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Monday, February 19, 2018

A Model for Analysis of Work Done by a Special Force

This model is an example as application one of theorems mentioned in article of “The Change Depends on the Direction of the Motion: The Symmetric Group Action (2)” posted on link:


Work Done by Force

When we are speaking about the work done by a force, we have usually two options:

1. The work done on a particle by a constant force in magnitude and direction which is calculated by below equation:


2. The work done on a particle by a varying force in magnitude and direction in which we should use the line integral to calculate it as follows:


But, in this article, I am willing to introduce you a special force which has the constant magnitude but varying in direction. In this case, I have made a model that it is able to show us the location of the particle after being displaced by this special force.

Suppose a particle P (0, 0, 0) just like below figure being displaced by a force F:

Theorem: The force with below vector has the constant magnitude equal to 1.22474487139159 but varying in direction:


In this case, we can still use option (1) and below equation to calculate the work done:


Below figure shows the components of this model:



Let me explain you about the components of above model as follows:

1. In left side on cells C2:D3, we have inputs including the angle φ (degree) of force vector and amount of displacement of particle “P” which is “r”.

2. In right side on cells F3:J8, we have outputs including:

   2-1) on cells G3:I3, a special force vector can be calculated by using above theorem.

   2-2) on cells G4:I4, the direction of special force can be obtained by using below formula:


The magnitude of this force is the constant and equal to 1.22474487139159.

 2-3) on cell G5, we have the work done which is calculated by below formulas:


  2-4) on cells G7:I7, we have the location particle P (x, y, z) which is derived by this model.

  2-5) on cells G8:I8, the direction of motion particle P can be calculated by using below formula:


 2-6) The change of direction particle P will be done by click on cell A1 and also this change will again go back by click on cell B1 (Go & Back).  

You can see below screenshot as the examples for this model:


All researchers and individual people, who are interested in having this model, don’t hesitate to send their request to below addresses:


Friday, February 9, 2018

A Model to Solve a System of Nonlinear Equations by Using Microsoft Excel

This is an example for solving a system of nonlinear equations by using the Big Data Science Analysis with Microsoft excel plus VBA.




Definitely, there are many complicated problems that can be solved by using this analytic model. For instance, one of the applications of this model is to solve problems related to optimization, such as Hill Climbing.

For example, what is the maximum area of a rectangle inscribed in a circle with a radius of 2 units?

You can find the answers for “x” and “y” in this rectangle in the below data analysis:






I would like to inform you that this model is able to solve all linear and nonlinear system of equations in which it gives us less error than traditional methods such as Newton – Raphson , Gauss – Seidel, Jacobian and so on. Don't worry, if you have not any knowledge about traditional methods. Dive into new technology.


Example: A Model to Track the Location of a Particle in the Space

 One of the most important needs for many projects is, quickly to track the motion of a particle when
it is tripping between two points in the space. For instance, suppose particle P is moving by velocity “V” with accelerate equal to zero from point A (x1, y1, z1) to point B (x2, y2, z2). We want to know the coordination of P (x, y, z) at the moment of “t” when it is moving in direction of A to B. This model gives us the opportunity to obtain the coordination P (x, y, z) only by click. In fact, if we have the coordination P (x, y, z), it will be as an index for us that other particles which are moving from A to B such as P1, P2, …. are not tripping in the exact direction of A to B. In this case, we have to solve a system of two nonlinear equations as follows:

 First of all, we should define amount of “a” and “b” as follows:

The distance of point A to point B is calculated by using below formula and also “X” is the distance that particle P will travel after time “t” in direction of A to B:









Therefore, we should solve below system of two nonlinear equations







Below figure as well as shows you the components of this model











Let me explain you about the components of above model as follows:

1. In right side on cells G4:J10, we have inputs including the coordination points A and B, velocity of particle P (v), the time after staring to travel particle P (t) and Error which is the difference between the results. So, X = v.t, d = distance between point A and B and on cell G10, I have included a logic formula (=if (H8 >= H9, H8, 0)). It means, if X = 0 then the particle has passed point B

2. In left side on cells C5:E6, we have other inputs including the ranges to reach to the answer for coordination particle P (x, y, z) which is the solution for system of two nonlinear equations. 
 Here, there are lower and upper ranges which are changed by click on cell A2 and also this change will again go back by click on cell B2 (Go & Back).

