Company's Market Value Capital Structure - Assignment Example

Question 1 Determine the company’s existing market value capital structure.     Weight Market value Preference Shares $1,200,000 $0.3 $389,189 Ordinary Shares $1,000,000 $0.3 $270,270 Debentures $1,500,000 $0.4 $608,108 Total equity $3,700,000 $1.0 $3,700,000 B The Company adjusts the cash-flows for flotation costs. Discuss the appropriateness of this approach. The flotation cost is deducted from the cost of equity (Ke) of the business in determining the intrinsic value of a share and consequently the approach above is immaterial because, it will have an upshot on the real worth of the company’s intrinsic value. C Calculate the after tax cost of each source of financing KD= {Dividend (net of tax/Market price} Ks= {Dividend (1+growth}/value of equity} +growth Growth rate Dividends per share over the past five years Dollar change Percentage change] 2007 $1.00 $0.00 0% 2006 $0.98 $0.02 10% 2005 $0.92 $0.06 35% 2004 $0.83 $0.09 51% 2003 $0.83 $0.01 4% $0.18 25% Preference Shares {19.2-1.76/$19.2-0.5}=0.94 Ordinary Shares 1.76(0.99)/15.4)+10}=0.12 Debentures 12.5+(204-200/7 years/404/2}=0.07 Retained earning (1.75/19.21)=0.09 D Discuss an alternative manner to calculate the Cost of Equity Cost of equity is arrived at by adding the value of risk free rate and the beta then dividing by average market return and less the risk free rate of return. E Discuss an alternative calculation of the growth factor used in the dividend growth model Growth rate is computed by taking the square root of dividend last paid divided by forecasted dividend and Lessing one from the result. Growth rate= 1-n√ (Do/D1} F Assuming that the company maintains this optimum market value capital structure calculates the breaking points associated with each source of capital. Column1 Column2 Column3 Column4 Column5 Column6 Column7 Kp 0.94 Ks 0.12 Kd 0.07 Kr 0.009 G Using the breaking points developed determine each of the ranges of total new financing over which the company’s WACC remains constant. Range of financing Source of capital Weight cost of capital Weighted cost 0-1000000 Preference Shares 0.3 0.9 0.3   Ordinary Shares 0.2 0.1 0.0   Debentures 0.4 0.1 0.0   Retained earning 0.1 0.1 0.0   Weighted average cost   0.1           1000000-1200000 Preference Shares 0.3 0.9 0.3   Ordinary Shares 0.2 0.1 0.0   Debentures 0.4 0.1 0.0   Retained earning 0.1 0.1 0.0   Weighted average cost   0.1           1000000-1200000 Preference Shares 0.3 0.9 0.3   Ordinary Shares 0.2 0.1 0.0   Debentures 0.4 0.1 0.0   Retained earning 0.1 0.1 0.0   Weighted average cost   0.1 H Calculate the WACC for each range of total new financing. Range of financing Source of capital Weighted cost 0-1000000 Weighted average cost 9.2% 1000000-1200000 Weighted average cost 9.4% 1000000-1200000 Weighted average cost 12% Complete the Investment Opportunities Schedule by estimating the missing Internal Rates of Return           Project Project Estimated Estimated Internal Identification Cost Annual Life Rate     Cash Inflows (in years) of Return A $400,000.0 $82,650.0 $ 6.00 15% B $200,000.0 $52,840.0 $ 5.00 15% C $800,000.0 $163,400.0 $ 7.00 14% D $500,000.0 $115,240.0 $ 6.00 13% E $300,000.0 $61,600.0 $ 7.00 12% F $500,000.0 $158,120.0 $4.00 11% G $500,000.0 $131,270.0 $5.00 11% Company’s weighted marginal cost of capital (WMCC) function and IOS. Column1 Column2 Column3 Column4 Column5 Column6 Column7 15 Kp 0.94 14 Ks 0.12 14 Kd 0.07 13 12 Kr 0.009 11 10 400 200 300 400 500 New Financing Which, if any, of the available investments would you recommend the firm accept? The business must undertake an investment F in view of the fact that, the internal rate of return is the same as the cost of equity which is implies that the business will generate utmost returns in this investment alternative. What effect would a shift to a more highly levered capital structure consisting of 60 per cent long-term debt, 20 per cent preference capital and 20 per cent ordinary equity have on your findings above? 15 Kp 0.94 14 Ks 0.12 14 Kd 0.07 13 12 Kr 0.009 11 10 400 200 300 400 500 New Financing The growth in the level of leverage to 60% debt and 2% preference may lead to a reduction in value of the company due to the fact that an increase in cost of capital as observed above. Which capital structure - the original one or this one - seems better? Why The optimal capital structure from the above analysis is the original one due to the fact that it portray a low cost of capital with high worth that is optimal in terms of debt and equity of the Question 2 PART A Scatter graph Visually fit a cost line to the scatter diagram Estimate the variable and fixed components of the department’s cost behaviour pattern using the visually fit cost line Regression line (Y0= bX+a) bX is the variable cost per unit while (a) is the fixed cost and hence a regression equation is derived as follows. Y=-8.8x+3657 Where -8.x is the variable cost per unit 3657 the fixed cost I. Estimate the cost behaviour using the high-low method. Use an equation to express the results of this estimation method. High low method Maintenance hours(X) Maintenance cost(Y) High 480 4470 Low 300 2820 Maintenance cost= {4470-2820) = 