Fuzzy Method for Decision Making (CON): Application of Pascal’s Triangular plus Monte Carlo Analysis

Following to the articles of “Fuzzy Method for Decision Making:  A Case of Asset Pricing Model” posted on link: http://www.emfps.blogspot.com/2013/04/fuzzy-method-for-decision-making-case.html?m=1 and “Fuzzy Method for Decision Making (CON):  A Case of Newton's Law of Cooling” posted on link: http://www.emfps.blogspot.com/2013/05/fuzzy-method-for-decision-making-con.html?m=1, we can find many cases in different fields which can be analyzed by using of this method (Fuzzy method). For instance, in Financial Management, there are many theories for Dividend Policy. One of the best ways to make decision for dividend payment is to utilize fuzzy method. In fact, if I change the topic of “Fuzzy Method for Decision Making:  A Case of Asset Pricing Model” to “Fuzzy Method for Decision Making:  A Case of Dividend Policy” and I replace dividend payment instead of second car’s price, I will be able to have a new analysis on case of GAINESBORO MACHINE TOOLS CORPORATION (please see “Case Analysis of GAINESBORO MACHINE TOOLS CORPORATION: The Dividend Policy” posted on link: http://emfps.blogspot.com/2012/03/case-analysis-of-gainesboro-machine.html and “Case Analysis of GAINESBORO MACHINE TOOLS CORPORATION (CON): A New Financial Simulation Model” posted on link: http://emfps.blogspot.com/2012/04/case-analysis-of-gainesboro-machine.html).

The purpose of this article is to apply Pascal’s Triangular plus Monte Carlo Analysis instead of the method used in above articles where the template is the same and referred to fuzzy set theory. Then I will compare the final results in which we can say that both of them (methods) are compatible with together while the method of Pascal’s Triangular plus Monte Carlo Analysis is very easier and more reasonable than the method applied in previous articles.

I had illustrated the method of Pascal’s Triangular plus Monte Carlo Analysis on below links:  

“Application of Pascal’s Triangular plus Monte Carlo Analysis to Find the Least Squares Fitting for a Limited Area” posted on link: http://emfps.blogspot.com/2012/05/application-of-pascals-triangular-plus_23.html

“Pascal’s Triangular Plus Monte Carlo Analysis to Appraise the Wisdom of Crowds” posted on link: http://emfps.blogspot.com/2012/05/application-of-pascals-triangular-plus_08.html.

“Application of Pascal’s Triangular Plus Monte Carlo Analysis to Design a Strategic Plan” posted on link: http://emfps.blogspot.co.uk/2012/07/application-of-pascals-triangular-plus_10.html

“Application of Pascal’s Triangular plus Monte Carlo Analysis to Find the Least Squares Fitting for a Limited Area: The Case of Constant – Growth (Gordon) Model” posted on link: http://emfps.blogspot.co.uk/2012/07/application-of-pascals-triangular-plus.html

“Application of Pascal’s Triangular Plus Monte Carlo Analysis to Calculate the Risk of Expected Utility” posted on Link: http://emfps.blogspot.com/2012/05/application-of-pascals-triangular-plus.html

Therefore, I directly start to analyze previous cases again by using the method of Pascal’s Triangular plus Monte Carlo Analysis as follows:

 Case Study: Asset Pricing Model

In the reference with the article of “Fuzzy Method for Decision Making:  A Case of Asset Pricing Model” posted on link: http://www.emfps.blogspot.com/2013/04/fuzzy-method-for-decision-making-case.html?m=1, we had two examples:

Example (1)

To find probability distribution inferred from Pascal’s triangular, I chose:

 n = 200 for X 1 = 5000 and X2 = 10000

Then, we have below probability distribution:

