Categories
Business Mathematic

Stochastic Process

A stochastic or random process, a process involving the action of chance in the theory of probability. It is a mathematical entity that is typically known as a random variable family. Time series can be used to describe several stochastic processes. Notwithstanding, a stochastic process is commonly ceaseless while a period arrangement is a bunch of perceptions recorded by whole numbers. A stochastic cycle may include a few related arbitrary factors. For example, every atom is subject to a fixed probability of breaking down at any given time interval in radioactive decay. The growth of a bacterial population, an electrical current fluctuating due to thermal noise, or the movement of a molecule of gas are typical examples. As mathematical models of structures and phenomena that tend to differ in a random way, stochastic processes are commonly used. They are used in many areas, such as biology, chemistry, ecology, neuroscience, physics, image processing, signal processing, control theory, information theory, computer science, cryptography, and telecommunications.

Albeit stochastic process hypothesis and its applications have gained extraordinary ground as of late, there are still a ton of new and testing issues existing in the regions of hypothesis, examination, and application, which cover the fields of stochastic control, Markov chains, recharging measure, actuarial science, etc. By using more advanced theories and instruments, these problems warrant further research. In order to refer to a stochastic or random process, the term random function is often used, since a stochastic process can also be viewed as a random element in a function space. The terms stochastic process and arbitrary interaction are utilized reciprocally, regularly with no particular numerical space for the set that records the irregular factors.

A stochastic process more commonly refers to a family of random variables indexed against some other variable or variable set. It is one of the most general subjects of probability analysis. Some fundamental sorts of stochastic processes incorporate Markov measures, Poisson measures, (for example, radioactive rot), and time arrangement, with the record variable alluding to time. For example, by its state space, its index set, or the dependency among the random variables, a stochastic process can be categorized in various ways. The cardinality of the index set and the state space are one common method of classification. At the point when deciphered as time, if the file set of a stochastic interaction has a limited or countable number of components, for example, a limited arrangement of numbers, the arrangement of whole numbers, or the regular numbers, at that point the stochastic cycle is supposed to be in discrete time. If the index set is any real line interval, then time is said to be continuous.

Example of Stochastic Process

The two types of stochastic processes are called stochastic processes of discrete-time and continuous-time, respectively. Discrete-time stochastic processes are considered easier to study because more advanced mathematical techniques and expertise are needed by continuous-time processes, particularly because the index set is uncountable. On the off chance that the file set is the whole numbers, or some subset of them, at that point, the stochastic process can likewise be known as an arbitrary grouping. The number of customers (X) in a checkout line is one example of a stochastic process that evolves over time. As time shifts, so do X customers, one or more at a time, come and go. If time is sampled at short intervals, X can fluctuate a little (say, one second). For longer periods, variations in X would be even larger. For example, if t is measured every hour, then the number of customers could change by tens, hundreds, or even thousands at a time.

The word stochastic in English was originally used as an adjective from a Greek word meaning “pertaining to conjecturing” with the description “to aim at a mark, guess” and the Oxford English Dictionary gives the year 1662 as its earliest occurrence. In a 1934 paper by Joseph Doob, the term stochastic process first appeared in English. The concept of a stochastic process differs, but historically a stochastic process is defined as a set of random variables indexed by some set. The terms irregular cycle and stochastic process are viewed as equivalent words and are utilized conversely, without the file set being exactly determined. Both “collection” and “family” are used, while the terms “parameter set” or “parameter space” are often used instead of “index set.”

In particular, the emphasis will be on applications of stochastic processes as key technologies in different fields of study, such as chains of Markov, renewal theory, control theory, nonlinear theory, queuing theory, risk theory, engineering of communication theory, and traffic engineering. The law of a stochastic process or an irregular variable is likewise called the likelihood law, likelihood circulation, or dissemination. With regards to the numerical development of stochastic processes, the term routineness is utilized while examining and expecting certain conditions for a stochastic process to determine conceivable development issues. For example, it is assumed that the stochastic process adheres to some sort of regularity condition, such as the sample functions being continuous, to research stochastic processes with uncountable index sets.

 

Information Sources:

  1. britannica.com
  2. hindawi.com
  3. wikipedia
Categories
Business Mathematic

Dynamic Network

Dynamic network analysis is surely an emergent scientific field that includes traditional social network analysis, link investigation, social simulation along with multi-agent systems within network science along with network theory. You can find two aspects in this field. The first is the statistical analysis involving DNA data. The second is the utilization of simulation to address issues of network dynamics.

Categories
Business Mathematic

Discuss on Law of Sines

Broad objective of this article is to Discuss on Law of Sines. Here explain Law of Sines with examples. You can use the Law of Cosines discussed within the last section to solve standard triangles, but only within certain conditions. The formulas which is to be developed in this part provide more flexibility within solving these general triangles. Law of Sines is appropriate for obtuse triangles as well as acute and right triangles, considering that the value of the sine is positive in both first and second quadrant—that will be, for angles less as compared to 180°.

