain: The value of the stock at the end of
2015 is (25)( ₱3,650)= ₱91,250.
Thus, in value, Sheena gained ₱91,250 - ₱80,000 =
₱11,250.
*Total Gain=Dividends + Capital Gain = ₱2,550 +
₱11,250 = ₱13,800

Types of Stocks
1. Preferred Stocks – have a par value and dividend rate. In terms of
rights, the claim of preferred stockholders is above that of the
claim of common stockholders but below that of the entities the
company has debt.
2. Common Stocks – the company issuing common stocks may or
may not pay dividends to its stockholders. Also, future dividends
are uncertain. The amount of dividends is mainly dependent on
the economic performance of the company. In terms of rights, they
have higher voting rights than preferred stockholders. However,
in case of bankruptcy, they are the last to be paid.

Example 2
As a new trader, Argie bought a common stock, 18
shares of CEPALCO at ₱350 per share. He heard
from some investors that being a utility company,
CEPALCO is estimated to be growing in value by
2.9% every year. Hence, he expects to receive
dividends which also grow at 2.9% each year. What
is the value of Argie’s CEPALCO stocks? If ₱25 was
paid as dividend today, find the amount of dividend
Argie will be expecting in the next 5 years.

Year Dividends (₱)
1 (₱25)(1.029)= ₱25.725
2 (₱25.725) (1.029)= ₱26.471025
3 (₱26.471025) (1.029)= ₱27.23868473
4 (₱27.23868473) (1.029)= ₱28.02860658
5 (₱28.02860658) (1.029)= ₱28.84143617
Solution:
The value of the stock is (18)( ₱350)= ₱6,300.
Since the expected dividends grow by 2.9% each year, then
the dividends for the next 5 years are:

Mutual Funds
A mutual fund is a collection of stocks, bonds,
money market instruments, and other assets.
When an investor buys a mutual fund, he owns
part of the mutual fund company and its assets.
The price of each mutual fund units or share is
expressed as the Net Asset Value (NAV).

To find the value of NAV, the formula is given:
NAV =
𝐹−𝐿
𝑂
Where:
F = Fund Assets L = Fund Liabilities O= Outstanding Shares
To determine the number of shares in an investment, the formula
is given:
No. of Shares =
𝑇𝑜𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
𝑁𝐴𝑉

Example 1
A mutual fund has 58,000,000 worth of stocks,
25,000,000 worth of bonds, and 35,000,000 in other
assets. The fund’s total liabilities amount to 10,000,000
and there are 7,000,000 outstanding shares. An
investor plans to invest 30,000 in this mutual fund.
(a)What is the Net Asset Value (NAV) of the mutual
fund?
(b)How many shares will the investor have?

Solutions: (a)
NAV =
𝐹−𝐿
𝑂
=
58𝑀+25𝑀+35𝑀 −10𝑀
7𝑀
= ₱15.42
(b)
No. of Shares =
𝑇𝑜𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡
𝑁𝐴𝑉
=
₱30,000
₱15.42
= 1,945 shares

5.5 Introduction to Linear
Programming

It is a known fact that resources around us are
limited. Hence, it is everyone’s goal to make the most
out of them. For example, we all want to use our
time productively or be able fully utilize the little
that we have.
Linear programming (LP) is (Eiselt & Sandblom, 2007)
is a subset of mathematical programming which is part of
operations research (OR). OR, on the other hand, is also
known as management science (MS) is a discipline that
deals with the optimization and control of systems. Thus,
OR is synonymous to optimization, and hence, is very
much used in economics.

In LP, we either maximize the output for some
given input, or minimize the input for some
required output. For example, you use LP when
you want to find the shortest path to take when
you are driving home to school, and thus
minimize travel time. Businessmen also make use
of LP when they try to maximize their profit given
their limited resources including capital.
LP can be used to solve various problems around us.
Aside from being used to solve business problems, it can
also be used to solve optimization problems in food,
agriculture, engineering, education and transportation.

Formulating an LP problem.
Mathematical models need to be formulated in order to solve real-
world problems. In what follows, we define some of the terminologies
used in linear programming (Kolman and Beck, 1995).
a. Decision variables or unknowns. These are mathematical symbols
which represent the values which describe the optimal allocation of the
scarce resources represented by the model. It is from which decisions
are based.
b. Parameters. These are numerical values that are included in the
objective functions and constraints.
c. Objective function. This function represents the goal of the problem
being solved. This could either be maximizing or minimizing some
value and is usually denoted by Z.
d. Constraints. This is a function that defines the condition or restriction
that the decision variables can assume.

