BUSF_A01.qxd

Cost of individual capital elements

Whenever the current market price of a share, pE 0 , is used in any equation relating
to dividends and the cost of equity, we must use the ex-dividend price. If a dividend
is soon to be paid and the shares are being traded cum dividend, we need to deduct
the amount of the forthcoming dividend per share to arrive at the equivalent ex-
dividend price.

Deriving the cost of equity
To say that equities should be valued on the basis of future dividends is not to assume
that any particular investor intends to hold a particular share forever, because the pro-
ceeds of any future disposal of the share will itself depend on expectations, at the date
of disposal, of future dividends.
If a business is expected to pay a constant dividend dper share, at the end of each
year indefinitely, the value of each ordinary share will be:

pE 0 =

to the investor who intends to hold the share forever.
Suppose, however, that the investor intends to sell the share at time t; its value at
that time will be:

This would mean that the value to our shareholder would be:

pE 0 =+ ×

• that is, the value depends on the dividends to be received until time t(suitably
  discounted) plus the market value at time t[discounted at 1/(1 +kE)t] to bring this mar-
  ket value to the present value.
  This expression reduces to:

pE 0 =

Hence, irrespective of whether disposal at some future date is envisaged or not,
provided that valuation always depends on dividends, the current value will be
unaffected.
Unlike loan interest, which is usually fixed by the contract between borrower and
lender, and preference dividends, which businesses usually endeavour to pay (and for
which there is a defined ceiling), ordinary share dividends are highly uncertain as
to amount. This poses a major problem in the estimation of the cost of equity. The pre-
diction of future dividends is a daunting task. One of two simplifying approaches may
be taken, however:

l assume that dividends will remain as at present; or

l assume some constant rate of growth in them.

dn

(1 +kE)n

∞

∑

n= 1

D

F

1

(1 +kE)t

dn

(1 +kE)n

∞

∑

n=t+ 1

A

C

dn

(1 +kE)n

t

∑

n= 1

dn

(1 +kE)n

∞

∑

n=t+ 1

dn

(1 +kE)n

∞

∑

n= 1