Investment decision in conditions of certainty
1/ Calculate the NPV for project A and project B for k = 10% (the cost of capital)
C1
C2
C3
C4
C5
C6
NCFt
1  K 
1
times
0
1
2
3
4
5
6
7
NCF of
Project A
-725
100
250
250
200
100
100
100
NCF of
Project B
-850
100
200
200
200
200
200
200
t
1  K 
t
NPV
n
!!! Recall that NPV   I 0  
t 1
(C2×C4)
-725
90.90909
206.6116
187.8287
136.6027
62.09213
56.44739
51.31581
1
0.909091
0.826446
0.751315
0.683013
0.620921
0.564474
0.513158
NCFt
1  K t

NCFt
)
1  K t
(C3×C4)
-850
90.90909
165.2893
150.263
136.6027
124.1843
112.8948
102.6316
66.80739 32.77467
RV n
n
1  K n

t 0
NCFt
1  K t

RV n
1  K n
Investment decision in conditions of uncertainty
1/ Calculate the NPV with Certainty Equivalent Method for project A and project B for
Risk Free Rate of Return Rf = 2.5% and Certainty Equivalent Coefficients αt in the table
below.
C1
C2
C3
E(CFt)
of
Project
A
E(CFt)
of
Project
B
C4
C5
Risk free
2.5%
1
times
0
-725
-850
Certainty
Equivalent
Coefficients
αt
1
1
2
100
250
100
200
3
4
5
6
7
250
200
100
100
100
200
200
200
200
200
C6
C7

 t  E CFt
1  R 
1  R f 

t
t
f

 t  E CFt
1  R 

t
f
(C4×C3×C5)
1
(C4×C2×C5)
-725
0.91928251
0.84508034
0.97561
0.951814
89.6861
201.0899
89.6861
160.8719
0.77686757
0.71416077
0.65651551
0.60352323
0.55480835
0.928599
0.905951
0.883854
0.862297
0.841265
180.3497
129.3989
58.0264
52.04162
46.6741
144.2798
129.3989
116.0528
104.0832
93.34819
32.2667
-12.2791
NPVadj1
n
NPVadjusted   I 0  
t 1

 t  E CFt
1  R 
t
f

-850
1
!!! Recall that
2/ Calculate the NPV with Risk Adjusted Discounted Rate Method for project A and project
B for Risk Free Rate of Return Rf = 2.5% , Risk premium of project α1 = 5% and Risk
premium of institution α2 = 4% in the table below. k*=2.5%+5%+4%
C1
C2
C3
E(CFt) of
Project A
E(CFt) of
Project B
-725
100
250
250
200
100
100
100
-850
100
200
200
200
200
200
200
C4

E CFt
1 K
1
1  K 
* t
times
0
1
2
3
4
5
6
7
C5
n
NPVadjusted   I 0  
t 1

E CFt
1 K

1  K


-12.2790972
K *  R f  1   2
* t
and


* t
(C3×C4)
-850
89.68609865
160.8719258
144.2797541
129.3988826
116.0528095
104.0832372
93.34819483
32.2667
E CFt
NPVadj2 = NPVadj1
Some remarks:


* t
(C2×C4)
-725
89.6861
201.0899
180.3497
129.3989
58.0264
52.04162
46.6741
1
0.896861
0.8043596
0.7213988
0.6469944
0.580264
0.5204162
0.466741
NPVadj2
!!! Recall that
C6
1  R 

1  K 
t
t
f
* t
3/ Calculate the expected NPV with Statistical Techniques of Risk Analysis for project A and
project B with Risk Free Rate of Return Rf = 2.5% per annum and the probability Pt over the
period in the table below. We assume total independence between CF over the period.
C1
C2
C4
C5
C6
C7

1
1  R f 
E CFt
times
0
1
2
3
4
5
6
7
NCF of
Project A
-725
1000
2500
2500
2000
1000
1000
1000
n
E ( NPV )   I 0  
t 1
probability
Pi
1
0.1
0.25
0.25
0.2
0.1
0.05
0.05

E CFt
1
0.97561
0.951814
0.928599
0.905951
0.883854
0.862297
0.841265
E(CFt)
(C2×C4)
-725
100
625
625
400
100
50
50
1225
1
  1083.763
1  Rf 
t
n
V ( NPV )  
 ( NPV )  V ( NPV )  2690349  1640.228
t 1

1  Rf t
t
(C6×C5)
-725
97.561
594.834
580.375
362.380
88.385
43.115
42.063
1083.763

E CFt 2
1  Rf 
C8

E CFt 2
1  Rf 
(C2×C2×C4×C5×C×5)
525625
95181.43962
1415547.882
1347338.853
656597.2567
78119.84017
37177.79425
35386.35979
4190974.43
  1083.763
2t
CV ( NPV ) 

2t
2
 2690349
2
1640.228
 1.34
1083.763
!!! To choose between projects we must calculate the risk associated for each them and
the coefficient of variation. We repeat the same procedure for project B
C1
C2
C4
C5
C6
C7

E CFt
t
1  R f 
t
times
0
1
2
3
4
5
6
7
total
NCF of
Project B
-850
1000
2000
2000
2000
2000
2000
2000
n
E ( NPV )   I 0  
t 1
probability
Pi
1
0.1
0.25
0.25
0.2
0.1
0.05
0.05
1

E CFt
1
0.97561
0.951814
0.928599
0.905951
0.883854
0.862297
0.841265
  897.2752
1  Rf t

1  Rf 
1
E(CFt)
(C2×C4)
-850
100
500
500
400
100
100
100
1050
n
V ( NPV )  
t 1
 ( NPV )  V ( NPV )  2742762  1656.129
(C6×C5)
C7

E CFt 2

1  Rf 2t
(C2×C2×C4×C5×C×5)
-850
97.561
475.9072
464.2997
362.3803
176.7709
86.22969
84.12652
897.2752

E CFt 2
1  Rf 
95181.44
905950.64
862296.87
656597.26
312479.36
148711.18
141545.44
3845262
  897.2752
2
2t
CV ( NPV ) 
722500
 2742762
1640.228
 1.577
1083.763
Summarize
E(NPV)
σ(NPV)
CV(NPV)
Project A
1083.763
1640.228
1.34
Project B
897.2752
1656.129
1.577
!!! for two criteria we choose the project A
4/ Calculate the expected NPV with Statistical Techniques of Risk Analysis for project A and
project B. We assume that we have 3 alternatives for A, 5 alternatives for B and the
associated probabilities for each project and the life of them is one period with Risk Free
Rate of Return Rf = 2.5% . We assume total independence between CF between
alternatives.
Project B
I0 = 20000
pj
CF1
0.2
30000
0.6
40000
0.2
55000
Project A
I0 = 50000
pj
CF1
0.1
50000
0.2
65000
0.3
70000
0.3
80000
0.1
90000
3
Solution:
We can use the same techniques above (question 3). We change t by i the ith alternative. But
we can use another easier method based on properties of variance.
Project A
pj
0.1
0.2
0.3
0.3
0.1
CF1
50000
65000
70000
80000
90000
pj×CF1
5000
13000
21000
24000
9000
pj×(CF1)2
250000000
845000000
1470000000
1920000000
810000000
Total
72000
5295000000
ECF1   72000 V CF1   5295000000  720002  111000000
E NPV    I 0 
E CF1 
1 R f

E NPV   50000 

V NPV  
and
72000
 14285.714
1  0.12
 NPV   88488520.408  9406.834
V CF1 
1  R f 2
V NPV  
CV  NPV  
111000000
1  0.122
 88488520.408
9406.834
 0.658
14285.714
Project B
pj
0.2
0.6
0.2
CF1
30000
40000
55000
pj×CF1
6000
24000
11000
pj×(CF1)2
180000000
960000000
605000000
Total
41000
1745000000
ECF1   41000 V CF1   1745000000  410002  64000000
E NPV    I 0 
E CF1 
1 R f

E NPV   20000 

and V  NPV  
41000
 16607.143
1  0.12
 NPV   51020408.16  7142.857
V CF1 
1  R f 2
V NPV  
CV  NPV  
64000000
1  0.122
 51020408.16
7142.857
 0.430
16607.143
4
Summarize
Project A
14285.714
9406.834
0.658
E(NPV)
σ(NPV)
CV(NPV)
Project B
16607.143
7142.857
0.430
!!! for Two criteria we choose the project B
5/ Calculate the expected NPV with Statistical Techniques of Risk Analysis for Multi-period
Project. We assume that we have 3 alternatives for each period and the associated
probabilities. The life 3 periods with Risk Free Rate of Return Rf = 10% . We assume total
independence between CF over time.
Period 1
CF1
pj
12000
0.2
15000
0.3
17000
0.5
I 0  100000 .
period 2
CF2
50000
60000
70000
period 3
pj
0.4
0.3
0.3
CF3
60000
85000
95000
pj
0.1
0.7
0.2
To calculate the expected value of NPV, the variance and coefficient of variation, we
calculate the expected value of each cash flow and its variance like question 4 because the
covariance between different CF is zero.
Period 1
period 2
period 3
CF1
pj
pjCF1
pj CF1^2
CF2
pj
pjCF2
pj CF2^2
CF3
pj
pjCF3
pj CF3^2
12000
0.2
2400
28800000
50000
0.4
20000
1000000000
60000
0.1
6000
360000000
15000
0.3
4500
67500000
60000
0.3
18000
1080000000
85000
0.7
59500
5057500000
17000
0.5
8500
144500000
70000
0.3
21000
1470000000
95000
0.2
19000
1805000000
total
1
15400
240800000
total
1
59000
3550000000
total
1
84500
7222500000
ECF1  15400
ECF 2  59000
V CF 2  69000000
V CF1  3640000
ECF 3  84500
V CF 3  82250000
E NPV   100000 
15400
59000
84500


 26246.431
2
1  0.1 1  0.1 1  0.13
V NPV  
69000000
3640000
1  0.1
2

1  0.1
4

82250000
1  0.16
 96564173.63
 NPV   96564173.63  9826.707
5
6/ Calculate the expected NPV with Statistical Techniques of Risk Analysis for Multi-period
Project. We assume that we have 3 alternatives and the associated probabilities for each
them. The life is 3 periods with Risk Free Rate of Return Rf = 10% . We assume perfect
dependence between CF over time. I 0  100000
Period 1
CF1
pj
12000
0.2
15000
0.3
17000
0.5
period 2
period 3
CF2
50000
60000
70000
pj
0.4
0.3
0.3
CF3
60000
85000
95000
pj
0.1
0.7
0.2
To calculate the expected value of NPV, the variance and coefficient of variation, we
calculate the expected value of each cash flow and its variance like question 4 or 5
because the coefficient of correlation is equal 1  CF 1, CF 2    CF 1, CF 3   CF 2, CF 3  1 .
ECF1  15400
ECF 2  59000
V CF 2  69000000
V CF1  3640000
E NPV   100000 
 NPV  
V CF 3  82250000
15400
59000
84500


 26246.431
2
1  0.1 1  0.1 1  0.13
3640000
69000000
82250000


 15413.221
2
1  0.1
1  0.1
1  0.13
‫شرح‬
Explanations
  X ,Y  
ECF 3  84500
Cov X , Y 
 1  Cov X , Y     X  Y 
  X  Y 
V NPV  
V NPV  
V CF1

V CF 2

V CF 3
1  R f  1  R f  1  R f 
2
4
6

2CovCF1, CF 2
1  R f 
3

2CovCF1, CF 3
1  R f 
4

2CovCF 2, CF 3
1  R f 5
 2 CF1  2 CF 2  2 CF 3 2 CF1 CF 2 2 CF1 CF 3 2 CF 2 CF 3


1  R f 2 1  R f 4 1  R f 6
  CF1  CF 2  CF 3 

V NPV   


2
3
 1 R f
1

R
1

R
f
f



 
 


2
1  R f 3

1  R f 4

1  R f 5


and  NPV     CF1   CF 2   CF 3 
2
3

 1 R f

 1  R f  1  R f 


6
7/ Risk analysis in the event of a partial correlation between cash flows over time.
Suppose project investment cost $ 70,000 and the distribution of cash flows as
follows:
Calculate the expected value, the variance and coefficient of variation of NPV. We assume
that the risk-free rate of return is fixed at 7%.
To resolve this issue we prefer to use the decision tree:
Total =
1
25636.30
124751740.73
E(NPV) = 25636.30
V(NPV) = 124751740.73
σ (NPV) =11169.232
7
!!! The manager maintain the project if PNPV  0  0.95
If we assume that NPV  N ENPV ;V NPV  then we can calculate this probability.
Step 1 : NPV  N E NPV ;V NPV   Z 
NPV  E NPV 
 N 0,1
V NPV 
Step 2: PNPV  0  0.95  PNPV  0  0.05

0  E NPV 
0  25636.3 

PNPV  0  P  Z 
 PZ  2.295  1.1%
  P Z 
11169.232 
V NPV  


Step3 : we maintain the project because 1.1% is less than 5%.
8
9