Example 1
Geran Go, a budding businessman, is embarking on a housing rental
business project. These consists of building apartments and single-
family homes in Cagayan de Oro City. Demand for apartment and
single-family homes by potential renters is projected at 10 units each
for every month. Geran only has 14.5 million pesos to finance the
entire construction. A unit of apartment requires 500,000 pesos while a
unit of single-family home needs 950,000 pesos. Once rented, each
apartment makes 10,000 pesos in profit while a unit of single-family
home give a profit of 15,000 pesos.
Geran would want to find out how many units of apartments
and single-family homes should be constructed to maximize profit.

Demand Cost of construction Profit
x 10 500,000 10,000
y 10 950,000 15,000
Available resource 14.5 million
Solution:
Let x and y denote the number of units of apartments and single-
family homes, respectively, to be constructed. These correspond to
the decision variables.
To better understand the problem, we tabulate the given
information as follows.

The relationship between the cost of construction
and the available budget can be described as
500000𝑥 + 950000𝑦 ≤ 14500000 (since 14.5
million is equivalent to 14,500,000)
while demand for both apartment and single-family
homes is formulated as
𝑥 ≤ 10, and 𝑦 ≤ 10.
An additional constraint requires that all decision
variables be nonnegative, that is
𝑥 ≥ 0 and 𝑦 ≥ 0.

The profit that Geran might receive from the housing
rental business, which is to be maximized, is given
by
𝑍 = 10000𝑥 + 15000𝑦.
Thus, our mathematical model is
maximize 𝑍 = 10000𝑥 + 15000𝑦
subject to
500000𝑥 + 950000𝑦 ≤ 14500000
𝑥 ≤ 10
𝑦 ≤ 10
𝑥 ≥ 0
𝑦 ≥ 0

The standard form of an LP model is

Protein (g) Carbohydrates (g) Fat (g) Cost per serving (PhP)
A 6 2 3 10
B 3 3 4 12
Example 2
A nutritionist is planning a lunch menu consisting of two main foods,
A and B, for a 19-year old student. These foods contain the following
nutrients.
The nutritionist wants the meal to provide at least 24 grams of protein,
at least 12 grams of carbohydrates and at most 15 grams of fat.
Moreover, the student only has a budget of 100 pesos for the meal.
The nutritionist wants to determine how many servings of each
food the student should take to meet the required nutrients while
minimizing total cost.

Solution:
Let
x be the number of servings of food A that the student should take
y be the number of servings of food B that the student should take.
Then the LP model for this problem can be formulated as
minimize 𝑍 = 10𝑥 + 12𝑦
subject to
6𝑥 + 3𝑦 ≥ 24
2𝑥 + 3𝑦 ≥ 12
3𝑥 + 4𝑦 ≥ 15
10𝑥 + 12𝑦 ≤ 100
𝑥 ≥ 0
𝑦 ≥ 0

Graphical Method
A linear programming problem involving
only two decision variables can be solved using a
graphical solution procedure.
To plot sets of constraints,
a.Change inequalities to equalities.
b.Plot the equalities.
c. Identify the correct side of the line for the
original inequalities.

Various graphing software applications are now available in
order to find optimal solutions of LP problems. In the
discussion that follows, solutions are obtained using
https://www.desmos.com/calculator.
To solve the problem
maximize 𝑍 = 10000𝑥 + 15000𝑦
subject to
500000𝑥 + 950000𝑦 ≤ 14500000
𝑥 ≤ 10
𝑦 ≤ 10
𝑥 ≥ 0
𝑦 ≥ 0

using desmos, we obtain the following
The set of solutions for this problem are those
values which are in the intersection of the lines
defined by the constraints. Since the problem is a
maximization problem, we find the values of (x,
y) which makes Z maximum. This is shown in
Figure 5.1.

Figure 5.1. Graphical
solution to problem in
Example 1.

The points which are on the corners of the region
determined by the intersection of the lines
defining the constraints are the candidate optimal
solution for this problem, namely (0,10), (10,10),
(10,0) and (0,0). Since the problem is looking for
the values of x and y which will maximize the
profit, we substitute these values to 𝑍 =
10000𝑥 + 15000𝑦. It can be seen that (10,10) will
give the largest value for the profit Z.

Chapter 5 Mathematics of Finance and Optimization.pdf

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