Calculating NPV Of Overhaul And Replacement Of Vital Spark

Particulars Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	Value
Sale of Vital Speak (A)	200,000
Book value (B) Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	140,000
Tax {C=[(A-B) * 35%]}	21,000
Sale proceeds [D = (A-C)]	179,000
Investment (E)	820,000
Initial Investment [F = (E-D)]	641,000

Question 1: Calculating NPV of Overhaul with and without New Machine

Treatment of Depreciation:

Particulars	1	2	3	4	5	6	7	8
Depreciation rate (A)	14.29%	24.49%	17.49%	12.49%	8.93%	8.93%	8.93%	4.45%
Int investment (B)	820,000
Tax rate (C)	35%
Depreciation (D=B*A)	(117,178)	(200,818)	(143,418)	(102,418)	(73,226)	(73,226)	(73,226)	(36,490)
Tax shield (E=-D*C)	41,012	70,286	50,196	35,846	25,629	25,629	25,629	12,772

Calculating operating cash flows:

Overhaul without New Machine:

Particulars	0	1	2	3	4	5	6	7	8	9	10	11	12
Initial investment	(820,000)
Revenue (A)		1,400,000	1,435,000	1,470,875	1,507,647	1,545,338	1,583,971	1,623,571	1,664,160	1,705,764	1,748,408	1,792,118	1,836,921
Operating cost (B)		(1,181,000)	(1,210,525)	(1,240,788)	(1,271,808)	(1,303,603)	(1,336,193)	(1,369,598)	(1,403,838)	(1,438,934)	(1,474,907)	(1,511,780)	(1,549,574)
PBT (C=A-B)		219,000	224,475	230,087	235,839	241,735	247,778	253,973	260,322	266,830	273,501	280,339	287,347
After tax Profit [D=C*(1-35%)]		142,350	145,909	149,556	153,295	157,128	161,056	165,082	169,209	173,440	177,776	182,220	186,776
Depreciation (E)		(117,178)	(200,818)	(143,418)	(102,418)	(73,226)	(73,226)	(73,226)	(36,490)
Tax shield (F)		41,012	70,286	50,196	35,846	25,629	25,629	25,629	12,772
Sale proceeds (G)	179,000
Cash Flow Cash Flow [H=D+F]	(641,000)	183,362	216,195	199,753	189,142	182,757	186,685	190,711	181,981	173,440	177,776	182,220	186,776

Overhaul with New Machine: 

Particulars	0	1	2	3	4	5	6	7	8	9	10	11	12
Initial investment	(1,420,000)
Revenue (A)		1,400,000	1,435,000	1,470,875	1,507,647	1,545,338	1,583,971	1,623,571	1,664,160	1,705,764	1,748,408	1,792,118	1,836,921
Operating cost (B)		(1,020,000)	(1,045,500)	(1,071,638)	(1,098,428)	(1,125,889)	(1,154,036)	(1,182,887)	(1,212,459)	(1,242,771)	(1,273,840)	(1,305,686)	(1,338,328)
PBT (C=A-B)		380,000	389,500	399,238	409,218	419,449	429,935	440,683	451,701	462,993	474,568	486,432	498,593
After tax Profit [D=(C*(1-35%))]		247,000	253,175	259,504	265,992	272,642	279,458	286,444	293,605	300,946	308,469	316,181	324,085
Depreciation (E)		(202,918)	(347,758)	(248,358)	(177,358)	(126,806)	(126,806)	(126,806)	(63,190)	–	–	–	–
Tax sheild (F)		71,021	121,715	86,925	62,075	44,382	44,382	44,382	22,117
Sale proceeds (G)	179,000
Cash Flow [H=D+F]	(1,241,000)	318,021	374,890	346,430	328,067	317,024	323,840	330,826	315,722	300,946	308,469	316,181	324,085

Calculating NPV:

Overhaul without New Machine:

Particulars	0	1	2	3	4	5	6	7	8	9	10	11	12
Cash Flow [H=D+F]	(641,000)	183,362	216,195	199,753	189,142	182,757	186,685	190,711	181,981	173,440	177,776	182,220	186,776
Discounted (I=H/((1+14%)^n)	(641,000)	161,553	167,824	136,617	113,973	97,027	87,324	78,597	66,078	55,486	50,109	45,252	40,867
NPV (Total I)	459,708

Overhaul without New Machine:

Particulars	0	1	2	3	4	5	6	7	8	9	10	11	12
Cash Flow [H=D+F]	(1,241,000)	318,021	374,890	346,430	328,067	317,024	323,840	330,826	315,722	300,946	308,469	316,181	324,085
Discounted (I=H/((1+14%)^n)	(1,241,000)	280,195	291,013	236,934	197,688	168,311	151,480	136,342	114,640	96,278	86,947	78,520	70,910
NPV (Total I)	668,258

From the overall evaluation use of Overhauling measures and New machine is the better option for the company, as it provides the highest NPV. In addition, the second option also has highest income due to reduced cost involved in operations. However, the extra burden in initial investment is due to the purchase of new machinery, which is used in reducing the overall cost of company. In addition, the NPV of Overhauling measures with New machine is at the levels of 668,258. On the other hand, the project with only overhauling has a NPV of 459,708. This relevantly indicates that the use of Overhauling measures with New machine will substantially increase the chance of the organisation to generate higher returns from investment. In this context, Baum & Crosby (2014) stated that with the help of investment appraisal techniques the financial viability of the project is detected, which helps the company to increase its firm value in future. On the contrary, Gotze, Northcott & Schuster (2016) argued that without adequate research the results obtained from investment appraisal techniques are impractical, as it does not depict the actual financial capability of the project or investment option. 