Cut Offs	X
0	5025
7.72E-45	7100
0.01	7275
0.1	7450
0.4	7725
0.9	8550
1	10000

I assigned the formulas of Rand and Vlookup for a1 and a2 as follows:

x	0
Left Side			Right Side
Random	0.626742		Random	0.58275
a1	7725		a1	7725
a2	10000		a2	10000
x	0		x	0
(Formula)1	0		(Formula)1	1
(Formula)2	-3.3956		(Formula)2	4.395604
(Formula)3	1		(Formula)3	0
Alpha -cut	0		Alpha -cut	1

Then, I got two ways data table for (x) between 5000 and 10000 with 400 iterative calculations for left side and right side as follows:

Finally, I calculated the average for α –Cut (Left) and α –Cut (Right) as follows:

x	8400	8600	8800	9000	9200
α –Cut (Left)	0.301045	0.366571	0.471798	0.557612	0.642259
α –Cut (Right)	0.696507	0.622444	0.535629	0.441788	0.349999
			α - Cut	0.503714

As we can see, α –Cut (Left) and α –Cut (Right) are approximately close around x = 8800 and α –Cut = 0.5

Of course, I chose ∆x = 200. If we consider ∆x = 100, the final results will be more accurate.

In the previous article, the best price of the first try to advertise was equal to $8883.33 with confidence level of 0.52 (α = 0.52). We can see that the final results are compatible.

Example (2)

To find probability distribution inferred from Pascal’s triangular, I chose:

 n = 200 for X 1 = 4000 and X2 = 6000 (Right side)

n = 200 for Y1 = 6000 and Y2 = 8000   (Left side)

Then, we have below probability distribution:

Cut Offs	X
0	4010
7.72E-45	4840
0.01	4910
0.1	4980
0.4	5090
0.9	5420
1	6000

Cut Offs	Y
0	6010
7.72E-45	6840
0.01	6910
0.1	6980
0.4	7090
0.9	7420
1	8000

I assigned the formulas of Rand and Vlookup for a1 and a2 as follows:

x	4000		y	6000
Right Side			Left Side
Random	0.858574		Random	0.701062
a1	5090		a1	7090
a2	10000		a2	10000
x	4000		y	6000
(Formula)1	0		(Formula)1	1
(Formula)2	-0.222		(Formula)2	1.37457
(Formula)3	1		(Formula)3	0
Alpha -cut	0		Alpha -cut	1

Then, I got two ways data table for (x) between 4000 and 10000 and for (y) between 6000 and 10000 with 400 iterative calculations for left side and right side as follows:

(Of course, the best range for both of them is (x) and (y) between 6000 and 10000)

  Finally, I calculated the average for α –Cut (Left) and α –Cut (Right) as follows:

For ∆x = 100, we have:

(x, y)	7900	8000	8100	8200	8300	8400
α –Cut (Right)	0.574654	0.593947	0.614103	0.63463	0.65446	0.675476
α –Cut (Left)	0.71642	0.681869	0.649753	0.61348	0.58049	0.547297
			α –Cut	0.62405

For ∆x = 200, we have:

(x, y)	7600	7800	8000	8200	8400	8600
α –Cut (Right)	0.512874	0.553806	0.594213	0.634706	0.67472	0.715272
α –Cut (Left)	0.820029	0.753847	0.681916	0.614954	0.548512	0.479935
			α –Cut	0.62483

As we can see, α –Cut (Left) and α –Cut (Right) are approximately close around x = 8200 and α –Cut = 0.62

In the previous article the best price of the first try to advertise was equal to $8183.33 with confidence level of 0.4 (α = 0.4). We can see that the final results are compatible.

Case Study: Predictions in temperature transferring 

In the reference with the article of “Fuzzy Method for Decision Making (CON):  A Case of Newton's Law of Cooling” posted on link: http://www.emfps.blogspot.com/2013/05/fuzzy-method-for-decision-making-con.html?m=1, we had below algorithm:

Warming to Cooling			Cooling to Warming
Ta	5		Ta	100
T0	100		T0	5
t	34		t	1
k	0.055		k	0.0106
T (t)	19.641748		T (t)	6.001681708
µ	0.1964175		µ	0.060016817
α-cut	0.1964175		α-cut	0.060016817

Firstly, I chose the same random range for “t” and also “k”. For instance, t1 = 0 to t2 = 100 and k1 = 0 to k2 = 100.

