Example Analysis

As mentioned in the chapter 3- Methodology, the analysis was conducted at a single firm level then the same proceed with each of the consider firms will be taken into analysis. In order to better understand the proposed model, the writer develops a numerical example below with an assumption that analyze a set of F = 15 firms, for five years. In the analysis, the existence of nineteen symptoms as it follows:

Table 5.1: Selected variables and their default probability correlation

No. Symbol Ratio Sign

1 VAR 1 Current Ratio +

2 VAR 2 Quick Ratio +

3 VAR 3 Net Working Capital to Total Assets + 4 VAR 4 Current Assets to Net Assets + 5 VAR 5 Total Liabilities to Net Worth -

6 VAR 6 Retained Earnings to Sales +

7 VAR 7 Debt Ratio -

8 VAR 8 Times Interest Earned +

9 VAR 9 Revenues to Net Working Capital + 10 VAR 10 Accounts Receivable Turnover + 11 VAR 11 Accounts Payable Turnover -

12 VAR 12 Sales to Net Worth +

47

5 year history data consequence of one sample firm are arrange into the table 5.2. The data of another 14 companies are show later in the list of figure.

Table 5.2: 5 year history data of firm No.1

Firm No.1

(ACMTA) Year I Year II Year III Year IV Year V Actual Year 1972 1973 1974 1975 1976 VAR1 X1 3.4330 3.5270 1.4545 2.5931 4.1683 VAR2 X2 2.8325 2.9562 1.2634 2.3115 3.4158 VAR3 X3 0.6487 0.5266 0.2332 0.3637 0.3850 VAR4 X4 1.2481 0.9285 1.5324 0.7672 0.5765 VAR5 X5 0.3963 1.0242 3.5241 1.6003 0.9285 VAR6 X6 0.2081 0.2121 0.0398 0.0607 0.1558 VAR7 X7 0.2838 0.5060 0.7790 0.6154 0.4815 VAR8 X8 37.7037 13.8194 -9.5519 1.1197 1.4519 VAR9 X9 3.6344 3.4677 3.5581 4.1834 2.1663 VAR10 X10 4.9930 3.3645 2.2740 2.0995 1.9437 VAR11 X11 26.4538 17.4515 11.1902 9.9414 10.1431 VAR12 X12 3.4485 3.4444 3.7805 4.1834 1.8246 VAR13 X13 16.7764 11.9245 14.4978 10.9723 6.8751 VAR14 X14 2.5708 1.9802 1.3505 1.1765 0.8079 VAR15 X15 40.3190 9.9322 5.2065 3.7887 1.8149 VAR16 X16 0.1528 0.0954 -0.2322 0.0503 0.0517 VAR17 X17 0.2030 0.1702 -1.2452 0.0080 0.0192 VAR18 X18 0.0704 0.0540 -0.2042 0.0019 0.0107 VAR19 X19 0.2242 0.1597 -1.1804 0.0085 0.0259

1. Absolute Degree of Grey Incidence

48 The sequence to compute the absolute matrix of incidence as the consequence X0 = X1 for firm 1 as follow:

 Step 1: Compute image of zeroing starting point:

Xj = (x1j- x1j, x2j - x1j, x3j - x1j, x4j - x1j, x5j - x1j) (5.1)

Whereas j = 1, ... , n. n = number of symptom. In this example, n = 19.

For example:

X1 = {x(1); x(2); x(3); x(4); x(5)}

= (3.4330; 3.5270; 1.4545; 2.593; 4.1683) X1 = (x1,1- x1,1; x2,1 - x1,1; x3,1 - x1,1;x4,1 - x1,1; x5,1 - x1,1)

= (0; 0.094; -1.978; -0.840; 0.735) X2 = {x(1); x(2); x(3); x(4); x(5)}

= (2.8325; 2.9562; 1.2634; 2.3115; 3.4158) X2 = (x1,2- x1,2; x2,2 - x1,2; x3,2 - x1,2; x4,2 - x1,2; x5,2 - x1,2)

= (0; 0.124; -1.569; -0.521; 0.583; -1.675) This process will be continued until X19.