Overhaul without New Machine:

Particulars	0	1	2	3	4	5	6	7	8	9	10	11	12
Initial investment	(820,000)			6
Operating cost (A)		(1,181,000)	(1,210,525)	(1,240,788)	(1,271,808)	(1,303,603)	(1,336,193)	(1,369,598)	(1,403,838)	(1,438,934)	(1,474,907)	(1,511,780)	(1,549,574)
After tax cost [B=A*(1-35%)]		(767,650)	(786,841)	(806,512)	(826,675)	(847,342)	(868,526)	(890,239)	(912,495)	(935,307)	(958,690)	(982,657)	(1,007,223)
Depreciation (E)		(117,178)	(200,818)	(143,418)	(102,418)	(73,226)	(73,226)	(73,226)	(36,490)
Tax sheild (F=-E*35%)		41,012	70,286	50,196	35,846	25,629	25,629	25,629	12,772
Cash Flow [H=B+F]	(820,000)	(726,638)	(716,555)	(756,316)	(790,829)	(821,713)	(842,896)	(864,610)	(899,723)	(935,307)	(958,690)	(982,657)	(1,007,223)
Discounted (I=H/((1+14%)^n)	(820,000)	(640,209)	(556,234)	(517,268)	(476,540)	(436,255)	(394,275)	(356,327)	(326,695)	(299,221)	(270,221)	(244,032)	(220,382)
NPV (Total of H)	(5,557,659)

Calculating equivalent annual cost:

Solving for equivalent cost	Amount	Discounted	Years
Overhaul cost	5,557,659	13.50%	12
EAC	960,426.44	5.79

Solving for equivalent cost	Amount	Discounted	Years
Overhaul cost with NM	5,349,110	13.50%	12
EAC	924,386.74	5.79

Solving for equivalent cost	Amount	Discounted	Years
Replacing the boat	6,024,157	13.50%	20
EAC	883,449.43	6.82

From the overall evaluation Replacing the boat is much more beneficial for the company, as it portrays the least cost for improving its operations. Furthermore, from the overall evaluation it could be identified that equivalent annual cost of only overhauling is at 960,426.44, which is relevantly higher in comparison with other two options. Moreover, the equivalent annual cost of overhauling and new machine is at the levels of 924,386.74, which significantly better than only overhauling. However, equivalent annual cost of changing the boat to new one is only the at the levels of 883,449.43, which significantly better in comparison with other options presented to the company. However, Li & Trutnevyte (2017) argued that without the evaluation of actual cost of capital the results obtained from equivalent annual cost cannot help the company make adequate investment decision. Replacing the boat might help in generating high level of profits for the organisation, as it reduces the overall cost from operations. In addition, the equivalent annual cost incurred by the company on each year is relevant lower while replacing the boat, while comparing it with other cost and options. Upton et al., (2015) mentioned that detection of equivalent annual cost allows the organisation to identify value on opportunities with different levels of tenure. 

The data is mainly obtained from authentic data source such as yahoo finance, which provides all the relevant information regarding company and index price change. In addition, the relevant model helps in depicting the alpha and beta of the stock, which could be used by investor in making adequate decisions. These information is vital, which could be useful when drafting adequate portfolio that brings in more returns than investment. However, Brisley et al., (2016) stated that the statistical data using the regression is obtained on certain limited period, which could not help in detecting the actual risk involved in the investment. 

Question 2: Comparing Equivalent Annual Costs of Overhauling and Operating the Vital Spark and Buying and Operating a Replacement Vessel

Graph and Summary output of Myer Holding Ltd: 

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.371264626
R Square	0.137837423
Adjusted R Square	0.137409975
Standard Error	0.008068388
Observations	2019
ANOVA
	Df	SS	MS	F	Significance F
Regression	1	0.020992173	0.020992173	322.4659578	5.25038E-67
Residual	2017	0.131304445	6.50989E-05
Total	2018	0.152296618
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	0.000179214	0.00017957	0.998019819	0.318389461	-0.000172947	0.000531376	-0.000172947	0.000531376
Return MYR.AX	0.146548322	0.008160916	17.95733716	5.25038E-67	0.130543616	0.162553028	0.130543616	0.162553028

The above table mainly helps in identifying the overall beta and alpha of the stock, which would be used by investors in detecting its risk attributes. In addition, this also helps in detecting whether the investment opportunity within a stock is present for an investor. The above summary mainly depicts the results of MYR company, which is relatively in favour of the investors. The oral beta is relatively at the level of 0.146548322, which is relatively lower than 1 or 0.5. This relatively indicates that risk attributes of the organization is minimal in comparison to other stocks, which is an adequate investment opportunity for the investor. The investor could use the company in their portfolio to reduce the risk from investment while increasing portfolio return. However, Zhu et al., (2016) argued that during an economic crisis the actual returns and risk of the stock changes due to the high and volatility present within the capital market. 