After that, I found probability distribution inferred from Pascal’s triangular for:

 n = 200 for k1 = 0 and k2 = 100

Where we have below probability distribution:

Cut Offs	k
0	0.5
7.716E-45	42
0.01	45.5
0.1	49
0.4	54.5
0.9	71
0.9999999	100

Then, I used the formulas of Rand and Vlookup for “k” as follows:

t	0
Warming to Cooling			Cooling to Warming
Ta	5		Ta	100
T0	100		T0	5
t	0		t	0
Rand	0.4782881		Rand	0.0027308
k	54.5		k	42
T (t)	100		T (t)	5
µ	1		µ	0.05
α-cut	1		α-cut	0

I got two ways data table for “t” and α –Cut as follows:

As we can see, all results are same and equal to zero. Therefore, I decreased the range of “k” to [0, 10] and I repeated above steps where I found the same data table as follows:

Finally, I decreased the range of “k” to [0, 0.1] and I repeated again above steps where I found below data table:

Cut Offs	k
0	0.0005
7.72E-45	0.042
0.01	0.0455
0.1	0.049
0.4	0.0545
0.9	0.071
0.9999999	0.1

Thus we can consider the range of “k” between 0 and 0.1.

In the next step, I did a sensitivity analysis for “t” in the range of [0, 100] and α –Cut (Left) and α –Cut (Right) as follows:

t	α-cut	α-cut	mean	STDV	CV
0	0.949611	0.115111	0.532361	0.59008	1.108421
1	0.949611	0.100389	0.525	0.60049	1.143791
2	0.917367	0.148106	0.532736	0.543949	1.021048
3	0.856708	0.171214	0.513961	0.484718	0.943102
4	0.813919	0.219088	0.516504	0.420609	0.814338
5	0.806696	0.2766	0.541648	0.374835	0.692026
6	0.758013	0.276962	0.517487	0.340154	0.657319
7	0.740877	0.351305	0.546091	0.275469	0.504437
8	0.664287	0.461675	0.562981	0.143268	0.254481
9	0.631704	0.418296	0.525	0.150902	0.287433
10	0.631995	0.44915	0.540573	0.129291	0.239174
11	0.571632	0.424084	0.497858	0.104332	0.209562
12	0.543964	0.594767	0.569365	0.035923	0.063094
13	0.552433	0.497567	0.525	0.038796	0.073897
14	0.528407	0.521593	0.525	0.004818	0.009178
15	0.50553	0.580543	0.543036	0.053042	0.097676
16	0.447209	0.602791	0.525	0.110013	0.209549
17	0.42614	0.715861	0.571001	0.204864	0.35878
18	0.443256	0.735336	0.589296	0.206531	0.350471
19	0.424451	0.625549	0.525	0.142197	0.270852

As we can see, “t” between 12 min and 15 min has the least CV and STDEV. Therefore, I used Pascal’s Triangular plus Monte Carlo Analysis for “t” as follows:

n = 200, t1 = 12, t2 = 15 where the probability distribution is below cited:

Cut Offs	x
0	12.015
7.72E-45	13.26
0.01	13.365
0.1	13.47
0.4	13.635
0.9	14.13
0.9999999	15

I applied the Rand and Vlookup in excel in which the algorithm was changed as follows:

k	0
Warming to Cooling			Cooling to Warming
Ta	5		Ta	100
T0	100		T0	5
Rand	0.85862		Rand	0.978536
t	13.635		t	14.13
k	0		k	0
T (t)	100		T (t)	5
µ	1		µ	0.05
α-cut	1		α-cut	0

