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He following data relates to a company in the manufacturing sector in Kenya for the year ending 31 December 2019
He following data relates to a company in the manufacturing sector in Kenya for the year ending 31 December 2019.
Kshs. “000”
Sales
25,678
Total Assets
49,579
Total Liabilities
5,044
Retained Earnings
1,77
Working Capital
-1,777
Earnings before Interest and Taxes
2,605
Market value of Equity
10,098
Book value of Total Liabilities
5,044
The company is paying interest on a long-term debt instrument amounting to Kshs. 905,000 per year and that the company’s total liabilities is constituted in the ratio of 2:5 between current and non-current components.
Using the Springate model assess the financial health of the company
(3 Marks)
Taking each variable above explain how the management of the above company can improve the financial health of this company
Expert Solution
As per the given scenario,
| S. No | Particulars | Amount (in Kshs.) |
| 1 | Sales | 25678000 |
| 2 | Total Assets | 49579000 |
| 3 | Total Liability | 5044000 |
| 4 | Retail Earnings | 177000 |
| 5 | Working Capital | -1777000 |
| 6 | EBIT | 2605000 |
| 7 | MV of Equity | 10098000 |
| 8 | BV of Total Liability | 5044000 |
Profit before Tax = EBIT - Interest = 2605000 - 905000 = 1700000
Current Liability = Total Liability x 2/7 = 5044000 x 2/7 = 1441143
Non Current Liability = Total Liability x 5/7 = 5044000 x 5/7 = 3602857
Springate Model Formula.
Z = 1.03A + 3.07B + 0.66C + 0.4D
A = Working Capital / Total Assets = -1777000/49579000 = -0.03584
B = EBIT / Total Assets = 2605000/49579000 = 0.05254
C = EBT / Current Liability = 1700000/1441143 = 1.17961
D = Sales / Total Assets = 25678000/49579000 = 0.51792
Therefor,
Z = 1.03A + 3.07B + 0.66C + 0.4D
= (1.03 x -0.03584) + (3.07 x 0.05254) + (0.66 x 1.17961) + (0.4 x 0.51792)
= -0.03692 + 0.161305 + 0.77854 + 0.207168
= 1.110105
According to Springate Model, It has a standard calcuation in which if the company has a value Z > 0.0862 then it will be classified as a healthy company, While company's with a value of Z < 0.0862 then it will be classified as potential company bankrupty.
As per above scenario, Z = 1.110105. Were Z is greater than 0.0862. It means the company's financial health is good.
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