Categories
Business Mathematic

Define and Discuss on Angles

Principle objective of this article is to Define and Discuss on Angles. An angle is usually a measure of rotation. Angles are measured in degrees. One complete rotation is actually measured as 360°. Angle measure could be positive or negative, based on the direction of rotation. The angle measure is the volume of rotation between the two rays forming the perspective. Rotation is measured in the initial sideto the terminal side from the angle.

Categories
Business Mathematic

Analysis Functions of Acute Angles

Basic purpose of this article is to Analysis Functions of Acute Angles. The nain characteristics of similar triangles, originally formulated by Euclid, are the inspiration of trigonometry. Euclid’s theorems state if two angles of one triangle have the same measure as two perspectives of another triangle, then your two triangles are identical. Also, in similar triangles, position measure and ratios connected with corresponding sides are maintained. Because all right triangles contain a 90° angle, all right triangles that have another angle of equal measure should be similar.

Categories
Business Mathematic

Role of Business Mathematics

Introduction:

The role of mathematics in Business decisions has very important in the process of managerial decision models and algorithms. To turn to the specific aspects of the quantitative decision making process, it is possible to recognize three distinct phases in every decision situation. First is carefully defined the problem, second is a conceptual model to be generated and third is the selection of the appropriate quantitative model they may lead to a solution. Lastly a specific algorithm is selected. Algorithms are the orderly delineated sequences of mathematical operations that lead to a solution. The algorithms generate the decision which is subsequently implemented managerial action program. The entire process is shown below:

Defined problem ——> Conceptual model —> Quantitative Model —-> Algorithms—-> Decision —- Action programs.

You are not being ripped off? You need to use math to calculate compound interest rates (to see how much your savings can grow). You also need to use math to understand the monthly percentages, which are added to your credit cards or bank loans, or you could end up paying Rs10,000 in 5 year’s time for borrowing Rs2,000 today! This is a good reason to understand mathematics

Criticism on mathematics

Mathematics is not a closed intellectual system, in which everything has already been worked out. There is no shortage of open problems. Mathematicians publish many thousands of papers embodying new discoveries in mathematics every month.

Mathematics is not numerology, nor is it accountancy; nor is it restricted to arithmetic. Pseudo mathematics is a form of mathematics-like activity undertaken outside academia, and occasionally by mathematicians themselves. It often consists of determined attacks on famous questions, consisting of proof-attempts made in an isolated way (that is, long papers not supported by previously published theory). The relationship to generally accepted mathematics is similar to that between pseudoscience and real science. The misconceptions involved are normally based on misunderstanding of the implications of mathematical rigor; attempts to circumvent the usual criteria for publication of mathematical papers in a learned journal after peer review, often in the belief that the journal is biased against the author; lack of familiarity with, and therefore underestimation of, the existing literature. The case of Kurt Heegner’s work shows that the mathematical establishment is
neither infallible, nor unwilling to admit error in assessing ‘amateur’ work. And like astronomy, mathematics owes much to amateur contributors such as Fermat and Mersenne. ‘Modern mathematics’ is indeed boring and devoid of meaning in most papers. That’s why I stick to reading the works of the greats like von Neumann, Wiener, Einstein, and hosts of others. What made them great? They explain things, and then go on to carry out tremendously complicated calculations. I think the advent of calculators and numerical methods has hurt the advance and
understanding of math to a large degree. However, on the flip side, new symbolic methods allow computations, which are a great, help and lead to new relations not previously seen, so maybe there is hope yet. But, apart from that fact, I wouldn’t say modern mathematics is particularly boring. In fact, I think it’s the second most exciting thing in the universe! The first is, of course, the mathematics of the future. The problem is just that most writers of mathematics succeed, against all odds, at making the subject seem boring. They’ve developed a lot of methods for
doing this. One is to make the results hard to understand.

Another is to not provide enough contexts for people to see why the results are interesting. A third is to write in a style that has all the drama and flair of overcooked porridge.

Role of mathematics in business:

 

Mathematics is used in most aspects of daily life. Many of the top jobs such as business consultants, computer consultants, airline pilots, company directors and a host of others require a solid understanding of basic mathematics, and in some cases require a quite detailed knowledge of mathematics. It also play important role in business, like Business mathematics by commercial enterprises to record and manage business
operations. Mathematics typically used in commerce includes elementary arithmetic, such as fractions, decimals, and percentages, elementary algebra, statistics and probability. Business management can be made more effective in some cases by use of more advanced mathematics such as calculus, matrix algebra and linear programming.

Commercial organizations use mathematics in accounting, inventory management, marketing, sales forecasting, and financial analysis.

