1
Equity Valuation
Dividend Discount Method with constant
earnings growth rate
Balance Sheet
FAIRWAY CORPORATION
Balance Sheet
As of December 31, 2010, and 2011
(in thousands)
Assets
2
2010
2011
Current assets:
Cash and cash equivalents…………………………………………….
$ 230
$ 326
Accounts receivable………………………...………………………………….
$ 586
$ 673
Inventories……………………...…………………………………………………..
$ 610
$ 657
Total current assets………………………………………………….
$ 1,426
$ 1,656
Long-term assets:
Property, plant, and equipment, at cost…………………………………….
$ 2,000
$ 2,350
Accumulated depreciation……………………….………………………………..
$ (1,000)
$ (970)
Property, plant, and equipment, net………………………………………………
$ 1,000
$ 1,380
Investment securities…………………………………………..…………………..
$ 450
$ 400
Total noncurrent assets……………………………………………
$ 1,450
$ 1,780
Total assets……………………………………..………………………………………………..
$ 2,876
$ 3,436
Liabilities and Shareholders' Equity
Current liabilities:
Accounts payable…………………………….…………………………………….
$ 332
$ 388
Income taxes payable………………………….…………………………………….
$
9
$ 10
Short-term debt……………………………...….…………………………………….
$ 147
$ 126
Total current liabilities………………………………………………..
$ 488
$ 524
Change
$ 96
$ 87
$ 47
$ 230
$
$
$
$
$
$
350
30
380
(50)
330
560
$
$
56
1
$ (21)
$
36
Long-term debt…………………………...…………………………………………………….
$ 500
$ 835
$ 335
Deferred taxes………………………………………………………………………..
$
65
$
Total liabilities……………………………...……………………………..
$ 1,053
$
70
5
$ 1,429
$ 376
Shareholders' equity:
Common stock ($1 par) …………………………………………...…………….
$
50
$ 60
Additional paid-in capital…………….…………………………………….
$ 133
$ 167
Retained earnings…………………………………..………………………………
$ 1,640
$ 1,780
Total shareholders' equity…………………………..………………..
$ 1,823
$ 2,007
Total liabilities and shareholders' equity………………………………….………..
$ 2,876
$ 3,436
$ 10
$ 34
$ 140
$ 184
$ 560
Balance Sheet
Assets
3
FAIRWAY CORPORATION
Balance Sheet
As of December 31, 2011
(in thousands)
2011
Current assets:
Cash and cash equivalents…………………………………………….
$ 326
Accounts receivable………………………...………………………………….
$ 673
Inventories……………………...…………………………………………………..
$ 657
Total current assets………………………………………………….
$ 1,656
Long-term assets:
Property, plant, and equipment, at cost…………………………………….
$ 2,350
Accumulated depreciation……………………….………………………………..
$ (970)
Property, plant, and equipment, net………………………………………………
$ 1,380
Investment securities…………………………………………..…………………..
$ 400
Total noncurrent assets……………………………………………
$ 1,780
Total assets……………………………………..………………………………………………..
$ 3,436
Liabilities and Shareholders' Equity
Current liabilities:
Accounts payable…………………………….…………………………………….
$ 388
Income taxes payable………………………….…………………………………….
$ 10
Short-term debt……………………………...….…………………………………….
$ 126
Total current liabilities………………………………………………..524
$
Long-term debt…………………………...…………………………………………………….
$ 835
Assets
Liabilties and
Shareholders' Equity
Current Assets
CE
$
326
AR
$
673
INV
$
657
CA
$ 1,656
Current Liabilitites
AP
$
388
ITP
$
10
SD
$
126
CL
$
524
Long-term Assets
Long-term Liabilities
GC
AD
NC
IS
LA
$
$
$
$
$
2,350
(970)
1,380
400
1,780
LD
T
LL
TA
$
3,436 PAR
APC
RE
EB
$
$
$
835
70
905
Shareholders' Equity
Deferred taxes………………………………………………………………………..
$ 70
Total liabilities……………………………...……………………………..
$ 1,429
Shareholders' equity:
Common stock ($1 par) …………………………………………...…………….
$ 60
Additional paid-in capital…………….…………………………………….
$ 167
Retained earnings…………………………………..………………………………
$ 1,780
Total shareholders' equity…………………………..………………..
$ 2,007
Total liabilities and shareholders' equity………………………………….………..
$ 3,436
LE
$
$
$
$
60
167
1,780
2,007
$
3,436
Equity Value
4

For a simple firm

Liabilities + Equities = Debt Capital + Equity Capital + “Non-Capital”

Capital = Debt Capital + Equity Capital
C = DB + EB



Fair value of firm = Fair value of firm’s capital
V=D+E

Fair value of equity, E, is computed from the discounted cash flow
to the equity holders discounted at the rate cost of equity capital

Assume that the cash flow to equity holders is only dividends
Firm Value
N
E=å
i=1
N
V=å
DIVi
(1+kE )i
FCFi
i
(1+k)
i=1
5
Constant Growth Rate Equity Value
6

The value of an financial entity is the present value of future cash flows
discounted at the rate cost of the cash flows

Now assume that the dividend is growing at a constant rate, gDIV
N
DIVi
E0  
i
(1

k
)
i1
E

DIVi  DIVi-1  (1  gDIV )
Write as a infinite sum as follow
 1
(1  gDIV ) (1  gDIV )2
(1  gDIV )1 
E0  DIV1 


 ... 
2
3
 
1

k
(1

k
)
(1

k
)
(1

k
)
E
E
E
E


(1  gDIV )i1
E0  DIV1 
(1  kE )i
i1

E0 =
DIV1
kE - gDIV
Constant Growth Value
7

As the spread
between the rate
cost, k, and cash flow
growth, g, narrows,
the convergence
slows considerably