The Alpha of the stock is at the levels of 0.000179214, which is relatively lower and indicates no movement of the organization in share market if the market index does not move.  this relatively indicates that the volatility of the capital market does not come back but the actual returns of the organization, which is a relatively a positive indication for the investor to reduce the risk from investment. In this context, Yang, Zheng & Zaheer (2015) mentioned that low beta stocks could allow investors to obtain minimum returns while nullifying the risk involved an investment.

Graph and Summary output of Rio Tinto Ltd: 

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.718207447
R Square	0.515821937
Adjusted R Square	0.515581888
Standard Error	0.00604637
Observations	2019
ANOVA
	df	SS	MS	F	Significance F
Regression	1	0.078557937	0.078557937	2148.822767	0
Residual	2017	0.073738682	3.65586E-05
Total	2018	0.152296618
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	3.0349E-05	0.000134589	0.225493385	0.821618315	-0.000233599	0.000294297	-0.000233599	0.000294297
Return RIO.AX	0.364382078	0.007860618	46.35539631	0	0.348966298	0.379797857	0.348966298	0.379797857

The above table provides relevant data regarding alpha and beta of Rio Tinto stock, which could be used by investors in the portfolio to reduce risk from investment. The stock involves a negative Alpha, which relatively indicates that if the market index is not moving then the returns provided by the stock will be negative. This is relatively due to the high beta, which is valued at 0.364382078. However, the beta is relatively not that prominent and close to 1 where the risk involved in investment is the highest. From the evaluation of above table, Rio Tinto is considered to be one of the least riskiest investment opportunity, which could allow investors to reduce the risk from portfolio. Furthermore, the involvement Rio Tinto stock would also indicate a reduced expected return that is generated from a Portfolio. Nevertheless, the overall combination of low beta stocks and high beta stocks could eventually help in hedging the portfolio, while generating high returns from investment (Pevzner, Xie & Xin, 2015).

The chart mainly represents the overall return of the company in comparison to the market Returns, which depicts the overall risk involved in Investments.  The returns of the company have recently increased or decline providing a higher risk attributes to the investor, which could hamper their investment capability. According to Jenter & Kanaan (2015), regression analysis needs to be conducted with adequate data or else it would not provide the relevant outcomes to the investor, which are mainly used in formulating the portfolio.

Question 3: Writing a Report Based on the Market Model as Given in Equation 1

Graph and Summary output of Commonwealth Bank of Australia: 

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.813812531
R Square	0.662290836
Adjusted R Square	0.662123405
Standard Error	0.005049677
Observations	2019
ANOVA
	df	SS	MS	F	Significance F
Regression	1	0.100864655	0.100864655	3955.594816	0
Residual	2017	0.051431964	2.54992E-05
Total	2018	0.152296618
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-0.000210171	0.00011253	-1.867689682	0.061950259	-0.000430857	1.0516E-05	-0.000430857	1.0516E-05
Return CBA.AX	0.634438221	0.010087497	62.89351967	0	0.614655218	0.654221224	0.614655218	0.654221224

The output range depicted in the above table and graph represents the results for Commonwealth Bank of Australia, which has the highest risk in comparison with other two stocks. This is the main reason why Alpha of the company is negative having a value of -0.000210171, which indicates that during non-movement of capital market the company will in core loss and reduce returns of the investor. The negative beta indicates the high volatility and risk in Commonwealth Bank of Australia for investors who are risk averse. Moreover, the beta of Commonwealth Bank of Australia is relatively at the levels of 0.634438221, which is close to 1 and indicates a high-risk involvement of the company. Both alpha and beta presented in the above table indicates that Commonwealth Bank of Australia has the highest risk, which could increase chances of loss for the investor if fluctuations in capital market increases (Barberis, Mukherjee & Wang, 2016). The graphic valuation indicated that returns of the company sometimes are abrupt, which increases and declines the actual value of the stock due to high volatility. This is represented with the diverse dots that is detected in the graph.

Therefore, with the help of regression analysis and market model equation 1, relevant risk and return capability of the stock can be identified. this regression analysis helps investors in identifying the actual price movement of the stock if capital market decline. Moreover, the capability of the stock in generating high level of returns due to the change in volatility of capital market is also evaluated. This provides an overall just to the investors regarding the investments co presented within a stock, which could allow them to increase returns while reducing risk from investment. In addition, from the analysis of the regression relevant investment opportunities within the 3 stocks identify, which allow investors to increase the returns from investment. Commonwealth Bank is relatively identified to have the highest risk involved in investment followed by Rio Tinto and then Myer Holdings Limited. Consequently, investors could use an adequate portfolio comprising of all the three stocks with different we take to reduce the risk from investment while increasing high returns. 

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