1. LEARNING OBJECTIVES Adjust for risk by varying the discount rate
Present a sensitivity graph and discuss
break-even NPV
Undertake scenario analysis
Make use of probability analysis to describe
the extent of risk facing a project and thus
make more enlightened choices
Discuss the limitations, explain the
appropriate use and make an accurate
interpretation of the results of the four risk
techniques described in this chapter

2. What is Risk?
Certainty
Risk and uncertainty
Objective probabilities
Subjective probabilities

4. Subjective probabilities

5. ADJUSTING FOR RISK THROUGH THE DISCOUNT RATE
Level of risk
Risk-fr
ee rate (%)
Risk pr
emium (%)
Risk-adjusted rate (%)
Low 9 +3 12
Medium 9 +6 15
High 9
+10 19
The pr
oject cur r
ently being consider
ed has the following cash flows: T
ime (years) 1 2
Cash flow (£)
–100 55 70
If the pr
oject is judged to be low risk: 55 70
NPV = – +
= +£4.91
( ) 2
Accept.
If the pr
oject is judged to be medium risk: 55 70
NPV = –
= +£0.76
( ) 2
Accept.
If the pr
oject is judged to be high risk: 55 70
NPV = –
= –£4.35
( ) 2
Reject.

6. Drawbacks of the risk-adjusted discount rate method:
Adjusting for risk
Drawbacks of the risk-adjusted discount rate method:
Risk classification is subjective
Difficulty in selecting risk premiums

7. SENSITIVITY ANALYSIS
Sensitivity analysis identifies the extent to which NPV changes as key variables are changed
A “what-if” analysis:
Acmart plc
new product line – Marts
likely demand for 1,000,000 a year
price of £1
four-year life of product
initial investment £800,000
Cash flow per unit £
Sale price
Costs
Labour
Materials 0.40
Relevant overhead 0.10
0.70
Cash flow per unit
Required rate of return = 15 per cent

8. NET PRESENT VALUE (1) Annual cash flow = 30p  1,000,000 = £300,000.
Present value of annual cash flows =
300,000 annuity factor for 4 15%
= 300,000  = ,500 Less initial investment –800,000
Net present value ,500

9. What if the price achieved is only 95p for
sales of 1m units (all other factors remaining
constant)?
Annual cash flow = 25p  1m = £250,000.
250,000  ,750
Less initial investment ,000
Net present value –86,250
What if the price rose by 1 per cent?
Annual cash flow = 31p  1m = £310,000.
310,000  ,050
Net present value ,050

10. What if the quantity demanded is 5 per cent
more than anticipated?
Annual cash flow = 30p  1.05m = £315,000.
315,000  ,325
Less initial investment ,000
Net present value ,325
What if the quantity demanded is 10 per cent
less than expected?
Annual cash flow = 30p  900,000 =
£270,000.
270,000  ,850
Net present value –29,150

11. What if the appropriate discount rate is 20 per
cent higher than originally assumed (that is, it
is 18 per cent rather than 15 per cent)?
300,000  annuity factor for 4 18%
300,000  ,030
Less initial investment ,000
+7,030
What if the discount rate is 10 per cent lower
than assumed (that is, it becomes 13.5 per
cent)?
300,000  annuity factor for %.
300,000  ,230
+83,230

12. % deviation of variable from expectation
100 50
NPV (£000)
Discount
rate
Sale
–50
volume
Sale
price
–100
–20
–15
–10 –5 5 10 15 20 25
% deviation of variable from expectation
Sensitivity graph for Marts

13. Break-even NPV calculations
Advantages of using sensitivity analysis:
Information for decision making
To direct search
To make contingency plans
Drawbacks of sensitivity analysis:
No formal assignment of probabilities
Each variable is changed in isolation

14. SCENARIO ANALYSIS Observing NPV when numerous factors change
Worst-case scenario
Sales
900,000 units
Price
90p
Initial investment
£850,000 Pr
oject life
3 years
Discount rate
17%
Labour costs
22p
Material costs
45p
Over
head
11p
Cash flow per unit
Sale price
0.90
Costs
Labour
0.22
Material
0.45
Over
head
0.11
0.78
Cash flow per unit
0.12
Annual cash flow = 0.12 
900,000 = £108,000 Pr
esent value of cash flows ,000  =
238,637
Less
initial investment
–850,000
Net pr
esent value
–611,363
Acmart plc: Project proposal for the production of Marts