In academia, “Business Mathematics” includes mathematics courses taken at an undergraduate level by business students. These courses are slightly less difficult and do not always go into the same depth as other mathematics courses for people majoring in mathematics or science fields. The two most common math courses taken in this form are Business Calculus and Business Statistics. Examples used for problems in these courses are usually real-life problems from the business world.An example of the differences in coursework from a business mathematics course and a regular mathematics course would be calculus. In a regular calculus course, students would study trigonometric functions. Business calculus would not study trigonometric functions because it would be time- consuming and useless to most business students, except perhaps economics majors. Economics majors who plan to continue economics in graduate school are strongly encouraged to take regular calculus instead of business calculus, as well as linear algebra and other advanced math courses.

Other subjects typically covered in a business mathematics Curriculum include:

Matrix

Algebra
Linear  programming
Probability theory

Another meaning of business mathematics, sometimes called commercial math or consumer math, is a group of practical subjects used in commerce and everyday life. In schools, these subjects are often taught to students who are not planning a university education. In the United States, they are typically offered in high schools and in schools that grant associate’s degrees.

A U.S. business math course might include a review of elementary arithmetic, including fractions, decimals, and percentages. Elementary algebra is often included as well, in the context of solving practical business problems. The practical applications typically include checking accounts, price discounts, markups and markdowns, payroll calculations, simple and compound interest, consumer and business credit, and mortgages. The emphasis in these courses is on computational skills and their practical application, with practical application predominating. For example, while computational formulas are covered in the material on interest and mortgages, the use of prepared tables based on those formulas is also presented and emphasized. Mathematics can provide powerful support for business decisions. In their later business careers, this will motivate them to consult with mathematicians and employ effective quantitative methods.

Mathematics provides many important tools for economics and other business fields. However, our discipline does not profit from this work when students (who later become part of the general public) are unaware of its existence. Presenting trivial mathematical applications only makes matters worse, since they are clearly recognizable as being of little importance. This actually diminishes our subject in the eyes of students. Using computers to bring the underlying structure of significant mathematics to undergraduates allows them to appreciate the role that our Subject can play in their academic work and later lives. The recognition of its importance by many students each year will certainly strengthen the position of mathematics in our society. Why do business consultants and directors need to know math?” you may ask. Business is all about selling a product or service to make money. All transactions within a business have to be perfectly measure through the mathematics terms.

Role of Business Mathematics

Categories
Business Mathematic

Applications of Mathematics in Business

Introduction

Mathematicians, economists, statisticians and others have applied their skills to manage problems for many years. The first concerted and dramatically successful effort in their areas occurred during World War II when specialists were formed into operations analysis groups to assist in planning military operations. These analysts used mathematics and statistics extensively in their studies, and their resulting recommendations were an important contribution to the war effort. Following the war, analysts, soon joined by others, turned their attention and their new techniques to attacking problems of manage mental operations. They accomplished major improvement in inventory control, quality control, warehouse location, industrial operations, agriculture, purchasing decisions, scheduling complex tasks and a variety of others.

A Real World Case

Food is related to the primary needs of human being. To fulfill this need we cultivate different types of crops and make a variety of foods such as bread, biscuits, soft drinks, tea, coffee, and different sorts of junk foods. For making these foods many restaurants, hotels, bakeries etc. are established in a large number.

A few days back, we visited such a type of food making factory to gather different information of their production to apply mathematical calculations which we learnt from our course- ‘Business Mathematics I.’ When we visited the factory we were welcomed warmly by its’ owner and workers. They helped us by giving a variety of information we asked them.

A brief description of the factory and information supplied by its’ owner and workers are given below:

  1      Name of the factory: Rana Bread

  2      Address: 10/B Vagalpur Lane, Hazaribag, Dhaka-1209

  3      Owner: Md. Ariful Qader Rana

  4      Date of establishment: November 2

5       Type of business: Small industry

  6       Fixed Cost:

Description

 

Amount (Tk.)

Land 50000
Machinery 40000
Equipment 20000
Vehicle 24000
Building 14000
Gas line  5200
Others 1300
Total 154500

 7   Variable cost per unit: Tk. 5

  8   Selling price per unit: Tk. 15

  9   Financing method:

Description Amount (Tk.)
Own 54500
Loan from relatives 40000
Bank loan 60000
Total 154500

 → Loan from relatives is @ 10% simple interest

→ Bank loan is @ 12% compound interest

  10   Units sold (last 6 months):

Name of the month Units
November 2009 16000
December 2009 17500
January 2010 18000
February 2010 20000
March 2010 22000
April 2010 25000
Total 118500

 → 8% of the volume of sales is selling expense.