I obtained the sensitivity analysis for “k” and α –Cut (Left and Right). The final results are as follows:

k	α-cut	α-cut	mean	STDEV	CV
0.039	0.608189	0.441811	0.525	0.117647	0.224089
0.04	0.600629	0.460166	0.530398	0.099323	0.187261
0.041	0.593173	0.450781	0.521977	0.100686	0.192893
0.042	0.589543	0.464183	0.526863	0.088642	0.168246
0.043	0.57856	0.47144	0.525	0.075746	0.144278
0.044	0.571402	0.478598	0.525	0.065623	0.124996
0.045	0.568174	0.496989	0.532582	0.050336	0.094513
0.046	0.557376	0.488759	0.523067	0.04852	0.09276
0.047	0.550505	0.499495	0.525	0.036069	0.068703
0.048	0.543727	0.506273	0.525	0.026483	0.050445
0.049	0.53704	0.509006	0.523023	0.019823	0.037901
0.05	0.530445	0.515575	0.52301	0.010514	0.020103
0.051	0.527943	0.526062	0.527003	0.00133	0.002524
0.052	0.52413	0.53248	0.528305	0.005904	0.011176
0.053	0.515239	0.532164	0.523702	0.011968	0.022852
0.054	0.509015	0.545057	0.527036	0.025486	0.048357
0.055	0.498782	0.547127	0.522954	0.034185	0.065369
0.056	0.492704	0.557296	0.525	0.045673	0.086996
0.057	0.486709	0.559164	0.522937	0.051234	0.097973
0.058	0.484937	0.565063	0.525	0.056657	0.107918
0.059	0.474961	0.575039	0.525	0.070766	0.134793
0.06	0.473377	0.580794	0.527085	0.075956	0.144105
0.061	0.467712	0.586471	0.527092	0.083976	0.159319

Above table shows us, for k = 0.051 and α – cut = 0.53 we have the least CV and STDEV.

Note:  “All spreadsheets and calculation notes are available. The people, who are interested in having my spreadsheets of this method as a template for further practice, do not hesitate to ask me by sending an email to: soleimani_gh@hotmail.com or call me on my cellphone: +989109250225. Please be informed these spreadsheets are not free of charge.”
 

In the previous article, we had the answer for “k” approximately equal to 0.055 per min and

α – cut = 0.54

Therefore, we can see both methods are compatible with together.

But why am I following the constant points or values, intersections and distance among fuzzy numbers? Because, if we are willing to utilize Fuzzy Logic Control (FLC) as the methodology to analyze and solve some complex cases, the most important step is to evaluate the rules in FLC. In classic or theory physics, if we have to solve the complex cases, the evaluation of the rules will be easier than other cases because we usually encounter the universal laws or constant points (referred to article of “The Constant Issues, Universal Laws and Boundaries Conditions in Physics Theory” posted on link: http://emfps.blogspot.com/2011/10/constant-issues-universal-laws-and.html). In the other complex cases such as strategic management or financial management, the tracking the intersections and consequently the distance among fuzzy numbers to evaluate the rules in FLC methodology is very important. On the other hand, if we are using FLC as the methodology to analyze the complex cases in the field of strategic management or designing a strategic plan, we should know that the rules are changing the moment to moment and we have to replace new rules instead of old rules rapidly. How? I think that the article of “EMFPS: How Can We Get the Power Set of a Set by Using of Excel?” posted on Link: http://emfps.blogspot.com/2012/08/emfps-how-can-we-get-power-set-of-set.html) let us increase our speed to replace new rules.

 

The New Theorems to Measure the Distance among Triangular Fuzzy Numbers

In the reference with the article of “Fuzzy Method for Decision Making: A Case of Asset Pricing Model”, you can find the important pionts about the final part of above article which described new definitions and laws of the distance measurement among fuzzy numbers.

 The purpose of this article is to demonstrate the new theorems for measuring the distance among triangular fuzzy numbers where these theorems have been inferred from the method modified by Hsieh and Chen (1999).

