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Wednesday, November 21, 2012


FIN622 Assignment 1 Fall 2012 Solution

Tuesday, November 20, 2012 Edit This
Net Present Value (NPV)
NPV = -Io + CF1 / (1+i)^1 + CF2 / (1+i)^2 + CF3 / (1+i)^3 + CF4 / (1+i)^4 +…….+ ∞
For Alpha:
Io= 20,000
i = initial investment for alpha = 14%
For Beta:
Io= 20,000
i = initial investment for Beta = 12% 
......................................
PV of Future Cash Flows of project ALPHA = 53428+ 63435 + 52956+ 37725
= 207547

PI= 207547/200000

ye calculation apni kr laina...main ny rounded kiya howa hy.... same isi tarah BETA ki b calculate krni hy.....just intial investment ki amount less nahi krni....balky intial investment ko divide krna hy
......................
Net Present Value (NPV)
NPV = -Io + CF1 / (1+i)^1 + CF2 / (1+i)^2 + CF3 / (1+i)^3 + CF4 / (1+i)^4 +…….+ ∞
For Alpha:
Io= 20,000
i = initial investment for alpha = 14%
For Beta:
Io= 20,000
i = initial investment for Beta = 12%
...................... 
Profitability Index (PI):
Profitability Index = PV of Future Net Cash Flows / Initial Investment Required 

...................... 
Net Present Value (NPV)
NPV = -Io + CF1 / (1+i)^1 + CF2 / (1+i)^2 + CF3 / (1+i)^3 + CF4 / (1+i)^4 +…….+ ∞
For Alpha:
Io= 20,000
i = initial investment for alpha = 14%
For Beta:
Io= 20,000
i = initial investment for Beta = 12%

Profitability Index (PI):
Profitability Index = PV of Future Net Cash Flows / Initial Investment Required
Year Project Alpha
Years
End Rs.(000)
FVIF 
12.3%=1/(1.123)^t
PV
PI
-200,000
1.000
-200000
-0.85
1
60,000
1.123
67380
2.83
2
80,000
1.261
100890.32
2.12
3
75,000
1.416
106218.59
2.27
4
60,000
1.590
95426.78128
2.83
Total
75,000
169915.6913
Project Beta
Years
End Rs.(000)
FVIF 
12.3%=1/(1.123)^t
PV
PI
-200,000
1.000
-200000
-0.851
1
55,000
1.123
61765
3.093
2
65,000
1.261
81973.385
2.617
3
70,000
1.416
99137.35069
2.430
4
80,000
1.590
127235.7084
2.126
Total
NPV
170111.4441
PV of Future Cash Flows of project ALPHA = 53428+ 63435 + 52956+ 37725
= 207547

PI= 207547/200000 
...................... 
Answers are
1: Net present values (NPV):
Alpha: NPV = 7545.63
Beta: NPV = 243.8
Profitability index (PI):
Alpha: PI = 1.0377
Beta: PI = 1.0012
2) NPV of Alpha is higher, it is favorable
3) Profitability index of Alpha is higher
4) IRR of Alpha is higher so it is favorable project. 
...................... 
Answer 4:
PROJECT ALPHA
Year End
Project Alpha
IRR
IRR+1
(IRR+1)^T
NPV of Cash Flow
0
-200000
14.00%
1 + 14% = 1.14
(1.14)^0
=
1.000
(200,000.00)
1
60000
14.00%
1 + 14% = 1.14
(1.14)^1
=
1.140
52,631.58
2
80000
14.00%
1 + 14% = 1.14
(1.14)^2
=
1.300
61,557.40
3
75000
14.00%
1 + 14% = 1.14
(1.14)^3
=
1.482
50,622.86
4
60000
14.00%
1 + 14% = 1.14
(1.14)^4
=
1.689
35,524.82
TOTAL NPV
336.66
PROJECT Beta
Year End
Project Alpha
IRR
IRR+1
(IRR+1)^T
NPV of Cash Flow
0
-200000
12.00%
1 + 12% = 1.12
(1.12)^0
=
1.000
(200,000.00)
1
55000
12.00%
1 + 12% = 1.12
(1.12)^1
=
1.120
49,107.14
2
65000
12.00%
1 + 12% = 1.12
(1.12)^2
=
1.254
51,817.60
3
70000
12.00%
1 + 12% = 1.12
(1.12)^3
=
1.405
49,824.62
4
80000
12.00%
1 + 12% = 1.12
(1.12)^4
=
1.574
50,841.45
TOTAL NPV
1,590.81

Using IRR criterion, project Beta IRR is higher than Project Alpha. So, Select Project Beta. Answer 4:
PROJECT ALPHA
Year End
Project Alpha
IRR
IRR+1
(IRR+1)^T
NPV of Cash Flow
0
-200000
14.00%
1 + 14% = 1.14
(1.14)^0
=
1.000
(200,000.00)
1
60000
14.00%
1 + 14% = 1.14
(1.14)^1
=
1.140
52,631.58
2
80000
14.00%
1 + 14% = 1.14
(1.14)^2
=
1.300
61,557.40
3
75000
14.00%
1 + 14% = 1.14
(1.14)^3
=
1.482
50,622.86
4
60000
14.00%
1 + 14% = 1.14
(1.14)^4
=
1.689
35,524.82
TOTAL NPV
336.66
PROJECT Beta
Year End
Project Alpha
IRR
IRR+1
(IRR+1)^T
NPV of Cash Flow
0
-200000
12.00%
1 + 12% = 1.12
(1.12)^0
=
1.000
(200,000.00)
1
55000
12.00%
1 + 12% = 1.12
(1.12)^1
=
1.120
49,107.14
2
65000
12.00%
1 + 12% = 1.12
(1.12)^2
=
1.254
51,817.60
3
70000
12.00%
1 + 12% = 1.12
(1.12)^3
=
1.405
49,824.62
4
80000
12.00%
1 + 12% = 1.12
(1.12)^4
=
1.574
50,841.45
TOTAL NPV
1,590.81

Using IRR criterion, project Beta IRR is higher than Project Alpha. So, Select Project Beta
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Monday, November 19, 2012

STA301 Assignment 2 Solution Fall 2012



Question no 02 part a

If coefficient of skewness is zero then distribution is symmetric or zero skewed.\
\

question no 2 part b
The least squares regression line

The least squares regression line is the line which produces the smallest value of the sum of the squares of the residuals. A residual is the vertical distance from a point on a scatter diagram to the line of best fit. Therefore the least squares regression line can be seen as the best line of best fit.
The equation of the least squares regression line of y on x is:


Where b is:


As you can see there are 3 different ways to calculate the value for b, based on what information you are given in the question
Example

Calculate the least squared regression line of y on x from the following data:
x
20
30
40
50
60
70
y
2.49
2.41
2.38
2.14
1.97
2.03
Firstly draw up a table of the values you need and fill it out. In this example we'll use the third equation for b so we need all the values of x2and xiyi:
x
y
x2
xiyi
20
2.49
400
49.9
30
2.41
900
72.3
40
2.38
1600
95.2
50
2.14
2500
107
60
1.97
3600
118.2
70
2.03
4900
142.1
Sum:
270
13.42
13900
584.7
Next step is to calculate the means of the x and y values:




Now the value of b can be calculated:
Hence the equation is therefore:


Cleaning it up:

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PHY101 1st Assignment Solution nov 2012




Q 1: A football player kicks a football straight up in to the air with an initial velocity of 25m/s. What is the approximate maximum height that the ball reaches? Use acceleration due to gravity (g) as 9.8m/s2. Marks = 7

Ans:-

2as = Vf2 – Vi2
Where a = g = 9.8m/s2
Vf2 = 0
Applying the values:
2as = Vf2 – Vi2
2(9.8m/s2) s = 0 – (25m/s)2
(19.6m/s2) s = 625m2/s2
s =625m2/s2 / (19.6m/s2)
s = 31.8m


Q 2: In the classic horse and cart problem, a horse is attached to a cart that can roll along on a set of wheels. Which of the following statement is true? Also explain the reason of your selected choice as well.
a) The horse pulls on the cart with a force, and the cart pulls, back on the horse with an equal force.
b) The horse pulls on the cart with a force greater than the force that the cart pulls back on the horse.
c) The horse pulls the cart forward because the force of friction forward is greater than the force of the wagon backward.
d) The horse pulls the cart forward, because of a time delay between the action and reaction forces.
e) The horse pulls the cart forward, but only if the horse has a mass greater than the mass of the cart. Marks = 6 (2+4)

Ans:-
c) The horse pulls the cart forward because the force of friction forward is greater than the force of the wagon backward.

Reason:-
The forces on the cart include the forward force the horse exerts on the cart and the backward force due to friction at the ground, acting on the wheels. At rest, or at constant velocity, these two are equal in size, because the acceleration of the cart is zero. The forces on the horse include the backward force the cart exerts on the horse and the forward force of the ground on its hooves. At rest, or at constant velocity, these two are equal in size, because the acceleration of the horse is zero. The force the horse exerts on the cart is of equal size and opposite direction to the force the cart exerts on the horse, by Newton's third law. The horse places its feet so as to change the angle of the force its hooves exert on the ground, thereby increasing the backward force of its hooves on the ground.

Q 3: Is it true to say that a projectile has zero acceleration at its peak? If yes explain if not give a reason. Marks = 4

Ans:-
The acceleration due to gravity is constant at 9.8 m/sec^2. One of the variables in projectiles is its speed --- it slows down as it goes up and gains speed as it goes down. When the projectile reaches its maximum height, its velocity is always zero. Acceleration has nothing to do with instantaneous velocity. Acceleration shows the relationship between the velocities of the projectile at 2 different points in time. If acceleration was zero, the difference in velocity of the projectiles between any two points of its trajectory would be 0. However, with an acceleration, even if the velocity of the projectile at one point was 0, but the acceleration would not be zero because the acceleration of a projectile is always equal to the acceleration of gravity.



Q 4: (a) Anna is pushing but the box does not move. What is the force of static friction shown in below figure? Marks = 3

Ans:-
Fs=mu*N
=1*7N
=7N

(b) What is the maximum static friction force shown in below figure? The coefficient of static friction for these materials is 0.31. Marks = 5

Ans:-

=0.31*27N

27 nai only 7 hai
or answer hai 2.17
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