1650 Maintenance hours= (480-300} =180 Variable cost per unit= {1650/180} =9.2 per unit Total fixed cost {2820+ (9.2*300) Fixed cost =2820+2750=$5570 Y=5570+9.2x Compute the least-squares regression estimate of the variable and fixed-cost components Regression line (Y) Y= {a+bx} x y X'Y X^2 January 470 4050 1903500 220900 February 350 3300 1155000 122500] March 340 3160 1074400 115600 April 320 3030 969600 102400 May 520 4470 2324400 270400 June 490 4260 2087400 240100 July 300 2820 846000 90000 September 310 2960 917600 96100 October 480 4200 2016000 230400 November 320 3000 960000 102400 December 400 3600 1440000 160000 4320 38840 15694900 1638300 Y=695.647.26x PART B: Units produced and maintenance costs 1. Draw a scatter diagram of the cost data. 2. Visually fit a cost line to the scatter diagram. 3. Estimate the variable and fixed components of the department’s cost behaviour pattern using the visually fit cost line and specify an equation to express the department’s cost behaviour. Fixed cost= 3648 Variable cost per unit=19.3x Y= {19.3X+3648 4. Estimate the cost behaviour using the high-low method. Use an equation to express the results of this estimation method. Maintenance hours(X) Maintenance cost(Y) High 90000 4470 Low 55000 2820 Maintenance cost {4470-2820) =1650 Maintenance hours (90000-55000} =35000 Variable cost per unit = {1650/35000} =0.03 per unit Total fixed cost {2820+ (0.03*300) Fixed cost= {2820+9=$2829} = 2829 Y=2829+0.03x Compute the least-squares regression estimate of the variable and fixed-cost components x y X'Y X^2 January 80000 4050 324000000 220900 February 68000 3300 224400000 122500] March 69000 3160 218040000 115600 April 64000 3030 193920000 102400 May 90000 4470 402300000 270400 June 85000 4260 362100000 240100 July 55000 2820 155100000 90000 September 62000 2960 183520000 96100 October 82000 4200 344400000 230400 November 60000 3000 180000000 102400 December 70000 3600 252000000 160000 784000 38950 2839680000 1638300 Regression equation Y=224.8+0.06 PART C Use multiple regression analysis to estimate the behaviour of the maintenance costs. SUMMARY OUTPUT Regression Statistics Multiple R 0.98 R Square 0.96 Adjusted R Square 0.96 Standard Error 2383.21 Observations 11.00 ANOVA   df SS MS F Significance F Regression 1 1.2300E+09 1.2300E+09 2.1611E+02 1.3400E-07 Residual 9 5.1117E+07 5.6797E+06 Total 10 1.2800E+09         Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 6952.4 4440.1 1.6 0.2 -3091.7 16996.5 -3091.7 16996.5 X Variable 1 18.2 1.2 14.7 0.0 15.4 21.0 15.4 21.0 1. Interpret all significant statistics. The above table provides the whole goodness of fit test R2=0.96 Correlation between Y and Y-Hat is 0.98 where the squared provides 0.96 Adjusted R2 = R2 - (1-R2 )*(k-1)/(n-k) = .96 - .57*2/2 = 0. 0.96 The standard error here is the estimated standard deviation of the error term. The R squared means that 0.96 of the variation of yi around ybar (its mean) is explained by the regressors x2i and x3i. Interpreting the ANOVA test The ANOVA table divides the sum of squares into its constituent . Total sum of square= Residual (or error) sum of squares + Regression sum of squares. Thus Σ i (yi - ybar)2 = Σ i (yi - yhati)2 + Σ i (yhati - ybar)2 where yhati is the value of yi  approximated from the regression line ybar is the sample mean of  y. Since 1.57> 0.05, we don’t reject H0 at 0.05 significance level From the ANOVA table the F-test statistic is 1.35with p-value of 0.152 since the p-value is 0.152>0.05 we don’t reject the null hypothesis that the regression parameters are zero at 0.05 significance level It might consequently be realized that the parameters are equally statistically insignificant at 0.05 significance level. Calculate the total variable cost and the fixed cost per maintenance hour at 600 hours of maintenance and 87 000 photocopies. Explain the problems might occur when using fixed cost per hour in decisions At 600 hours of maintenance Y=696+7.3(600) =5051 Total variable cost 4356$ 87 000 photocopies Y=224.8+0.054(870000} =4837.6 Total variable cost 4611$ Question 3 Maximize p = x1600+960x + 350y subject to the constraints x + y = 960 x >= 0, y >= 350 x >= 0, y >= 0 Solve the related equations for y y = 2560 - x A x = 960 B y = 0.5x C Filling the table and graph X Y 0 25 30 -5 10 0 10 30 0 0 30 15 Here is the augmented matrix of the system of equations: 1 1 25 16.7 -1 2 0 8.3 Do all the corner points at once: Corner Point Value of P = 0.12x + 0.10y 1 1 24 16.7 2.8 -1 3 0 8.4   1 2 24 11 2.8 1 0 11 14   1 1 11 12 1.8 -1 3 0 6   1025 Maximum x and y values 10.3 Difference between "shade" lines Points of intersection -650 2250 650 -2250 -1 -650 2250 650 -2250 -1 -650 2250 -650 2250 1 1) What effect would the following production changes have on your solution to 1 above? Outlet Strip Labour hours to manufacture one unit: 2 Machine hours to assemble one unit: 3 Surge Protector Contribution margin per unit: $9 Labour hours to manufacture one unit: 3 Machine hours to assemble one unit: 2 Constraints Total labour time available each month: 1500 hours Total machine time available each month: 900 hours Production of outlet strips cannot exceed 380 units per month Objective function; 3x+2y>9 3x Read More