Before releasing these theorems, let me divide the debate of the distance among triangular fuzzy numbers into two types as follows:

1)      All triangular fuzzy numbers have been limited into interval [x1, x2] in which “xm” referred to membership function µ(x) is the same for all fuzzy numbers and equal to 1.   For instance, if we consider the fuzzy numbers of “A”, “B”, “C” and “D”, we will have:

A = (x1, xm, x2)

B = ((x1< x < xm), xm, (xm < x < x2))

C = ((x1< x < xm), xm, (xm < x < x2))

D = ((x1< x < xm), xm, (xm < x < x2))

Fig (1) shows us the concept of type (1):

2)      In this case, there is not the limitation for independent variable of “x” and also “x” assigned to membership function µ(x) = 1 is not the same for all fuzzy numbers in which fuzzy numbers A, B, C and D can be considered as follows:

      A = (x1, x2, x3)

B = (x4, x5, x6)

C = (x7, x8, x9)

D = (x10, x11, x12)

Fig (2) illustrates the concept of type (2):

 

The method Modified by Hsieh and Chen (1999)

According to Fu (2006), there are many approaches to measure the distance between two triangular fuzzy number such as Bortolan and Degani (1985), Liou and Wang (1992) and Heilpern (1997) who applied a geometrical distance measuring and his method was modified by Hsieh and Chen (1999). This modified method is described as follows:
If we have two triangular fuzzy numbers below cited:

 

                   

Then, we can calculate the distance by below function:

The New theorems for measuring the distance among triangular fuzzy numbers

In the reference with above function, new theorems can be listed as follows:

Theorem (1)

Assume, we have three triangular fuzzy numbers A, B and C where:

Theorem (2)

According to theorem (1), we can define theorem (2) as follows:

Theorem (3)

If A = (a1, a2, a3) and B = ((n + a1), (n + a2), (n + a3)) are triangular fuzzy numbers so that “n” is real numbers (n ϵ R), Then we can define the distance between A and B as follows:

d (A, B) = n

Theorem (4)

If triangular fuzzy number “Ж” is the sum all triangular fuzzy numbers A, B, C, D, E…

We have:

Ж = A + B+ C + D + E+…    and    Д = B+ C + D + E+…   

Then we will have:

Д = Ж – A

Therefore, in the reference with Theorem (1), we can define below mentioned:

d ((Ж – A), Ж) = d (A, 2A)

Or

d (Д, Ж) = d (A, 2A)

To be continued…….

Reference

- Fu, Guangtao. (2006). A fuzzy optimization method for multicriteria decision making:

An application to reservoir flood control operation. Expert Systems with Applications, 34, 145- 149.

Newton's Law of Cooling and Fuzzy Method for Decision Making

Following to article of “Fuzzy Method for Decision Making:  A Case of Asset Pricing Model”, the purpose of this article is more practice of this method by presenting a case in physics which is about Newton’s law of cooling. At the first, a case has been explained then reviewing Newton’s law of cooling and finally the application of method mentioned in above article to analyze the case is demonstrated.  Of course, there are many cases to analyze by this method such as the circuit electric of R – L in which we have:  i = V/ R [1- (e^ (-Rt/L))] in the field of electrical engineering, Benjamin Franklin (1706-1790)’s devise about composite interest rate in finance, Compressor Performance Control to track the distance between the operating point and set point, Head Generated by the Circulation Pumps to find out the operating point which is the intersection of pump head and circuit losses in mechanical engineering and so on.

I have chosen Newton’s law of cooling because it is the general case for all engineers.

Case Study: Predictions in temperature transferring 

Mr. X is willing to decrease the temperature of a pot filled of water boiled at 100 degrees C. He puts this pot into a sink full of water at 5 degrees C in which water temperature into sink will be stayed the constant at 5 degrees C during the period of the time. He would like to predict the temperature of water into pot after one hour. How?