15. SCENARIO ANALYSIS
Best-case scenario
Sales
1,200,000 units
Price
120p
Initial investment
£770,000 Pr
oject life
4 years
Discount rate
14%
Labour costs
19p
Material costs
38p
Over
head 9p
Cash flow per unit
Sale price
1.20
Costs
Labour
0.19
Material
0.38
Over
head
0.09
0.66
Cash flow per unit
0.54
Annual cash flow = 0.54 
1,200,000 = £648,000 Pr
esent value of cash flows 648,000  =
1,888,078
Less
initial investment
–770,000
Net pr
esent value
1,118,078
Acmart plc: Project proposal for the production of Marts (continued)

16. PROBABILITY ANALYSIS Return Probability of return occurring Pr oject 116
1.0 Pr
oject 2 20
1.0 Pr
oject 3
–16
0.25 36
0.50 48
0.25 Pr
oject 4 –8
0.25 16
0.50 24
0.25 Pr
oject 5
–40
0.10
0.60
100
0.30
Pentagon plc: Use of probability analysis

17. EXPECTED RETURN
The expected return is the mean or average outcome calculated by weighting each of the possible outcomes by the probability of occurrence and then summing the result. - x = x p + x p
+ ... x p 1 1 2 2 n n or
i=n - x =  ( x p ) i i i =1 -
wher e x
= the expected r
etur n i
= each of the possible outcomes (outcome 1 to n ) p
= pr
obability of outcome i
occur
ring n
= the number of possible outcomes

18. Pentagon plc Expected returns Pr oject 1 16  1 16 Pr oject 2 20  1
–16 
0.25 = –4 36 
0.50 = 18 48 
0.25 = 12 26 Pr
oject 4 –8 
0.25 = –2 16 
0.50 = 8 24 
0.25 = 6 12 Pr
oject 5
–40  
0.1 = –4 
0.6 =
100 
0.3 = 30 26
Pentagon plc: expected returns

19. Pentagon plc: Probability distribution for projects 3 and 5
0.7
0.6
0.5
0.4
Probability
0.3
0.2
0.1
–100
–80
–60
–40
–16 20 36 48 80
100 £m
Project 3
Project 5
Pentagon plc: Probability distribution for projects 3 and 5

20. STANDARD DEVIATION
The standard deviation is a statistical measure of the dispersion around the expected value. The standard deviation is the square root of the variance  2 - - - V
ariance of x =  2
= ( x – x ) 2 p
+ ( x – x ) 2 p
+ ... ( x – x ) 2 p x 1 1 2 2 n n
i=n - or  2 =  {( x – x ) 2 p } x i i i =1
Standar
d deviation
i=n -  =  2 or  {( x – x ) 2 p } x x i i i =1

21. Outcome Pr
obability
Expected
Deviation
Deviation
Deviation
(Return) r
eturn
squared
squared
times
probability - - - - Pr
oject xi pi x xi – x
(xi – x )2
(xi – x
)2pi 1 16
1.0 16 2 20
1.0 20 3
–16
0.25 26
–42
1,764
441 36
0.5 26 10
100 50 48
0.25 26 22
484
121 V
ariance =
612
Standar
d deviation =
24.7 4 –8
0.25 12
–20
400
100 16
0.5 12 4 16 8 24
0.25 12 12
144 36 V
ariance =
144
Standar
d deviation = 12 5
–40
0.1 26
–66
4,356
436
0.6 26
–26
676
406
100
0.3 26 74
5,476
1,643 V
ariance =
2,485
Standar
d deviation =
49.8
Pentagon plc: Calculating the standard deviations for the five projects

22. Expected r
etur
n x
Standar
d deviation x Pr
oject 1 16 Pr
oject 2 20 Pr
oject 3 26
24.7 Pr
oject 4 12 12 Pr
oject 5 26
49.8
Pentagon plc: Expected return and standard deviation

23. RISK AND UTILITY Diminishing marginal utility Risk averter Risk lover
Investment A
Investment B
Return Pr
obability
Retur n
Probability
Poor economic conditions
2,000
0.5
0.5
Good economic conditions
6,000
0.5
8,000
0.5
Expected r
etur n
4,000
4,000
Returns and utility
Risk averter
Risk lover

24. MEAN-VARIANCE RULE
Project X will be preferred to Project Y if at least one of the following conditions apply:
1 The expected return of X is at least equal to the
expected return of Y, and the variance is less
than that of Y.
2 The expected return of X exceeds that of Y and
the variance is equal to or less than that of Y.