11.     Usage of gas ( last 6 months):

Name of the month Cu.Feet Units
November 2009 1600 16
December 2009 1700 17
January 2010 1800 18
February 2010 2000 20
March 2010 2200 22
April 2010 2500 25
Total 11800 118

→ Rana Bread took the gas line from Titas Gas Co. which charges its’ customers- Tk. 350 as customer charge; Tk. 2.5 per unit (100 cubic feet) for the first 20 units; and Tk. 1.25 per unit each unit over 20.

12.  Depreciation of machinery: Rana Bread depreciates its’ machinery in value 10% applied to the original

13.  Depreciation of equipment: Rana  Bread depreciates its’ equipment in value, 15% in November 2009,               12.5%   in December 2009, 10% in January 2010 and so on; all percentages applying to the original cost.

After visiting Rana Bread we came to know about some important information on how one can make his/her desire come true. Rana Bread, though established a few months ago, is running its’ production and marketing activities quite impressively and making profit in an increasing percentage.

Application of mathematics in Rana Bread

There is currently a strong trend toward incorporating technology into the teaching of mathematics. Electronic tools include basic calculators, programmable calculators,  graphic  calculators and computer algebra system (CAS). The  effective use of this technology allows the individual to focus more on application, problem formulation, and the interpretation of numerical and graphical results.

As we said earlier, our main aim is to apply mathematics in business to implement our theoretical knowledge practically, so now we are trying our best to apply different mathematical terms in the information of Rana Bread which we got during the our visit to it.

Break-even interpretation I

From the case e came to know that,

                Fixed cost= Tk.154500

                Variable cost= Tk. 5 per unit

                Selling price= Tk. 15 per unit

We denote the number (quantity) of breads made and sold by q. Then since bread are sold at Tk.15 per unit, the revenue function is

                                                R (q) = 15q

Now since the variable cost per unit is Tk.5, the variable cost of making q units will be 5q. Adding the fixed cost of Tk. 154500, the cost function becomes

                                                C (q) = 5q+154500

Lastly, since profit is revenue minus cost, the profit function is

                                                      P (q) = R (q) – C (q)

                                                                 = 15q – (5q + 154500)

                                                                 = 10q -154500

Notice that the coefficient 10 of q represents the profit per unit and is precisely the difference between the selling price per unit of Tk.15 and the variable cost per unit of Tk. 5. Similarly, the constant term of

-154500 indicates that the manufacturer has to produce and sell enough units of breads to compensate for the fixed cost of Tk.154500 before breaking even.

One way for the owner of a business to make sure he/she is making a profit is to first analyze the circumstances necessary to break even (i.e. neither make a profit nor suffer a loss). Then the owner can develop a plan to do better than break-even (i.e. make a profit). Now to break-even simply means the revenues must equal costs, so since

                                                                Profit = Revenue – Cost

Then at break-even

Revenue = Cost

Means that

                                                                Profit = 0

 So, at break-even profit will be 0. Thus

                                                                     P (q) = 0

                                                                → 10q – 154500 = 0

                                                                → q = 154500÷10 units

                                                                → q = 15450 units

Finally, the break-even taka volume of sales  (revenue) is

                                                                R (15450) = 15×15450

                                                                                    = Tk. 231750

Break-even chart I

Returning to the revenue and cost functions of Rana Bread,

                R (q) = 15q and C (q) = 5q+154500

And plotting these in the usual manner, we obtain the graphs in Figure-1. Observe that the revenue line goes through the origin and the cost line has a y-intercept of Tk. 154500.

Figure-1

The dotted horizontal line shows that the fixed cost is constant at all levels of operation of the factory. The variable cost which is the 5q in the cost function is the vertical distance from the fixed cost line to the total cost line; it increases as more units are produced. To the left of the break-even point (15450, Tk. 231750), the cost line is above the revenue line; the vertical separation at any points represents the associated loss. On the other hand, to the right of the break-even point (15450, Tk.231750), the vertical separation represents the associated profit.

Break-even interpretation II

Production managers and other operations executives tend to think of break-even analysis in the way that we presented in the previous section. Controllers and other financial managers are more likely to think in accounting terms. To illustrate this latter way of viewing break-even, we shall consider that Rana Bread produces and sells its’ bread at a price that is presumably above the cost.

From the case of Rana Bread stated earlier we came to know that

                                Total cost = Fixed cost + Variable cost

                                                    = 154500+5(118500)

                                                    = 747000

                                Total selling price = 15×118500

                                                                   =1777500

As we know that mark-up is the difference between the selling price and the cost, so

                                Mark-up = Selling price – cost

                                                  =1777500-747000

                                                  =10305000

Now, as margin is the mark-up as a percentage of the selling price, so margin of Rana Bread will be

                                Margin  = Mark-up ÷ Selling price

                                                =10305000 ÷ 1777500

                                    =0.58 or 58%

This means that 58% of the selling price of Tk. 1777500 is margin, and the other 42% of Tk. 1777500 which is,

                                                0.42×1777500=746550,

                                                                                                is the cost.

