A Model to Derive Strategies in Strategic Management

One of the most important and also hardest step of a strategic analysis or designing strategic plan is to derive and to develop strategies by using of SWOT matrix. Because we have to compare and to match SO (strengths – opportunities) Strategies, WO (weakness – opportunities) Strategies, ST (strengths – threats) Strategies and WT (weaknesses – threats) Strategies to make new ones by thinking and using of our own ideas. Consequently, we need to take high level of focus and concentration on our job. Unfortunately, by using only SWOT matrix as a tool, we are not able to focus at ease on all strengths, opportunities, weaknesses and threats simultaneously. The purpose of this article is to increase students and strategic analysts’ convenience to have better concentration on content of SWOT matrix to make reasonable strategies. In this case, a new simple rule in set theory has been utilized to solve a case in excel and then by using of the result of excel case and also Monte Carlo method, a template on excel spreadsheet to derive strategies has been made and presented.

A new simple rule in set theory

In the reference with item 5 and rule 5 included in article of “Are These Rules new Conjectures in Set Theory?” posted on link:  http://emfps.blogspot.com/2015/11/are-these-rules-new-conjectures-in-set_17.html, we have below formula:

N' = n - r +1

One of the properties this rule is, to invert members of a set for instance:

 A = {1, 2, 3, 4}

 B = {(4-1+1), (4-2+1), (4-3+1), (4-4+1)}

 I used this property to solve a case in excel.

Case: How can we turn a matrix to 180 degree in excel?

We can easily turn a matrix to 90 degree by using of TRANSPOSE’ formula in excel. I am not a specialist in excel maybe there is another formula to turn a matrix to 180 degree. Here, I am willing to show you my solution to solve this case step by step by an example as follows:

- Assume we have a matrix of 16*8 including members of a, b, c…

- Copy matrix between cells N1:U1 and N16:U16

- On cell M1 to M16 fill numbers 1 to 16

- On cell A1 fill formula = 16 – M1 + 1 which is the same n - r +1 and copy to A16

- On cell B1 copy below formula and copy to K1 and A16 to K16

 =IFERROR(INDEX($M:$W,MATCH($A1,$M:$M,0),COLUMN(B1)),"")

I have enclosed new matrix below cited:

A template on spreadsheet to derive strategies in Strategic Management

Assume, we have made SWOT matrix and we will derive SO strategies. I used an example extracted from book of “Strategic Management Concepts and Cases” Thirteenth Edition by Fred R. David, Chapter 6 and Figure 6-3 as follows:

                 Strengths

S1. Inventory turnover up 5.8 to 6.7

S2. Average customer purchase up $97 to $128

S3. Employee morale is excellent

S4. In – store promotions = 20% increase in sales

S5. Newspaper advertising expenditures down 10%

S6. Revenues from repair/service in-store up 16%

S7. In-store technical support persons have MIS degrees

S8. Store’s debt-to-total assets ratio down 34%

                 Opportunities

O1. Population of city growing 10%

O2. Rival computer store opening 1mile away

O3. Vehicle traffic passing store up 12%

O4. Vendors average six new products/yr

O5. Senior citizen use of computers up 8%

O6. Small business growth in area up 10%

O7. Desire for Web sites up 18% by Realtors

O8. Desire for Web sites up 12% by small firms

I utilized from Monte Carlo method in which I had already explained this method in many articles such as below ones:

-A Monte Carlo Analysis on Case of Nike, Inc.: Cost of Capital (link: http://emfps.blogspot.com/2012/01/monte-carlo-analysis-on-case-of-nike.html

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

- Application of Pascal’s Triangular plus Monte Carlo Analysis to Design a Strategic Plan (link: http://emfps.blogspot.co.uk/2012/07/application-of-pascals-triangular-plus_10.html)

How should we make Cut Off tables?

First of all, we should divide the probability distribution in accordance with the number of Strengths (S) and Opportunities (O).

The most crucial thing to bear in mind is, to make two Cut Off tables for Strengths (S) in which the probability distribution is constant but Strengths (S) are inverse and also this is for Opportunities (O). In this case, we will cover all Strengths (S) and Opportunities (O) on template.