This part has been utilized by the method stated in article of “Can We Solve a Nonlinear Equation with Many Variables?” posted on link:
http://emfps.org/2016/10/can-we-solve-nonlinear-equation-with.html 

3. On cells C9:E9, we have outputs which are the answer to above 
system of two nonlinear equations.

4. On cells C12:E13, we have Control (1) where first equation is solved by using the coordination A 
and P and it is compared with amount of “X^2” on cell C14. So, second equation is solved by using the coordination B and P and it is also compared with amount of “(d-X)^2”on cell C15. The differences have been mentioned on cell G12 and G13 as the errors.

5. On cells E15:G17, we have Control (2) which is referred to track the directions in which we are willing to know if the motion of particle P is in direction from point A to point B. In this case, a control can be conducted by using below formulas:



















You can see below screenshot as the examples for this model:


Conclusion

.Above article gives you a simple example
In fact, I am willing to tell you that this model is able to solve all system of two or three nonlinear   
equations.

All researchers and individual people, who are interested in having this model, don’t hesitate to send 
their request to below addresses:



Monday, February 5, 2018

The Impact of Stochastic Matrix on Any Vector

A stochastic matrix shows us that the sum all elements on each column (or each row) is equal one or we can say that each column (or each row) of a stochastic matrix is just a probability vector.


Theorem (1): The sequence of operations a stochastic matrix on any vector will be finally obtained the constant vector. (The elements of vector are the members of Real Number)

V1 = V0 * M^1 , V2 = V1*M^2,  V3 = V2*M^3, …. Vn+1 = V n* M^k
Where vector Vn+1 always will be the constant.

A special state is when the vector is a probability vector which is named Markov Chain or Markov Process

Example (1):


The sample of our vector (1*4) is:

V0 =
                                                                        

Suppose we have below stochastic matrix M (4*4):


M =


The results of the sequences (V0 * M) are as follows:


As we can see, the vectors of V5 and V6 are approximately the same.

Example (2):

The sample of our vector (1*4) is:

V0 = 


The results of the sequences are as follows:


As we can see, the vectors of V5 and V6 are approximately the same.

An example for Markov Chain:

The sample of our probability vector (1*4) is:

V0 = 


I consider above stochastic matrix M (4*4).
Then, we will have below results:


As we can see, the vectors of V5 and V6 are approximately the same.


Is below conjecture is a theorem?

Conjecture (1): The sequence of operations for transpose of a stochastic matrix (n*n) with non - zero elements on any vector will be finally obtained a vector with the same elements. (The elements of vector are the members of Real Number)



Example (1):

Consider above stochastic matrix M (4*4):

M =



The transpose of matrix M is:

MT =


The sample of vector is:

V0 = 


The results of the sequences are as follows:


As we can see, the elements of vector V7 are the same.

Example (2):

The sample of vector is:

V0 =

The results of the sequences are as follows:


As we can see, the elements of vector V7 are the same.


Saturday, December 23, 2017

یک مدل برای پیدا کردن مرزهایی که سرعت زاویه ای را میتوان بردار در نظر گرفت











من فکر میکنم برای سرعت زاویه ای به جای گفتن اینکه بردار هست یا شبه بردار, بهتره بدنبال مرز هایی باشیم که رفتار بردار بودن رو از شبه برداری برای ما متمایز میکنه. چون سرعت زاویه ای هم بردار هست هم میتونه بردار نباشه. در تمام کتاب های رفرنس فیزیک آمده: در زاویه های دورانی بسیار کوچک, زاویه دوران میتواند بردار باشد در نتیجه فقط در این حالت سرعت زاویه ای میتواند بردار تلقی شود. اما مدلی که من ساختم میگه علاوه بر زاویه دوران, در کلیه حرکات ارتعاشی دامنه ارتعاش جهت نیل کردن به مرز بردار یا غیر بردار بودن سرعت زاویه ای خیلی مهم هست در حقیقت حساسیت به دامنه خیلی بیشتر از کوچک بودن زاویه دوران دارد. البته این موضوع مربوط به هفته پیش بود. اما دیشب با یک مدل جدید به این نتیجه رسیدم که:  علاوه بر کوچک بودن زاویه  و دامنه ارتعاش, جهت چرخش و تعداد درجه آزادی هم بسیار مهمه. 