We, now, suppose that Rana Bread uses a margin of 58% on all items it produces, so that if it sells Tk.s – worth of products, 58% of this amount is margin and 42% is the cost.

Thus,

                                Cost of goods sold = 0.42s

In the case

                                Selling expense  = 8% of sales

                                                                = 0.08

                                Fixed cost = 154500

So the total cost function is

                                                                C(s) = 0.42s+0.08s+154500

                                                                        =0.50s+154500

Now, if Rana Bread sells Tk.s –worth of products, its’ revenue will clearly be the same Tk.s. So that the revenue function is

                                                R(s) = s

Since profit is still revenue minus cost, the profit function is

                                                P(s)  = R(s)-C(s)

                                                        = s-(0.50s+154500)

                                                        = 0.50s-154500

Notice that the coefficient 0.50 of s represents the profit per taka of sales, it is precisely the difference between each Tk. 1.00 of sales and the associated Tk.0.50 cost per taka of sales. Alternatively, the 0.50 can be viewed as the difference between the Tk. 0.58 margin per taka of sales and the Tk. 0.08 selling expense per taka of sales. As before, the constant term of -154500 indicates that Rana Bread has to sell Tk. 154500 worth of products to compensate for the fixed cost of Tk. 154500.

Now, as before, break-even will occur when profit is 0. That is

                                                                    P(s) = 0

                                                                →0.50s-154500=0

                                                                →s=309000

 So, Rana Bread must target a sales of at least Tk. 309000 just to be assured of breaking even.

Break-even chart II

Returning to the revenue and cost functions of Rana Bread

                                                R(s)=s and C(s)=0.50s+154500

and plotting this in the usual manner, we obtain the graphs in Figure-2.

Revenue & CostFigure-2

The point of intersection (Tk. 309000, Tk. 309000) of revenue function R(s) and the cost function C(s) establishes the break-even level of sales. The separation of lines to the right of break-even indicates profit. On the other hand, the separation of lines to the left of break-even shows loss.

Piecewise linear functions

Sometimes an equation will consist of more than one section or piece, each of which is linear. For example,

                                          x    ;     x≤0

                                y=   2x   ;     x>0

describes a  function composed of two linear pieces, the first being x for all values of x≤0, and the second being 2x for all values of x>0. Such a function is a piecewise linear function and its’ graph is shown in Figure-3.

Figure-3

The charge for the first 20 units consists of the Tk. 350 customer charge plus the Tk. 2.5 charge per unit. Letting u represent the number of units used, this charge becomes

                                2.5u+350…………………….(1)

Now if Rana Bread uses more than 20 units, the charge from (1) for the first 20 units will be

                                2.5(20)+350=400

To this must be added the charge of Tk. 1.25 for each unit over 20, resulting in a charge of

                                   400+1.25(u-20)

                                = 400+1.25u-25

=1.25u+375……………………                                                                                                                                                                                                                  Putting (1) and (2) together, we get the cost function

                                                C(u) = 2.5u+350;     0≤u≤20

                             C(u)=   1.25u+375;     u>20

          Figure-4 graphs this function.Figure-4

Cost of November 2009, when 16 (<20) units of gas were used,

                                C(u)     = 2.5u+350

                                                =2.5(16)+350

                                                =40+350

                                                =390

Cost of December 2009, when 17 (<20) units of gas were used,

                                C (u)    =2.5u+350

                                                =2.5(17)+350

                                                =42.5+350

                                                =392.50

Cost of January 2010, when 18 (<20) units of gas were used,

                                C (u)    =2.5u+350

                                                =2.5(18)+350

                                                =45+350

                                                =395

Cost of February 2010, when 20 (=20) units of gas were used,

                                C (u)    = 2.5u+350

                                                =2.5(20)+350

                                                =50+350

                                                =400

Cost of March 2010, when 22 (>20) units of gas were used,

                                C (u)    =1.25u+375

                                                =1.25(22)+375

                                                =27.5+375

                                    =402.50

Cost of April 2010, when 25 (>20) units of gas were used,

                                C (u)    =1.25u+375

                                                =1.25(25)+375

                                                =31.25+375

                                                =406.25

Now potting these costs of last six months of Rana Bread we find the graphs in Fgure-5 which is a graph of piecewise linear functions.Figure-5

Simple interest  

In the case of Rana Bread stated earlier the owner has taken a loan of Tk. 40000 @ 10% in the month of November 2009 from his relatives.

Here

Amount of loan (P) = 40000

Interest rate (i) = 10% or 0.10

Time passed (n) = 6 months

                                =6/12

                                =0.50

So, the amount with interest to be paid if the loan is returned at the end of April 2010.