As cash flow growth
rate approaches the
rate cost, the series
does not converge
E0 =
DIV1
kE - gDIV
k=10%
Equity Value Management
8

Explore the relationships between





Earnings growth
Dividend payouts
Cost of equity (rate)
Fair value of equity
Based on the Dividend Discount Method with dividends
growing at a constant rate)

Gordon Growth Formula
Equity Value Per Share
9
DIV1
d1 
ns
d1
E[rE ]  rE  kE   g d
p0
E0
p0 
ns
d1
p0 
(k E  g d )
d1
kE   g d
p0
 d1
d1
d1 
p0 
 
 
kE
 (kE  g d ) kE 
p0  pvcy 
pvgo
g d  g DIV
d1
 dividend yield
p0
d1
 present value of first dividend as a perpetuity
kE
Share price v. Dividend Growth Rate
10
$50
$40
Share price, p0
p0
$30
pvgo
$20
$10
pvdy
$0%
1%
2%
3%
4%
5%
6%
dividend growth rate, gd
d1 = $0.50, kE = 10%
p0 = pvcy + pvgo
7%
8%
9%
Price/Earnings Ratio: pe
11
i=-1
Previous period
i=0
Next period
i=1
DC = DEB + DDB
= DRE + DPAR + DAPC + DDB
DC
DDB
DEB
DRE
DPAR
DAPC
= additional capital
= additional debt
= additional equity
= additional retained earnings (=NP1 – DIV1)
= additional common equity at par
= additional paid in common equity
Price/Earnings Ratio, pe
12
i=-1
i=0
e0
d0
eb-1
b = plowback ratio (assume
constant) thus ge = gd
(1-b) = dividend payout ratio
DRE = NP – DIV
= NP∙ b
DRE
NP
DIV
1b 
NP
b
i=1
e1
d1
eb 0
eb1
d1  (1  b)  e1
(1  b)  e1
p0 
(k E  g d )
RE
p0
(1  b)
pe 

e 1 (k E  g e )
DRE
NP
DIV
Price/Earnings Ratio: pe
13

In the case of no additional (external) investor financing



DDB = 0, DAPC = DPAR = 0
DC = DEB = DRE
And a scalable firm with a constant plowback, b
roe=
e1
eb0
Deb=b×e1 =b×roe×eb0
Deb
= b×roe = geb
eb0
b
= plowback ratio
reinvestment of earnings
(1-b)
= dividend payout ratio
eb
= equity book value per share
Long run assumption ge=geb
pe 
p
(1  b)

e (k E  b  roe )
Price/Earnings Ratio: pe
14
eb0
ge
b
e1
$
100
$
15%
0.8
2.00
Note: With this input, after ~40 years
geb -> ge
15
Source
PEG Ratio
16

The peg is a price measure normalized for earnings (e1) and
earnings growth rate (ge)

3 to 5 years estimated growth rate
pe
(1  b)
peg 

100  g e 100  g e  kE  g e 

Typical heuristic




> 2 relatively high valuation
= 1 pe ratio = earnings growth (converting ratio to percent)
< 1 relatively low valuation
Morningstar Table
Valuation Ratios
17
Forward PE Ratio
18
Critical Growth Rates
19


Internal growth rate, gint: the maximum growth rate that
does not require additional external financing
Sustainable growth rate, gsus: the maximum growth rate
that maintains the current capital structure, DB , with
EB
additional investor contributed debt
NA-1
NA0
NA1
IC-1
IC0
IC1
i=-1
i=0
i=1
DIV0 = NP0∙(1-b)
DIV1= NP0∙(1-b)∙(1+g)
DRE0 = NP0∙b
DRE1=NP0∙b∙(1+g)
Critical growth rate derivations
for core business operations
So use
•NA not TA and
•IC not C or LE
•NAIC
•roa is return on net book assets
•roe is return on book equity
Critical Growth Rates
20
DNA
= DIC
= IC1 – IC0
= DDB + DEB
= DDB + DAPC + DPAR + DRE
DNA
= g∙NA0
NA-1
NA0
NA1
DNA
= DRE  DDB
i=-1
i=0
i=1
= NP0∙b∙(1+g) + DDB
DRE0 = NP0∙b
DRE=NP0∙b∙(1+g)
DNA = DRE  DDB
Internal Growth Rate, gint
21
DNA  g int  NA  DRE  DDB
NA-1
NA0
NA1
DDB  0
i=-1
i=0
i=1
DRE0 = NP0∙b
DNA1 = DRE1  DDB1
DRE  (1  g int )  NP  b
roa 
gint  NA  (1  gint )  b·NP
g int  NA - g int  b·NP  b·NP
g int  (NA  b·NP)  b·NP
DRE1=NP0∙b∙(1+g)
NP
NA
g int (1 - b  roa)  b  roa
g int 
b  roa
(1  b  roa)
Sustainable
Growth Rate, gsus
22
DNA  g sus  NA  DRE  DDB
NA-1
NA0
NA1
i=-1
i=0
i=1
DRE0 = NP0∙b
DRE1=NP0∙b∙(1+g)
DNA1 = DRE1  DDB1
DB
DDB  DRE 
EB
g sus  NA  (1  g sus )  NP  b 
DRE  (1  g sus )  NP  b
g sus  (1  g sus )  b 
DB
g sus  NA  (1  g sus )  NP  b  (1  g sus )  NP  b 
EB
g sus  NA  (1  g sus )  NP  b  (1 
1
DB DB  EB NA


EB
EB
EB
DB
)
EB
roe 
NP
EB
NP
EB
g sus  (1  g sus )  b  roe
g sus 
b  roe
(1  b  roe )
NA
EB