Market Value Capital Structure - Financial Calculations - Assignment Example

Question 1 Determine the company’s existing market value capital structure. Weight Market value Preference Shares $1,200,000.0 0.324324 389189.2 Ordinary Shares $1,000,000.0 0.27027 270270.3 Debentures $1,500,000.0 0.405405 608108.1 Total equity $3,700,000.0 1 3700000 b The company adjusts the cash-flows for flotation costs. Discuss the appropriateness of this approach. The flotation cost is adjusted from the cost of equity of the company in arriving at the intrinsic value of a share and thus the approach above is irrelevant since, it will have an effect on the real value 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 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 19.21-1.75/$19.21-0.5}=0.933 Ordinary Shares 1.75(0.99)/15.37)+10}=0.11 Debentures 12.5+(204-200/7 years/404/2}=0.065 Retained earning (1.75/19.21)=0.091 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= 1-n(root(Do/d1} It is arrived at by less one form the number of years of dividend paid then taking the square root of dividend last paid divided by current dividend as depicted in the formula above. f Assuming that the company maintains this optimum market value capital structure, calculate the breaking points associated with each source of capital. Column1 Column2 Column3 Column4 Column5 Column6 Column7 Kp 0.93 Ks 0.11 Kd 0.065 Kr 0.0091 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 0.32 0.93 0.2976 Ordinary Shares 0.18 0.11 0.0198 Debentures 0.4 0.07 0.028 Retained earning 0.087 0.091 0.007917 Weighted average cost 9.1% 1000000-1200000 Preference Shares 0.32 0.93 0.2976 Ordinary Shares 0.18 0.11 0.0198 Debentures 0.4 0.07 0.028 Retained earning 0.087 0.098 0.008526 Weighted average cost 9.3% 1000000-1200000 Preference Shares 0.32 0.93 0.2976 Ordinary Shares 0.18 0.11 0.0198 Debentures 0.4 0.065 0.026 Retained earning 0.087 0.097 0.008439 Weighted average cost 11% H Calculate the WACC for each range of total new financing. Range of financing Source of capital Weighted cost 0-1000000 Weighted average cost 9.1% 1000000-1200000 Weighted average cost 9.3% 1000000-1200000 Weighted average cost 11% 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 $82,650 6 15 B $200,000 $52,840 5 15 C $800,000 $163,400 7 14 D $500,000 $115,240 6 13 E $300,000 $61,600 7 12 F $500,000 $158,120 4 11 G $500,000 $131,270 5 10 I Using your findings above with the Investment Opportunities Schedule (IOS), draw the company’s weighted marginal cost of capital (WMCC) function and IOS on the same set of total new financing or investment (x-axis) - WACC and IRR (y-axis) axes. Question 2 Column1 Column2 Column3 Column4 Column5 Column6 Column7 15 Kp 0.93 14 Ks 0.11 14 Kd 0.065 13 12 Kr 0.0091 11 10 400 200 300 400 500 New Financing J Which, if any, of the available investments would you recommend the firm accept? The company should consider taking an investment F since, the IIR equals the cost of equity which is an implication that the company will generate maximum returns in this investment option. k Assuming that the specific financing costs do not change 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 0.93 14 Ks 0.11 14 Kd 0.065 13 12 Kr 0.0091 11 10 400 200 300 400 500 New Financing The increase leverage level to 60% debt and 2% preference would to a decline in value of the firm and increase in cost of capital as observed above. L Which capital structure - the original one or this one - seems better? Why? In this regards, option one which is the original one is deem significant since, it depict a low cost with high value which is an optimal capital structure of the firm. 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 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 Variable cost per unit {4470-2820/480-300} = {1650/180}=9.2 per unit Total fixed cost {2820+ (9.2*300) =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 4300 38850 15693900 1628300 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= 3647.5 Variable cost per unit=19.27x 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 Variable cost per unit {4470-2820/90000-55000} = {1650/35000}=0.03 per unit Total fixed cost {2820+ (0.03*300) =2820+9=$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 785000 38850 2839780000 1628300 Regression equation Y=224.77+0.053 PART C 1. Use multiple regression analysis to estimate the behaviour of the maintenance costs. SUMMARY OUTPUT Regression Statistics Multiple R 0.979806 R Square 0.960019 Adjusted R Square 0.955577 Standard Error 2383.207 Observations 11 ANOVA   df SS MS F Significance F Regression 1 1.23E+09 1.23E+09 216.109 1.34E-07 Residual 9 51117064 5679674 Total 10 1.28E+09         Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0% Intercept 6952.37 4440.057 1.565829 0.15183 -3091.74 16996.48 -3091.74 16996.48 X Variable 1 18.23742 1.240587 14.70064 1.34E-07 15.43102 21.04383 15.43102 21.04383 2. Interpret all significant statistics. The above table provides the whole goodness of fit test R2=0.96 Correlation between Y and Y-Hat is 0.9798 where the squared provides 0.96 Adjusted R2 = R2 - (1-R2 )*(k-1)/(n-k) = .96 - .5679*2/2 = 0. 0.955577 The standard error here is the approximated standard deviation of the error term. The R squared implies that 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 worth of yi  forecasted from the regression line ybar is the sample mean of  y. Because 1.565829> 0.05, we do not reject H0 at 5% significance level From the ANOVA table the F-test statistic is 1.34with p-value of 0.15183 because the p-value is 0.1518>0.05 we do not reject the null hypothesis that the regression parameters are zero at 5% significance level It can therefore be Concluded that the parameters are mutually statistically inconsequential at significance 5% 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=695.64+7.26(600) =5051.4 Total variable cost 4356$ 87 000 photocopies Y=224.77+0.053(870000} =4837.55 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 16.66666667 -1 2 0 8.333333333 Do all the corner points at once: Corner Point Value of P = 0.12x + 0.10y 1 1 25 16.66666667 2.83333333 -1 2 0 8.333333333   1 1 25 10 2.7 1 0 10 15   1 0 10 10 1.7 -1 2 0 5   1024 Maximum x and y values 10.24 Difference between "shade" lines Points of intersection -640 2240 640 -2240 -1 -640 2240 640 -2240 -1 -640 2240 -640 2240 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