                                           A = P (1+in)

                                                =40000 (1+0.10×0.50)

                                                =40000 (1.05)

                                                =42000

Interest of last six months (Nov.’09-April ’10)

                                                   A-P

                                                = 42000-40000

                                                =2000

and interest of each month

                                                Interest ÷Time passed

                                                = 2000÷6

                                                            =333.33

Compound interest

In the case of Rana Bread stated earlier the owner has also taken a loan of Tk. 60000 @ 12% at the same time from the bank.

Here

                         Amount of loan (P) = 60000

                         Interest rate (i) = 12% or 0.12

 For the month of November 2009

                                           P = 60000

                                                i= 0.12

                                                n = 1/12 or 0.08

                Net amount, A = P (1+0.12)n

                                           =60000(1+0.12)0.08

                                               =60546

  Interest =60546-60000

                                    =546

For the month of December 2009

                                P = 60546

                                i = 0.12

                                n= 0.08

                Net amount, A = 60546 (1+0.12)0.08

                                               = 61097

                Interest = 61097-60546

                          = 551

For the month of January 2010

                P= 61097

             i=0.12

             n=0.08

             Net amount, A= 61097(1+0.12)0.08

                                           =61653

             Interest =61653-61097

                           =556

 For the month of February 2010

             P=61653

             i= 0.12

             n=0.08

            Net amount, A = 61653(1+0.12)0.08

                                           =62215

            Interest = 62215-61653

                          =562

 For the month of March 2010

            P=62215

            i=0.12

            n=0.08

            Net amount, A =62215(1+0.12)0.08

                                           =62782

            Interest =62782-62215

               =567

For the month of April 2010

            P=62782

            i=0.12

            n=0.08

            Net amount, A=62782(1+0.12)0.08

                                              =63354

           Interest = 63354-62782

                         =572

Salvage value

 Rana Bread uses machinery for its’ production. The machinery has a

                                     Cost (C) = 40000

                                     Depreciation rate (d) = 10% or 0.10

                                     Depreciation period (n) = 6 months

                                                                                    = 6/12 or 0.50

 So, after six months the salvage value of the machinery used by Rana Bread will be

                                                S = C (1-d)n

                                                =40000 (1-0.10)0.50

                                                =40000 (0.90)0.50

                                                =37947.33

                                                =37947

So, if the machinery of Rana Bread is sold right now (i.e. after being used for six months) the salvage value of the machinery would be Tk. 37947.

Arithmetic progression

An arithmetic progression is a sequence whose terms increase or decrease by a constant number called the common difference. A series in arithmetic progression thus becomes an additive series in which the common difference can be found by subtracting each term from its’ preceding one.

Thus if the first term and the common difference are known, the A.P. is completely known.

                The arithmetic progression

                                                                a, (a+d), (a+2d), (a+3d),…………………

whose first term is a the common difference is d, is designated as the standard form of an arithmetic progression.

                The corresponding arithmetic series

                                                                a+(a+d)+(a+2d)+(a+3d)+…………………….

is designated as the standard form of an arithmetic series.

A.P. in Rana Bread

Suppose the cost of the equipment used by Rana Bread is Tk. 100. Now the percentages of depreciation at the end of November 2009, December 2009, January 2010 and so on are respectively 15, 12.5, 10,…………. which are arithmetic progression with

                                                                a = 15

                                                                d = 12.5-15 = -2.5

                                                                n = 6

Hence, percentages of depreciation in April 2010,

                                                                a+ (n-1)d

                                                                = 15+(6-1)(-2.5)

                                                                = 15+5(-2.5)

                                                                = 15-12.5

                                                                = 2.5

So, the A.P. series of depreciation of equipment for six months

                                                15+12.5+10+…………+2.5

Also, total value of equipment depreciated in six months

                                                   n/2 (a+l)

                                                = 6/2 (15+2.5)

                                                = 3 × 17.5

                                                = 52.5

Hence, the value of equipment of Rana Bread at the end of April 2010

                                                   100-52.5

                                                = 47.5

Now the total cost being Tk. 20000, the value of Rana Breads’ equipment at the end of April 2010

                                                   (Tk. 20000÷100)×47.5

                                                = Tk. 9500

Conclusion

Mathematics, an ancient method of calculation, can be used by any one in solving different types of problems. As technology is developing very fast different applications of mathematics now can be done easily within a very short time using modern calculating devices. Whereas, the typical mathematics takes a longer time than the modern devices take. Although we can do our necessary calculations using modern calculating devices, we should try to know and use the typical mathematical terms to enrich our mathematical knowledge.