 Step 2: Find |s0| , |sj| , and |sj - s0| For X0 (= X1):

| | | | 5 (5.2)

= 0.094 1.978 0.840 0.735 2.357

For j = 2, 3, …, 19. Take j = 2 as an example, the same formula will be accounted for other value of j.

| | 5 (5.3)

49

= 0,124 1569 0. .521 0.583 1.675

| | 5 5 (5.4)

∑ 0.124 0.094 1.569 1.978 0,521

0.840 0.583 0.735 0.682

 Step 3: Attain the absolute degree of grey incidences:

| | | |

1 2.357 1.675 |1.675 2.357|

0.881

Table 5.3: The absolute j value of firm No.1

Firm 1

50 X₁₇⁰ 0 -0.033 -1.448 -0.195 -0.184 1.768 0.589 0.897 X180 0 -0.016 -0.275 -0.068 -0.060 0.389 1.968 0.656 X₁₉⁰ 0 -0.064 -1.405 -0.216 -0.198 1.784 0.573 0.900

2. Relative Degree of Grey Incidence

The relative degree of grey incidence is obtained using the following relations:

 Step 1: Compute the initial images of X0 and Xj

= (3.4330; 3.5270; 1.4545; 2.593; 4.1683)

, , , , j= [2,19]. (5.7) Take j = 2 for example:X2 = (2.8325; 2.9562; 1.2634; 2.3115; 3.4158)

1

51 0; 0.044; ‐0.554; ‐0.184; 0.206

 Step 3: Compute | |, and

| | ∑ 5 (5.10)

∑ 5 (5.11)

∑ 5 5 (5.12)

o | | 0.027 0.576 0.245 0.214 0.6865

o | | 0.044 0.554 0.184 0.206 0.5915

o | | 0.044 0.027 0.554 0.576 0.184

0.245 0.206 0.214 0.095

Table 5.4: The initial images value of firm No.1

Firm No.1

(ACMTA) X1’ X2' X3' X4' X5'

X10' 1 1.027 0.424 0.755 1.214

X20' 1 1.044 0.446 0.816 1.206

X30' 1 0.812 0.359 0.561 0.593

X40' 1 0.744 1.228 0.615 0.462

X50' 1 2.585 8.893 4.038 2.343

X60' 1 1.020 0.191 0.292 0.749

X70' 1 1.783 2.745 2.168 1.696

X80' 1 0.367 -0.253 0.030 0.039

X90' 1 0.954 0.979 1.151 0.596

X100' 1 0.674 0.455 0.420 0.389

X110' 1 0.660 0.423 0.376 0.383

X120' 1 0.999 1.096 1.213 0.529

X130' 1 0.711 0.864 0.654 0.410

X140' 1 0.770 0.525 0.458 0.314

X150' 1 0.246 0.129 0.094 0.045

X160' 1 0.624 -1.520 0.329 0.339

X170' 1 0.838 -6.133 0.040 0.095

X180' 1 0.768 -2.903 0.028 0.152

X190' 1 0.712 -5.266 0.038 0.115

52

 Step 4: Compute the relative degree of incidence

| |

3. Compute the synthetic degree of incidence: 

ρ1j = θ. 1j + (1- θ).r1j (5.14)

53 Whereas:

ρ1j: The synthetic degree of incidence

1j: The absolute degree of incidence r1j: The relative degree of incidence

With j = 2... n, (n = number of symptom. This example, n = 19).θ [0,1]

Table 5.6: The synthetic ρj value of firm No.1

Firm No.1 1j r1j ρ1i

X1 1 1 1

X2 0.8806 0.9599 0.9202

X3 0.7546 0.8009 0.7778

X4 0.7367 0.9984 0.8675

X5 0.5309 0.5174 0.5242

X6 0.6466 0.7794 0.7130

X7 0.5624 0.5478 0.5551

X8 0.5113 0.6546 0.5829

X9 0.6624 0.7603 0.7114

X10 0.6542 0.7630 0.7086

X11 0.5289 0.7525 0.6407

X12 0.5804 0.6985 0.6395

X13 0.5777 0.8788 0.7282

X14 0.8114 0.7839 0.7977

X15 0.5117 0.6691 0.5904

X16 0.6918 0.6349 0.6633

X17 0.8969 0.5644 0.7307

X18 0.6556 0.5984 0.6270

X19 0.8997 0.5701 0.7349

As mentioned in the chapter 3, generally researchers take θ = 0.5 to calculate the value of ρ. In this research, firstly, the author will take θ = 0.5 follow by previous researcher to consider the effect of other factor like as X0, the reasonable number of

54 collected data year. After that, the value of θ will be taking into account. So, the formula (5.14) can be converted as follow:

ρ1j = 0.5 1j + 0.5r1j

According to grey analysis theory’s principle, the greater value of ρi the larger effect of variable Xi into the firm’s default (or non-default) characteristics. Therefore, the result are presented in above table 5.6, the VAR2 play the most important role to the firm No.1’s default characteristics when X0 = X1.

After the analysis was conducted at a single firm level, the same process with each of the 15 firms (the first 15 firms) will be taken into analysis. By using an excel worksheet, we obtain a matrix of synthetic degree of grey

incidence:

While as: N = the total number of symptoms manifested at the level of considered firms (N = 19), and F = the total number of firms (F = 15).

