How can I use the value of WACC(weighted average cost of capital)? I have calculated the WACC of company to be 7%. What if the company had a smaller or bigger WACC? Which one would attract investment?
$\begingroup$
$\endgroup$
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
3
WACC is the weighted average cost of capital therefore from the business's standpoint, they would want to have a lower WACC because it is an average of the % cost of capital. From an investor's standpoint: it can be mixed. For a bondholder, they would want WACC to be a bit high but not by too much. For example, a higher WACC may mean the company is paying a higher % on the bonds. A higher WACC may sometimes mean the company has too much debt & therefore require a higher return on the bonds since it is now a bit risky. As the bondholder, a shareholders view may be mixed. A higher WACC may mean leverage for the company to expand so it may be attractive but if it is too high the profits the company makes will be used to pay off debt hence the $k-g$ in Gordon's Growth Model. But remember that bondholders get paid prior to shareholders!
-
$\begingroup$ "A higher WACC may mean leverage for the company to expand " what does that mean? also k−g in Gordon's Growth Model?? $\endgroup$– kinkajouCommented Dec 21, 2014 at 4:03
-
1$\begingroup$ @kinkajou a company may need to borrow money to expand its operation so they issue bonds at a % cost which is part of WACC. Since they are making money on borrowed money = leverage. Gordon's growth model can also be known as the dividend discount model which says that a stocks price = current dividend / (cost of capital - growth rate) .... Therefore the denominator = $k-g$ $\endgroup$– RimeCommented Dec 21, 2014 at 6:13
-
1$\begingroup$ The WACC can be used as a "hurdle rate" for investment. The company should accept projects (investments) that earn a higher return than the WACC. If a company has a low WACC it will find it worthwhile to undertake more projects and the company will expand (company managers love this). $\endgroup$– nbbo2Commented Oct 2, 2021 at 9:38