25. 30 Project 3 25 Project 5 Project 2 20 Project 1 Expected return 1510 5 12
24.7
49.8
Standard deviation
Pentagon plc: Expected return and standard deviations

26. EXPECTED NET PRESENT VALUES AND STANDARD DEVIATION
i=n
NPV =  (
NPV p ) i i i =1
wher e
NPV
= expected net pr
esent value
NPV
= the NPV if outcome i
occurs i p
= pr
obability of outcome i
occur
ring i n
= number of possible outcomes
The standar
d deviation of the net pr
esent value is
i=n  =  {(
NPV –
NPV ) 2 p }
NPV i i i =1

27. HORIZON PLC Purchase price, t0 £500,000 Refurbishment, t0 £200,000
£700,000
The Year 1 cash flows are as follows: Pr
obability
Cash flow at end of Y
ear 1
Good customer r
esponse
0.6
100,000
Poor customer r
esponse
0.4
10,000

28. HORIZON PLC Conditional probabilities for the second year
If the first year elicits a good response then:
Probability Cash flow at end
of Year 2
1 Sales increase in second year 0.1 £2m or
2 Sales are constant £1.6m
3 Sales decrease £0.8m
If the first year elicits a poor response then:
1 Sales fall further £0.7m
2 Sales rise slightly £1.2m
Note: All figures include net trading income plus sale of pub.

29. Cash
Probability
Cash
Conditional
Cash
Joint
Outcome
flow at p
flow at
probability
flow at
probability i T
ime 0 T
ime 1 T
ime 2
(£000s)
(£000s)
(£000s)
0.1
2,000
0.6 
0.1 = 0.06 a
100
0.7
1,600
0.6             b
0.6
0.2
800
0.6             c
–700
0.5
700
0.4
 0.5 = 0.20 d
0.4 10
0.5
1,200
0.4
0.5 = 0.20 e
1.00
An event tree showing the probabilities of the possible returns for Horizon plc

30. Outcome
Net pr
esent values
(£000 s)
NPV  Pr
obability
100
2,000 a – =
1044
1,044  0.06 = 63
1.1
(1.1)2
100
(1.1)2
1,600 b – =
713
713  0.42 =
300
1.1
100
800 c – = 52
52  0.12 = 6
1.1
(1.1)2 10
700 d – =
–112
–112  0.20 =
–22
1.1
(1.1)2 10
1,200 e – =
301
301  0.20 = 60
1.1
(1.1)2
407
Expected net pr
esent value
or £407,000
Expected net present value, Horizon plc

31. Exhibit 6.17 Standard deviation for Horizon plc2 )
ed times
NPV
367
Deviation
obability
24,346
39,327
15,123
53,872
2,247
134,915 –
£367,000
squar i pr
NPV = = ( or 2 )
Variance
134,915
– NPV
405,769
93,636
126,025
11,236
Deviation
squared
269,361 i
NPV (
d deviation =
– NPV
Deviation
637
306
–355
–519
–106 i
Standar
NPV
Exhibit Standard deviation for Horizon plc
Expected
NPV
NPV
407
407
407
407
407
Probability p i
0.06
0.42
0.12
0.20
0.20
713 52
1,044
301
–112
Outcome
£000s i
NPV a b c d e

32. Independent Probabilities
The initial cash out flow = £150,000

35. THE RISK OF INSOLVENCY 68.26 per cent 95.44 per cent 99.74 per cent X
Probability
68.26
per cent
95.44 per cent
99.74 per cent X –3  –2  –1   +1  +2  +3 
Standard deviations
Exhibit The normal curve

36. 34.13 per cent 68.26 per cent 95.44 per cent 99.74 per cent X –3  –2
Probability
34.13 per cent
Probability
68.26
per cent
95.44 per cent
99.74 per cent X –3  –2  –1   +1  +2  +3 
Standard deviations
Exhibit Probability of outcome being between expected return and one standard deviation from expected return

37.  is the mean of the possible outcomes
THE Z STATISTIC
Z =
where:
Z is the number of standard deviations
from the mean
X is the outcome that you are concerned about
 is the mean of the possible outcomes
 is the standard deviation of the outcome
distribution
X –  

38. V alue of the Pr obability that X lies Z statistic within Z standar
d deviations
above (or below) the
expected value (%)
0.0
0.00
0.2
7.93
0.4
15.54
0.6
22.57
0.8
28.81
1.0
34.13
1.2
38.49
1.4
41.92
1.6
44.52
1.8
46.41
2.0
47.72
2.2
48.61
2.4
49.18
2.6
49.53
2.8
49.74
3.0
49.87
Exhibit The standard normal distribution

39. Standard deviation = £6.5m
Probability
Probability
47.72
per cent X –3  –2  –1   +1  +2  +3 
£–5m
£+8m
Outcome
Exhibit Probability of outcome between and 2 from
Roulette plc
Maximum loss = £5m
Expected return = £8m
Standard deviation = £6.5m 
 

40. PROBLEMS OF USING PROBABILITY ANALYSIS
Too much faith can be placed in quantified
subjective probabilities
Too complicated for all managers to understand
Projects may be viewed in isolation

41. EVIDENCE OF RISK ANALYSIS IN PRACTICE