55 The figures above show the matrix of synthetic degree (ρ) with the number of symptoms is 19. This is the number of symptom in the previous thesis.

An essential emphasis point in the default analysis process is that different financial ratios which have contrary effect on firms’ financial statement. In the case X0

= Xi, while Xi stands for a positive symptom, it means that the smaller the level of an aggregated intensity, the more likely is that the firm to become bankrupt, and vice versa when Xi stands for a negative symptom.

The next step after obtaining a matrix of synthetic degree of grey incidence, we will establish intensity levels for each firm. By aggregation, we obtain a matrix of

intensity level for each symptom and firms, in the form:

11 1

The intensity of each firm was proposed to conduct form the synthetic degree of grey incidence as follow: in the case a symptom is positive, we attribute an intensity level q, equal to ρ; if negative symptom, the award will be in reverse order, q = 1 – ρ.

var 2 var 3 var 4 var 5 var 6 var 7 var 8 var 9 var 10 var 11 var 12 var 13 var 14 var 15 var 16 var 17 var 18 var 19

+ + + ‐ + ‐ + + + ‐ + + + + + + + +

1 0.78 0.87 0.48 0.71 0.44 0.58 0.71 0.709 0.359 0.639 0.728 0.798 0.59 0.663 0.731 0.627 0.735 1 0.75 0.62 0.42 0.67 0.35 0.51 0.58 0.646 0.211 0.535 0.517 0.53 0.506 0.693 0.554 0.663 0.822 1 0.86 0.98 0.04 0.8 0.17 0.54 0.66 0.749 0.392 0.746 0.607 0.696 0.835 0.654 0.623 0.656 0.611 1 0.73 0.65 0.17 0.65 0.31 0.59 0.55 0.661 0.194 0.81 0.562 0.663 0.694 0.646 0.631 0.625 0.598 1 0.99 0.94 0.01 0.9 0 1 0.69 0.735 0.331 0.687 0.591 0.883 0.663 0.928 0.915 0.903 0.903 1 0.67 0.61 0.46 0.65 0.46 0.52 0.53 0.544 0.465 0.539 0.528 0.541 0.594 0.537 0.536 0.54 0.533 Q =  1 0.64 0.64 0.38 0.58 0.44 0.63 0.52 0.579 0.317 0.534 0.522 0.538 0.809 0.693 0.713 0.723 0.762 1 0.72 0.55 0.46 0.74 0.45 0.52 0.53 0.542 0.214 0.535 0.528 0.538 0.536 0.543 0.541 0.542 0.537 1 0.78 0.83 0.1 0.64 0.25 0.64 0.61 0.645 0.326 0.799 0.941 0.949 0.711 0.652 0.691 0.639 0.623 1 0.7 0.58 0.43 0.7 0.42 0.56 0.56 0.745 0.463 0.561 0.601 0.579 0.559 0.587 0.588 0.677 0.591 1 0.9 0.73 0.47 0.65 0.09 0.53 0.59 0.608 0.442 0.813 0.585 0.613 0.602 0.613 0.525 0.62 0.575 1 0.88 0.7 0.28 0.83 0.09 0.6 0.64 0.712 0.332 0.694 0.619 0.941 0.651 0.941 0.94 0.859 0.874 1 0.79 0.57 0.46 0.65 0.45 0.67 0.56 0.638 0.219 0.57 0.729 0.683 0.82 0.57 0.565 0.576 0.559 1 0.79 0.85 0.24 0.64 0.33 0.62 0.72 0.634 0.406 0.87 0.823 0.851 0.656 0.734 0.703 0.671 0.706 1 0.87 0.75 0.13 0.91 0.04 0.55 0.65 0.676 0.147 0.646 0.626 0.693 0.666 0.711 0.662 0.696 0.611

56 On our numerical example, the default characteristics of firm in this case was presented by X0 = X1, the current ratio, it means that the smaller the level of intensity, the more likely is that the firm to become bankrupt. Among the considered firms, we can easily see that the firm nr.2, firm nr.6 and firm nr.8 or may be firm nr.4 record low level of intensity for most of the symptoms,comparatively, while firm nr.5 presents the best from this regard. This leads to a point that the firm nr.2, firm nr. 6 and firm nr.8 are the most weakness companies prone to default risk. Form the chart, it do not shows exactly default firm, it just shows which firm may be bankrupt. In this case, the firm nr.2, firm nr.6 and firm nr.8 were predicted bankruptcy.

Fig. 5.1: Sum of the registered intensity levels calculated based on matrix Q