In this step, we have to use to turn a matrix to 180 degree in excel.

I have attached the example for Cut Off tables as follows:

Now, we have four cell which we can change Strengths (S) and Opportunities (O) by each click on ENTER or F9 and also it cover all Strengths (S) and Opportunities (O) just like below cited:

Of course, for having a complete template, we should continue all above steps for WO (weakness – opportunities), ST (strengths – threats) and WT (weaknesses – threats).

Below video recorded shows you the components of this model:

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

https://t.me/Allisnumber

Are These Rules new Conjectures in Set Theory?

In the reference with article of “A Case of Accounting Control System Solved by a New Idea” posted on link: http://emfps.blogspot.com/2015/10/a-case-of-accounting-control-system.html, we can easily find out below rules in set theory:

1.  Let consider set “A” and power set of A which is set “C” as follows:

A = {a1, a2, a3, a4… an}

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by adding members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {a1+ a2}, {a1 + a3}, {a2 + a3 + a4)….{an}}

Rule 1: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

SB / SA = 2^ (n-1)               ,         n = number of members set A

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {1+2}, {1+3}, {1+4}, {2+3}, {2+4}, {3+4}, {1+2+3}, {1+2+4}, {1+3+4}, {2+3+4}, {1+2+3+4}}

SB = 0+1+2+3+4+3+4+5+5+6+7+6+7+8+9+10 = 80

SA = 1+2+3+4 = 10

SB / SA = 2^ (n-1) = 2 ^ (4-1) = 80 /10

2. Let consider A as set of Arithmetic Progression where:

d = 1,      a1 = 1   and    an = a1 + (n - 1) d, n = 1, 2, 3,…..

In this case, we have:

A = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1) d)}

And power set of A which is set “C” as follows:

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by multiplying members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {a1* a2}, {a1*a3}, {a2*a3*a4)….{an}}

Rule 2: If SB = total Sum of members of set B and SA = total sum of members of set A, we will have:

(SB+ 1) / SA = 2n!/n               ,         n = number of members set A

Or     SB = [(2n!/n)*SA] - 1 

Since for Arithmetic Progression of A, we have: SA= n (n+1)/2   then, this rule can be simplified to:

 SB=2n! n ∗n(n+1)2 −1=n!(n+1)−1=(n+1)!−1    

 SB = (n+1)!-1       

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {1*2}, {1*3}, {1*4}, {2*3}, {2*4}, {3*4}, {1*2*3}, {1*2*4}, {1*3*4}, {2*3*4}, {1*2*3*4}}

SA = 1+2+3+4 = 10   , SB = 1+2+3+4+2+3+4+6+8+12+6+8+12+24+24 = 119

(SB +1) / SA = (2*(1*2*3*4)/4) = 12 = (119 +1)/10

3. Let consider set “A” as subset of R and power set of A which is set “C” as follows:

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

We can find set “B” in which each member of set B is generated by average of members of each subset of power set C below cited:

B = {{}, {a1}, {a2}, {a3}, {a4}, {(a1+ a2)/2}, {(a1 + a3)/2}, {(a2 + a3 + a4)/3}….{an}}

Rule 3: If SB = total Sum of members of set B:

Then: SB = ((2^n)-1)*Average (A) , n = number of members set A

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

B = {{}, {1}, {2}, {3}, {4}, {(1+2)/2}, {(1+3)/2}, {(1+4)/2}, {(2+3)/2}, {(2+4)/2}, {(3+4)/2}, {(1+2+3)/3}, {(1+2+4)3}, {(1+3+4)/3}, {(2+3+4)/3}, {(1+2+3+4)/4}

SB = 0+1+2+3+4+1.5+2+2.5+2.5+3+3.5+2+2.333333+2.6666666+3+2.5 = 37.5

Average (A) = (1+2+3+4)/4 = 2.5

SB = ((2^ 4) -1)*2.5 = 37.5

4. Let consider set “A” and power set of A which is set “C” as follows:

A = {a1, a2, a3, a4… an}

C = {{}, {a1}, {a2}, {a3}, {a4}, {a1, a2}, {a1, a3},….{an}}

Rule 4: If N = number of sets included in C which are only contained sequence of members A except sets with one member