Mathematics in Business

Categories
Business Mathematic Finance

Report on Performance Evaluation of First Security Bank Limited from Different Point of View

Retained Earnings:

Retained earnings were placed at Tk. 106.14 million for the year 2004 as against Tk. 44.18 million in the previous year 2003. A year wise graphical prese

ntation of the same is provided on next page:

Year

2001

2002

2003

2004

Retained Earnings

Tk. 8.88 million

Tk. 11.63 million

Tk. 44.48 million

Tk. 106.14 million

Capture.1

Earnings Per share:

Earnings per share in year 2001 was Tk. 17.26 million but decreased in 2002 to Tk. 9.21 million, after that it started to increase and in year 2003 it stood Tk. 17.45 million

In year 2004 it was Tk. 31.71 million.

Year

2001

2002

2003

2004

EPS

Tk. 17.26 million

Tk. 9.21 million

Tk. 17.45 million

Tk. 31.71 million

Capture.JPG 2

 

Capital funds:

Authorized capital of the bank is tk. 1,000.00 million and paid up capital was tk. 200.00 million in the year 2004. Over the period 2001-2002 the paid –up capital was fixed at tk. 200.00 million and 320 over the period 2003-2004.Reserve Funds: Reserve fund of the bank was tk. 6476.86 million at 2003 and it stood at tk. 8500.00 million as on 31st December 2004 that means it increase at a increasing rate. The reserve fund of First Security Bank Limited for the period 2001-2004 was as follows-

2

 

Investments: First Security Bank Limiter’s investment portfolio mainly comprised investment in government securities, Treasury bills, Bonds etc.                 (Tk. In million)

3

q       Performance Evaluation of First Security Bank Limited from Different Point of View:

From owners viewpoint:

As an owner of a bank we will analyze three area of a bank:

1.  Profitability of the Bank

2.      Growth of the Bank

3.      Risk of the Bank

1.    Profitability of the Bank:

Operating Profit margin:

This ratio indicates operating income of the bank in relation to total revenue after taking into account operating cost.

Operating Profit margin = Operating profit / Total Revenue.

5

Interpretation: In case of First Security Bank Limited we observed that operating profit margin ratio increased gradually over the period this indicates that the bank is operationally efficient.

 

Net Profit margin: This ratio indicates net income of the bank in relation to total revenue after taking into account operating cost, financial expenses and corporate taxes.               Net Profit margin = Net income / Total Revenue.                 (Tk. in Million)

6

Interpretation: In case of First Security Bank Limited, we observed that net profit margin ratio increased over time except year 2002. This ratio has increased due to the decrease of administrative, general expense and income tax relative to revenue.

 

2. Growth of the Bank:

Return on assets (ROA): Return on assets measure the success of a firm in using assets to generate earning independent of the financing if those assets. This measure therefore separates financing activity from operating and investing activities. RAO is particularly useful in assessing the performance of business segment of firm when as is typical, financing for those segments comes form a central corporate pool of resources.                                          ROA = Net income / Total assets                

7

Interpretation: The ratio of return on asset of First Security Bank Limited is relatively low; it is lower than 1% over the years 2002 and 2003. It had increased little in 2004 .It indicates that total asset of the bank increased but net profit did not increase with the assets. This shows the management inefficiency in using the bank asset to generate profit.

 

Return on Equity (ROE): This ratio indicates the degree to which the form is able to convert operating income into an after income that eventually can be claimed by the shareholder. This is a useful ratio for analyzing the ability of the firm’s management to realize an adequate return on the capital invested by the owners of the firm.

                    ROE = Net income / Total equity                               (Tk. in Million)

89

Interpretation: From the above data we find that return on equity of FSBL increased in 2004 because the bank has increased its net income as well as the equity.

 

Provision for loan ratio:

Provision for loan loss / Total operating income                              (Tk. in Million)

11

Interpretation:

Provision for loan losses ratio of FSBL decreased in 2002, which shows the management efficiency to disburse and recovery of loan.:

Asset utilization: Asset utilization ratio = Total operating income / Total asset

    (Tk. in Million)

Year 2001 2002 2003 2004
Total operating income 173.27 263.99 385.38 591.67
Total asset 4459.52 7227.45 13369.50 16169.27
Asset utilization ratio 3.89% 3.65% 2.88% 3.66%

12

Interpretation: The data of asset utilization shows that the ratio of FSBL gradually decreased from 2001 to 2003. But in later period it increased in 2004. This indicates that the bank is most inefficient to mobilize profit from total assets.

 

Interest Income Ratio:

Interest Income Ratio = Interest Income / Total assets            (Tk. in Million)

Ratio/Year 2001 2002 2003 2004

Interest Income

321.40

496.62

818.29 1253.87
Total assets 4459.52 7227.45 13369.50 16169.27
Interest Income Ratio 7.21% 6.87% 6.12% 7.75%

13

Interpretation: From the above data we see that interest income ratio decreased overtime but little increased in 2004. It indicates that the bank is trying to keep the interest income more stables and they were not successful in this area.

 

Non-interest Income Ratio: = Non-interest Income / Total assets

                                                                                                                                                                                              (Tk. in Million)

14

Interpretation: Above table shows the non-interest income of FSBL increased in 2002 because of more activities other than deposit.