We as well as know that there are three mechanisms to transfer the temperature which are Conduction, Convection and Radiation. This case is referred to only Conduction.

Definitely he should use Newton’s law of cooling.

Newton’s law of cooling

In early 1701, Isaac Newton observed that the rate of heat loss a hot area during the period of the time has directly the relationship with the difference temperature between this area and adjacent area where the temperature of adjacent area (surrounding) will be the constant during the period of the time. Where we have:

T0 = the temperature of hot area in time zero

Ta = the temperature of adjacent area (the constant)

T (t) = the temperature of hot area at time “t”

T (t) > Ta, T = T (t)

We can replace above proportional relationship to a differential equation, if we include a heat transfer coefficient (k) which is positive. Since we have: (T – Ta) > 0, the trend of the curve will fall down. Therefore, we should consider a negative mark for this differential equation as follows:

Solution of differential equation:

We have:

In the reference with data included in the case study, the rate of heat loss will be as follows:

As we can see, Mr. “X” will be able to predict the temperature of water into pot after one hour, if he finds out a constant amount for “k”.

One the best ways to obtain “k” is to utilize Fuzzy set theory.

Fuzzy Method for Decision Making

Here, I am willing to apply fuzzy method mentioned in my previous article to calculate “k” step by step as follows:

Ø  On the spreadsheet of excel, we enter a algorithm to calculate T (t)

Ø  We use of the formula of µ = T (t) / T0 to change rate of temperature loss to Fuzzy membership function assigned to “t”

Ø  If we have the heat loss by falling down the trend ((T – Ta) > 0), it can be also considered vice versa. It is means, if the temperature of pot water is 5 degrees C and the temperature of sink water is 100 degrees C (the constant), we will have the increase heat into pot where the trend of the curve will go up ((T – Ta) < 0). Therefore, the differential equation will be the same and we can do above two steps for this situation (Please see below algorithms on my spreadsheet of excel).

Ø  The most important step is to find out the domain for rectangular k – t. As the first try, I start by method of try and error in which I reach a constant “µ” or where the rate of changes for “µ” will be negligible.

Ø  By Using of “µ” discovered from previous try, I utilize PLUG order in excel to limit “µ” to zero in which it is named α – cut.

Ø  As the second try, I use from the table of two way sensitivity analysis for rectangular k – t where the domain has been extracted from previous step.

(Please see Appendix I)

If we review Appendix I, we will see that the range of “k” is between 0.0106 and 0.1 (per min) while the range of “t” is between 1 (min) and 45 (min)

Ø  As the final try, I repeat to make the table of two way sensitivity analysis for rectangular k – t in accordance with above ranges. (Please see Appendix II and III)

Appendix II is for status of warming to cooling and Appendix III is for status of cooling to warming.

The final result has been presented in below diagram:

As we can see, the most density has been included into triangular ABC. Therefore, gravity centre of this triangular (ABC) is the answer for “k” where the coordination of gravity centre is:

t = 13.5 (min)

α – cut = 0.54

According to Appendix II or III, we can see that the answer for “k” is approximately 0.055 per min. Thus we have:

 

In the result, the temperature of pot water will be about 8.5 degrees C after one hour.

Now, you can do an experimental test at your home to prove this method.

Perhaps, we say that the solution of this case is very easy because Mr. X can measure the temperature of pot water after next 5 minute then he can solve above function to get “k”. Yes, it is true but we should consider that there are some adjacent areas into the space or even into the earth with different temperatures where we have not access to measure the temperatures during the period of the time.

 

Note:  “All spreadsheets and calculation notes are available. The people, who are interested in having my spreadsheets of this method as a template for further practice, do not hesitate to ask me by sending an email to: soleimani_gh@hotmail.com or call me on my cellphone: +989109250225. Please be informed these spreadsheets are not free of charge.”

Appendix I

 

Appendix II

Appendix III