Then, we will have:   N = C (n, 2)

Example: Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

{1,2}, {2,3},{3,4}, {1,2,3},{2,3,4}, {1,2,3,4}

N = C (4, 2) = 4! / [2! (4 -2)!] = 6

5. Following to item 4 and rule 4 (previous rule), if number of sets is equal to N = C (n, 2) +n, (added by sets with one member) then we will have below rule:

Rule 5: Number of sets in power set (C) which are included sequence of numbers and for each specific size of members (r) can be calculated by below formula:

N (n, r) = n - r +1

Where:   n = number of members set A, r = specific size of members in sets with sequence numbers

Example:

Assume, we have:

A = {1, 2, 3, 4} then

C = {{}, {1}, {2}, {3}, {4}, {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, {3,4}, {1,2,3}, {1,2,4}, {1,3,4}, {2,3,4}, {1,2,3,4}}

n = 4,

r = 1, N (4, 1) = 4 and sets are: {1}, {2}, {3}, {4}

r = 2, N (4, 2) = 3 and sets are: {1, 2}, {2, 3}, {3, 4}

r = 3, N (4, 3) = 2 and sets are: {1, 2, 3}, {2, 3, 4}

r = 4, N (4, 4) = 1 and set is: {1, 2, 3, 4}

6. Let consider “A1” as set of Arithmetic Progression where:

d = 1,      a1 = 1   and    an = a1 + (n - 1) d, n = 1, 2, 3,…..

In this case, we have:

A1 = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1) d)}

One of permutations of set A1 is to invert members of set A1 as follows:

A2 = {(a1 + (n - 1) d),….,(a2 +1), (a1+1), a1}      

We can generate many sets which are the periodicity of set A1 just like below cited:

A3 = {a1, (a1+1), (a2 +1),……. (a1 + (n - 1) d)}

A4 = {(a1 + (n - 1) d),….,(a2 +1), (a1+1), a1}


Finally, we will have set B:


B = {A1, A2, A3, A4, …….An}     


Rule 6: If size members of set A1 or A2 orA3 or An, is equal to size members of set B, we will have a square matrix (Mn*n) to be generated by sets A1, A2, A3,…….An in which Eigenvalue of this matrix will be calculated by using of Binomial Coefficient as follows:

Eigenvalue (Mn*n) = λ = C (k, 2)

Where:   k = n+1, nϵ N, λ > 0


Example:

A1 = {1, 2, 3, 4}

A2 = {4, 3, 2, 1}      

A3 = {1, 2, 3, 4}

A4 = {4, 3, 2, 1}  


Matrix (M 4*4) = 
  
1 2 3 4

4 3 2 1

1 2 3 4

4 3 2 1

λ = C ((4+1), 2) = 10

7. Let consider set “A” as follows:

A = {x | x ϵ R}

Rule 7: Each type square matrix which has been generated from set “A”  just like below matrix:

a1 0 0 0 0 0 0 0 0….

a1 a2 0 0 0 0 0 0…..

a1 a2 a3 0 0 0 0 0 ….

a1 a2 a3 a4 0 0 0 0……

It will show us the eigenvalues which are just equal members set A which have been included in this square matrix (λ = a1, a2, a3, a4, …)


Example:

A = {0.67, 2, 43, 5, -23, 9, -2.3}

Assume we have set B which is a subset of A:

B = {43, -23, -2.3, 9)

Matrix N will be:

43   0  0  0  0

43  -23  0   0

43 -23 -2.3 0

43 -23 -2.3 9

Eigenvalues of Matrix N are: λ = 43, -23, -2.3, 9

8. Consider square matrices 2*2 as follows:

a    -a

b    -b

Or 

 a      b

-a    -b

a , b ϵ R

Rule 8: Eigenvalues of above matrices are equal to: λ = 0 and

λ = a – b

9. Consider square matrices 3*3 as follows:

 a     -a      a

 b     -b      b

 c     -c      c  

Or

  a     b       c

 -a    -b     -c

  a      b      c  

a , b , c ϵ R

Rule 9: Eigenvalues of above matrices are equal to: λ = 0 and

λ = (a + c)-b

10. Consider square matrices 4*4 as follows:

a      -a      a     -a

b     -b      b     -b

c     -c       c     -c

d     -d      d     -d

Or

  a     b      c      d

 -a    -b    -c     -d

 a      b      c      d

-a    -b     -c     -d

 a, b , c , d ϵ R

Rule 10: Eigenvalues of above matrices are equal to: λ = 0 and

λ = (a + c -d)-b

11. Consider square matrix “A” as follows:

            a       0      0

A =      a       b      0

            a       b      c  

If we turn this matrix around first column (as axis) and then we turn it around first row (as axis), we will have matrix “B”:

          c      b     a

B =    0      b     a

          0      0     a  

a , b , c ϵ R

Rule 11: Eigenvalue of A + B is equal to λ = c

and eigenvalue of A – B or B - A is equal to zero (λ = 0)

Example:

If matrix A is:

            1.67     0        0

A =      1.67     4        0

            1.67     4       -3  

            -3     4      1.67

B =       0      4     1.67

             0      0     1.67  

                    -1.33     4        1.67

A + B =        1.67      8       1.67

                     1.67      4      -1.33  

λ =  -3

                4.67     -4       -1.67

A - B =    1.67      0       -1.67

                1.67      4       -4.67  

λ = 0

12. Here is another special matrix. Consider matrix A is a square matrix n*n just like below forms in which (a1, a3, a5, a7 a9…an) are diagonal of matrix as follows:

A =

a1   a2    0     0     0
0    a3    a4    0     0
0    0      a5   a6    0
0    0      0     a7   a8
0    0      0      0   a9
…………………....an

Or

A =

a1     0    0    0    0

a2    a3    0    0    0

0     a4   a5    0    0

0     0    a6   a7    0

0     0    0    a8   a9

…………………....an

a1, a2, a3,….an ϵ R  

Rule 12: Eigenvalues of matrix A are all members on diagonal of matrix A. In fact, property of matrix A to generate eigenvalues is just like Diagonal Matrix.

Example:

A =

-1.56    2     0
  0     -12     3
  0        0     4  

λ = -1.56, -12 and 4

Consequently, we can find out that inverse of previous special matrix (Rule 7) will have the same eigenvalues (members of diagonal). According to Rule 7, we had below matrix M:

M =

a1 0 0 0 0 0 0 0 0….

a1 a2 0 0 0 0 0 0…..

a1 a2 a3 0 0 0 0 0 ….

a1 a2 a3 a4 0 0 0 0……

Therefore, eigenvalues of M^-1 are all members on diagonal of M^-1.

13.

Consider "T" as set of vectors as follows:  

   

T = {V1, V2, V3, ….Vn}

Where:

V1 = a1i

V2 = a2 i + a3 j

V3 = a4 i + a5 j + a6 k

Vn = a7 i + a8 j + a9 k + a10 t …….. amtn   and     a1, a2, a3, a4, a5, …. am are members of R (real 

numbers)

Theorem 13:

If "A" is a square matrix n*n inferred from the set of above vectors just like below cited:

A = {(a1, 0, 0, 0,…), (a2, a3, 0, 0, 0, …), (a4, a5, a6, 0, 0, 0, …..), …....Vn}

In fact, Scalar amounts of V1, V2, V3, …Vn are respectively rows of matrix A.

Then:   Eigenvalues of (A^n) = all members on diameter of matrix A^n (n member of N).

A =

4	0	0	0	0
-6.5	1	0	0	0
5	2	7	0	0
3	9	-5	12	0
-3.45	9	43	15	72

A^4 =

256	0	0	0	0
-552.5	1	0	0	0
2210	800	2401	0	0
-13147.5	13995	-18335	20736	0
-249381	4740402	17264326	6713280	26873856

Eigenvalues are = 256, 1, 2401, 20736, 26873856

Questions:

1. Is this theorem a new one? Have you any counterexample or a proof?

2. If the answer to question (1) is positive, it means that we can produce many theorems every day. Therefore, I think these theorems are not very important but the most crucial thing is, what are the applications of these theorems in our real life?