Equity Multiplier: Equity multiplier measures the value of assets funded with each taka of equity capital. The higher the ratio, the more leverage or debt the bank is using to fund its assets. Thus the ratio is a measure of the degree of leverage the bank is using.

  Equity multiplier = Total assets / Total equity capital

                                                                                                                                                                                                                                                        (Tk. in Million)

15

Interpretation: Equity multiplier of FSBL was too high, which indicates that the bank was highly depended on leverage.

 

3. Risk of the Bank:

Interest expense ratio: Interest expense ratio is different method to evaluate the bank’s expense management.

Interest Exp. Ratio = Interest expenses / Total operating income

                                                                                                                                                                                    (Tk. in Million)

16

17

Interpretation: From the above data we see that interest expenses ratio was more over stable up to 2002 but increased more in 2003. It indicates that the bank was trying to keep the interest expenses more stables up to 2002but in later period they were unable to maintain the track.

 

Non-interest expense ratio:

Non interest expense /Total operating income                                      (Tk. in Million)

19

Interpretation: Above table shows the non-interest expenses of FSBL was decreasing  over time but it was too high in 2001 because of more activities other than deposit, which had to pay high cost.

 

Tax ratio:

Tax ratio= Income Taxes / Total operating income

(Tk. in Million)

21

Interpretation: From 2001 to 2002, the tax ratio decreased, because the total operating income of FSBL was increased.

From the view point of Depositors:

1.            Internal Liquidity

2.Operating Performance

 

Internal Liquidity: Current Ratio = Current assets / Current liabilities

(Tk. in Million)

22

23

2. Operating Performance: Capital to deposit ratio = Total Capital / Total deposit

(Tk. in Million)

24

Loan to deposit ratio: Loan to deposit ratio = Total Loan / Total deposit

                                                                                                                        (Tk. in Million)

25

26

Deposit to employee ratio=Total Deposit / Total number of employee

(Tk. in Million)

27

CAMEL Rating: CAMEL rating is a suitable way of analyzing bank’s financial performance. It is a technique, which is widely used by the central bank as well as financial analysts to identify bank’s financial position. Five alphabets of CAMEL means:

 

C = Capital adequacy

A = Asset quality

M = Management efficiency

E = Earnings

L = Liquidity

These cover all major aspects of financial indicators. To appropriately identify bank’s financial position, rating is used which is calculated by following way-

28

Elements of CAMEL rating are calculated step by step in the following:

 

1. Capital adequacy: Capital adequacy of First Security Bank Limited was 7.57% in year 2004.

29

Interpretation:

Since the capital adequacy ratio of the First Security Bank Limited in 2004 was 7.57%. Therefore, the capital adequacy of First Security Bank Limited was fair. That means rating was 3.

 

2. Asset quality: To identify of assets quality of First Security Bank Limited, classification of loans and advance and provision for loans and advances is required to know –

31

Interpretation:

Classified loan ratio of the First Security Bank Limited at 2004 was 16.72%. So, the asset quality of First Security Bank Limited is marginal. That means rating was 4.

 32

Interpretation: Return on assets of the First Security Bank Limited at 2004 was 1.23%. So, the Earnings of First Security Bank Limited were strong. That means rating was 1.

 

4. Liquidity: Liquidity ratio = total liquid assets / total times demanded liabilities

                                               = 7488.03 million / 60923.1 million = 35.38%

33

 

Interpretation: Liquidity ratio of the First Security Bank Limited at 2004 was 35.38%. So, the liquidity of First Security Bank Limited was strong. That means rating was 1.

 

5. Management efficiency:

The management rating is an average of the four ratings given to capital, assets quality, earnings and liquidity.

34

Composite Rating: Composite rating is the average of the all ratings.

Thus, Composite rating = (3+4+2+1+1) / 5   = 2.20

35

Categories
Business Mathematic

Lecture on Business Mathematics Equations

An equation of  the type ax+by+c = 0 is a linear equation. It shows relation between two algebraic expressions with the  sign of equality (=) An equation holds true for particular value or values of the variable/ variables. To find the value of a single variable, only one equation is needed. In order to find the values of two variables, two equations are
necessary. Accordingly, three equations are needed to find the values of three variables, and so on. There are various types of equations viz. Linear quadratic equation, Cubic equation etc.

Categories
Business Mathematic

Lecture on Annuity

Suppose, we want to make payment of Tk.5000 each year for 10 years to an account which earns 6 percent interest, compounded annually, and we want to calculate the total amount which will be in the account after 10 payments. This series of periodic payments is called an annuity. The Annuity Calculator is no longer available. Use the new Retirement Income Calculator to help you decide whether receiving monthly income is right for you by comparing the available options and features that might meet your needs: TSP monthly payments and a life annuity.