بطور مثال: فرض کنید یک آرمیچر روی یک صفحه دایره ای نصب شده و محور آرمیچر روی محور x دستگاه مختصات هست. حالا اگه آرمیچر در جهت مخالف عقربه های ساعت حول محور x بچرخه و صفحه دایره ای در جهت موافق عقربه های ساعت حول محور z بچرخه, آنالیز حرکت ارتعاشی با زمانیکه آرمیچر موافق عقربه های ساعت بچرخه و صفحه دایره ای مخالف عقربه های ساعت بچرخه فرق میکنه. حالا به این تحلیل کم یا زیاد شدن شعاع صفحه دایره ای و شعاع آرمیچر رو هم اضافه کنید که همون دامنه ارتعاش هست.
 حالا فرض کنید کل سیستم بالا روی یک چرخ و فلک فیکس و نصب شده. در این صورت ما یک درجه آزادی به سیستم اضافه کردیم و یک سیستم یا سه درجه آزادی داریم. در نتیجه سیستم آرمیچر و صفحه دایره ای حالا دارن حول محور y میچرخن که بازم جهت چرخش و شعاع چرخش, آنالیز جدیدی در خصوص بردار بودن و یا نبودن سرعت زاویه ای به ما نشان میدهد.

بطور کلی برای سیستم ها با دو درجه آزادی ما شش حالت داریم که هر کدام از این حالت ها در آنالیز و تحلیل حرکات ارتعاشی نتایج متفاوتی رو به ما نشان میدهند.
برای سیستم ها با سه درجه آزادی و چهار درجه آزادی ما دوازده حالت داریم که هر کدام از این حالت ها در آنالیز و تحلیل حرکات ارتعاشی نتایج متفاوتی رو به ما نشان میدهند. این یعنی کل مسیر آنالیز ما برای هر حالت از A تا Z عوض میشود. 



برای اطلاعات بیشتر در خواست خود را به آدرس تلگرام زیر بفرستید:

https://t.me/Allisnumber

Monday, November 13, 2017

A Model for Analysis of Bond Valuation By Using Microsoft Excel plus VBA







A bond is a type of asset in which a government or a company issues these securities as a long – term debt to borrow money from institutional investors (Banks) or public sector. Each bond has a time to maturity which is usually 5, 10, and 20 or 30 years. Initial value of bond is named par value or face value equal to $1000 with a coupon interest rate on the bond which is the percentage of par value and it will be paid annually or semiannually (two times in one year). In fact, the government or the company is committed regularly and continuously to pay these payments and also repayment of initial value (par value) at maturity time.
The purpose of this article is to present a model for analysis of bond value where there are seven independent variables including current bond price (bond value), YTM, coupon rate, purchased year, purchased month, purchased day and time period (n). This model simultaneously solves an equation with three independent variables accompanied by generating maximum and minimum of this function for given domain and range and also non simultaneously analyzes seven independent variables. One of the most crucial applications of this model is to obtain YTM (return rate) for current price equal to bond value without using any trial and error.

A coupon interest rate always stay the constant while the purchasers of bonds strongly look at and compare it with premium risk of market which is named the return rate or required rate on the capital or yield to maturity (YTM). This is why the price of bond varied with bond value. Of course, there are two factors for this discrepancy: (1) the difference between coupon interest rate and return rate (YTM) and (2) entering time (the time of purchasing). Below diagrams as well as show us the impact of these factors for time periods of 30, 20 and 10 years:




Above diagrams say to us, when YTM (rd) is greater than the coupon rate, the bond value will be less than its par value (Discount Bond), when YTM (rd) is less than the coupon rate, the bond value will be greater than its par value and when YTM (rd) is equal to coupon rate, the bond value will be equal to par value.
Generally the basic valuation model for any asset can be made by using below equation:

Where:

V0 = value of the asset at time zero

CFt = cash flow expected at the end of year t

r = appropriate required return (discount rate)

n = relevant time period

But for each specific asset such as Bonds, Stocks, Real estate and so on, we have to change a little bit above basic equation. For instance, the formula to evaluate the bond value can be as follows:



B0 = value of the bond at time zero

I = annual interest paid in dollars

r = appropriate required return (discount rate)

n = number of years to maturity

M = par value in dollars

rd = required return on a bond

I also used above equation to make this model for analysis of the bonds. Below figure as well as shows the features of this model:



As you can see in the figure above, there are seven independent variables (Inputs) which have been highlighted by red color. First, we enter the period of maturity which is “n”. Then, according to the issue date and the maturity date, we enter Year, Month and Day as current date. After that, we choose a range for Current price, YTM and Coupon rate (Low and High) and next we consider a specific current bond price which is into the range of current price. Finally, this model gives us the outputs which are the minimum and maximum bond value with  the appropriate  YTM and Coupon rate. In the meanwhile, you can see that specific bond price (input) is approximately equal to specific bond value (output) that it says to us about the appropriate YTM and Coupon rate for the specific bond price where we do not need to use any trial and error to obtain YTM for a specific bond price.
You can see below clips as the examples for this model:

The model for n = 30



The model for n = 10



All researchers, investors and individual people who are interested in having this model, don’t hesitate to send their request to below addresses: