assignment 1_adm2302 e
DESCRIPTION
Assignment 1_ADM2302 ETRANSCRIPT
Question 1
a)
Decision variables
How much to invest in Stock A (SA) and Stock B (SB)
s.t.
SA + SB <= 10
0.3*SA + 0.2*SB <= 2.5
SA >= 0
SB >=0
b)
Graph 1. Stock Investment Feasible Region and Constraints
Objective Function
MAX { 2SA +1.5SB
c)
Constraints, SA + SB <= 10 and 0.3*SA + 0.2*SB <= 2.5 meet at coordinates (5,5)
Constraints, SA + SB <= 10 and SB >=0 meet at coordinates (0,10)
Constraints, 0.3*SA + 0.2*SB <= 2.5 and SA >= 0 meet at coordinates (8.33, 0)
Using the corner point method, we can determine the optimal solution:
MAX { 2SA +1.5SB
= 2*5 + 1.5*5
=17.5
Or
MAX { 2SA +1.5SB
= 2*0 + 1.5*10
=15
Or
MAX { 2SA +1.5SB
= 2*8.33 + 1.5*0
=16.66
According to the corner point method, the optimal solution, if the investor is interested in
maximizing their profit, would be to invest $5 in Stock A and $5 in Stock B, which would yield
the optimal profit of $17.50.
Question 2
a)
Graph 2. Feasible Region, Constraints and Objective Function
Constraints, x+y>=1.5 and at y>=0 meet at (1.5, 0)
Constraints, 2x-y<=4 and y>=0 meet at (2, 0)
Constraints, 2x-y<=4 and 2x+3y<=12 meet at (3, 2)
Constraints, y<=2.5 and 2x+3y<=12 meet at (2.25, 2.5)
Constraints, y<=2.5 and y-x<=1 meet at (1.5, 2.5)
Constraints, x>=1 and y-x<=1 meet at (1, 2)
Constraints, x>=1 and x+y>=1.5 meet at (1, 0.5)
Using the corner point method, we can determine the optimal solution:
MAX {3x+y
= 3*1.5+0
= 4.5
Or
MAX {3x+y
= 3*2+0
= 6
Or
MAX {3x+y
= 3*3+2
= 11
Or
MAX {3x+y
= 3*2.25+2.5
= 9.25
Or
MAX {3x+y
= 3*1.5+2.5
= 7
Or
MAX {3x+y
= 3*1+2
= 5
Or
MAX {3x+y
= 3*1+0.5
= 3.5
The optimal solution is at point (3, 2) and the optimal value is 11. This is the optimal solution
according the corner point method. Using the coordinates of the intersections between the
constraint functions, we can determine the optimal value by plugging them into the objective
function.
b)
Graph 2. Feasible Region, Constraints and Objective Function
Objective Function: MIN {3x+y
Using the corner point method, we can determine the optimal solution:
MIN {3x+y
= 3*1.5+0
= 4.5
Or
MIN {3x+y
= 3*2+0
= 6
Or
MIN{3x+y
= 3*3+2
= 11
Or
MIN {3x+y
= 3*2.25+2.5
= 9.25
Or
MIN {3x+y
= 3*1.5+2.5
= 7
Or
MIN {3x+y
= 3*1+2
= 5
Or
MIN {3x+y
= 3*1+0.5
= 3.5
If we minimize the objective function, the optimal solution become (1, .05) with an optimal value
of 3.5. Similar to part a) of this problem, using the corner point method, it can be determined and
verified that (1, 0.5) is the optimal solution.
Question 3
Maximize:
4x+y=z
a)
Constraints:
1. 2x-y<=4
x=2
y=-4
2. 2x+10y<=100
x=100
y=10
3. x+y<=14
x=14
y=14
x>=3, y>=0
Objective function:
4x+y=20
x=5
y=20
Optimal solution:
Use corner point solution
Contraints 2&3
2x+10y<=100
x+y<=14
Intersection point is (40/9,86/9)
Optimal value:
4*40/9+86/9=28.4
By using corner point solution, contraints 2&3’s intersection ponit (40/9,86/9) maximizes the
optimal value.
b)
Constraints:
1. 2x-y<=4
x=2
y=-4
2. 2x+10y<=100
x=100
y=10
3. x+y<=14
x=14
y=14
4. x+y<=0
x=0
y=0
x>=3, y>=0
Optimal solution:
Use corner point solution
Contraints 3&4
x+y<=14
x+y<=0
Intersection point is (7,7)
Optimal value:
4*7+7=35
By using corner point solution, contraints 3&4’s intersection ponit (7,7) maximizes the optimal
value.
Question 4
We want to minimize 4x+2y
With constraints of
2x+y>=3 y>= 3-2x
y-2x<=1 y<=2x+1
x>=1
y<=4
y>=0
A) 4X + 2Y
To find the optimal solution we use the corner point solution
Intersection point of: 2x+y>=3 and y>=0 : Point 1
Y=Y
3-2x=0
X=3/2
Y=3-2x
Y=3-2(3/2)
Y=0
Therefor X, Y = (1.5, 0)
Intersection point of 2x+y>=3 and x>=1 : Point 2
Y=Y where x=1
2(1) + y=3
Y=3-2
Y = 1
Therefore X, Y (1, 1)
Intersection point of Y<=2x+1 and y<=4 : Point 3
Y=Y
2(x) +1=4
X=1.5
Y=2(1.5)+ 1
Y=4
Therefore X, Y= (1.5, 4)
Intersection point of Y<=2x+1 and x>=1 : Point 4
Y=Y where x=1
Y= 2(1)+1
Y= 3
Therefore (1, 3)
4X+2Y
Point 1: 4(1.5)+2(0) = 6
Point 2: 4(1)+2(1) = 6
Point 3: 4(1.5)+2(3) = 12
Point 4: 4(1.5)+2(4) = 14
The results have concluded that by using the corner point method, both points 1 and 2 minimize
the optimal function of 4x + 2y
B)
Demonstrated by the graph above; the optimal solution can not be maximized because the LP is
unbounded
Question 5
Maximize:
3x-y=z
a)
Constraints:
1. x-y=0
x=0
y=0
2. x+4y<=10
x=10
y=5/2
3. x-2y<=6
x=6
y=-3
4. x+3y<=9
x=9
y=3
5. x+5y<=12
x=12
y=12/5
x>=0, y>=0
Optimal solution:
Use corner point solution
Contraints 1&5
x-y=0
x+5y<=12
Intersection point is (2,2)
Optimal value:
3*2-2=4
By using corner point solution, contraints 1&5’s intersection ponit (2,2) maximizes the optimal
value.
b)
There are 2 redundant constrints, contraint 2&4, the instersection point doesn’t affect the optimal
solutions.
c)
Constraint 3 can be removed without changing the optimal solution, it doesn’t have
intersecion point with the other contraints.
Question 6
We want to maximize 2x+6y
With constraints of
x-y>=2 y>=x-2
x-3y<=2 y=((x-2)/3)
5x+4y =15 y= (15/4)-(5x/4)
x>=0
y>=0
A)
Intersection point of: x-y>=2 and Y>0 : Point 1
Y=Y
Y=0
Y=x-2
0=x-2
X=2
Therefor X, Y = (2, 0)
Intersection point of: 5x+4y = 15 and x-y>=2 : Point 2
Y=Y
X-2 = (15/4)-(5x/4)
x-2 = 3.75-1.25x
x=1.875-.625
x=2.556
Y= x-2
Y=2.556-2
Y=.556
Therefor X, Y = (2.556, .556)
Intersection point of: 5x+4y = 15 and x-3y<=2: Point 3
-1.25x+3.75=.33x-.67
3.75=1.58x-.67
4.42=1.58x
X=2.8
Y=3.75-1.25(2.8)
Y=3.75-3.5
Y=.25
Therefore X,Y =(2.8,.25)
2x+6y
Point 1: 2(2) + 6(0) = 4
Point 2: 2(2.556) + 6(.556) = 8.448
Point 3: 2(2.8) + 6(.25) = 7.1
B) For this objective function I used the corner point method because I found it to be most useful
for figuring out where the point is maximized on the function of 5x+4y = 15. At point 2
(intersection 5x+4y = 15 and x-y>=2 ) the function is maximized with a total value of 8,448
C) Using the corner point method the function was shown to be minimized at point 3. This is due
to the fact that 5x+4y = 15 which by default means that the minimizing point has to be located on
that line. 2x+6y was minimized at the intersection 5x+4y = 15 and x-3y<=2 which gave a total
value of 7.1