Linear Programming Exercise 01
Question:
Maximise Z=−x+2y, subject to the constraints: x≥3,x+y≥5,x+2y≥6,y≥0
Answer:

Rewrite the objective function as Z=x+2y

Identify the constraints: x≥3,x+y≥5,x+2y≥6,y≥0

Graph the constraints on a coordinate plane

Find the feasible region by shading the areas that satisfy the constraints

Find the coordinates of the vertices of the feasible region

Calculate the value of Z at each vertex

Choose the vertex with the highest value of Z as the optimal solution
Question:
Minimise Z=x+2y subject to 2x+y≥3,x+2y≥6,x,y≥0
Answer:
Step 1: Rewrite the objective function and constraints in standard form:
Minimise Z=x+2y
Subject to 2x+y3≥0 x+2y6≥0 x≥0 y≥0
Step 2: Introduce slack variables s1 and s2:
Minimise Z=x+2y+0s1+0s2
Subject to 2x+y+(3)s1≥0 x+2y+(6)s2≥0 x+0s1+0s2≥0 y+0s1+0s2≥0
Step 3: Introduce the artificial variables a1 and a2:
Minimise Z=x+2y+0s1+0s2a1a2
Subject to 2x+y+(3)s1+a1≥0 x+2y+(6)s2+a2≥0 x+0s1+0s2a1≥0 y+0s1+0s2a2≥0 a1,a2≥0
Question:
Minimize Z=−3x+4y, subject to x+2y≤8, 3x+2y≤12, x≥0, y≥0.
Answer:
Step 1: Write the given constraint equations in standard form.
x + 2y ≤ 8 —————> x  2y ≥ 8
3x + 2y ≤ 12 —————> 3x  2y ≥ 12
Step 2: Write the objective function in standard form.
Z = 3x + 4y —————> Z = 3x  4y
Step 3: Write the augmented matrix.
[ 1 2  8 ] [ 3 2  12 ] [ 3 4  0 ]
Step 4: Use Gaussian elimination to solve the system.
[ 1 2  8 ] —> [ 1 2  8 ] —> [ 1 0  4 ] [ 0 0  0 ] —> [ 3 2  12 ] —> [ 0 1  4 ] [ 3 4  0 ] —> [ 0 0  0 ]
Step 5: The solution is x = 4, y = 4, and Z = 12.
Question:
Minimise and Maximise of Z=5x+10y subject to x+2y≤120,x+y≥60,x−2y≥0,x,y≥0
Answer:
Step 1: Write the objective function as an equation: Z = 5x + 10y
Step 2: Write the constraints as equations: x + 2y ≤ 120 x + y ≥ 60 x − 2y ≥ 0 x ≥ 0 y ≥ 0
Step 3: Graph the constraints on a coordinate plane.
Step 4: Find the vertices of the feasible region. Vertex 1: (0, 60) Vertex 2: (60, 0) Vertex 3: (120, 0)
Step 5: Calculate the value of the objective function at each vertex. Vertex 1: Z = 300 Vertex 2: Z = 300 Vertex 3: Z = 600
Step 6: Determine the minimum and maximum values of the objective function. Minimum Value: Z = 300 Maximum Value: Z = 600
Question:
Minimise and Maximise Z=x+2y subject to x+2y≥100,2x−y≤0,2x+y≤200;x,y≥0
Answer:
Step 1: Write the objective function Z=x+2y
Step 2: Write the constraints: x+2y≥100,2x−y≤0,2x+y≤200;x,y≥0
Step 3: Minimise Z by solving the system of equations:
x+2y≥100 2x−y≤0 2x+y≤200 x,y≥0
Step 4: Use the method of substitution to solve the system of equations:
Substitute x=2002y in the first equation: 2002y+2y≥100 200≥100
Therefore, the constraint is satisfied.
Step 5: Substitute x=2002y in the second equation: 2(2002y)y≤0 4004yy≤0 4005y≤0
Therefore, the constraint is satisfied when 5y≤400 or y≤80.
Step 6: Substitute x=2002y in the third equation: 2(2002y)+y≤200 4002y+y≤200 400y≤200
Therefore, the constraint is satisfied when y≤200.
Step 7: Finally, the constraints are satisfied when 0≤y≤80 and 0≤x≤200.
Step 8: To minimise Z, substitute x=2002y in the objective function: Z=2002y+2y Z=200
Therefore, the minimum value of Z is 200.
Step 9: To maximise Z, substitute x=2002y in the objective function: Z=2002y+2y Z=400
Therefore, the maximum value of Z is 400.
Question:
Minimise Z=3x+5y such that x+3y≥3,x+y≥2,x,y≥0
Answer:

Rewrite the given constraints in standard form: x + 3y ≥ 3 → x  3y ≤ 3 x + y ≥ 2 → x  y ≤ 2

Write the objective function in the form of a maximisation problem (by multiplying the coefficients of x and y by 1): Maximise Z = 3x  5y

Set up the initial simplex tableau:
 Z  x  y  s1  s2  b   3  1  3  1  0  3   5  1  1  0  1  2 

Select a pivot element: Select the pivot element from the bottom row and the leftmost column (in this case, 3 from the Z column).

Perform row operations to create the new tableau:
 Z  x  y  s1  s2  b   1  1/3  1  1/3  0  1   0  2/3  2  1/3  1  2 
 Check for optimality: The optimal solution is x = 1, y = 2.
Question:
Maximize Z=5x+3y subject to 3x+5y≤15,5x+2y≤10,x≥0,y≥0
Answer:
Step 1: Identify the objective function to be maximized:
Z = 5x + 3y
Step 2: Identify the constraints:
3x + 5y ≤ 15 5x + 2y ≤ 10 x ≥ 0 y ≥ 0
Step 3: Rewrite the constraints in standard form:
3x + 5y ≤ 15 3x  5y ≥ 15
5x + 2y ≤ 10 5x  2y ≥ 10
x ≥ 0 x ≤ 0
y ≥ 0 y ≤ 0
Step 4: Set up the initial tableau:
x y s1 s2 s3 s4 Z
Step 5: Add slack variables to the constraints:
x y s1 s2 s3 s4 Z 3 5 1 0 0 0 0 5 2 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0
Step 6: Introduce the objective function:
x y s1 s2 s3 s4 Z 3 5 1 0 0 0 0 5 2 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 5
Step 7: Perform row operations to create a basic feasible solution:
x y s1 s2 s3 s4 Z 1 3 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 5
Step 8: Check for optimality:
The solution is optimal since all the coefficients in the objective row are zero except for the Z column.
Question:
Maximise Z=x+y, subject to x−y≤−1,x−y≥0,x,y≥0
Answer:
 Rewrite the given constraint equations in the standard form:
x  y ≤ 1 x + y ≥ 0 x ≥ 0 y ≥ 0
 Construct the feasible region:
The feasible region is the set of all points (x, y) that satisfies the constraint equations.
 Find the corner points of the feasible region:
The corner points of the feasible region are (0, 0), (0, 1), (1, 0), and (1, 1).
 Calculate the value of the objective function at each corner point:
At (0, 0): Z = 0 + 0 = 0 At (0, 1): Z = 0 + (1) = 1 At (1, 0): Z = 1 + 0 = 1 At (1, 1): Z = 1 + (1) = 0
 Select the corner point which maximises the objective function:
The corner point which maximises the objective function is (1, 0).
Hence, the maximum value of Z is 1.
Question:
Maximise Z=3x+2y subject to x+2y≤10,3x+y≤15,x,y≥0
Answer:
Step 1: Rewrite the objective function and the constraints in standard form.
Objective Function: Maximise Z = 3x + 2y
Constraints: x + 2y ≤ 10 3x + y ≤ 15 x, y ≥ 0
Step 2: Graph the constraints.
Step 3: Find the feasible region.
Step 4: Find the coordinates of the vertices of the feasible region.
Vertices: (0, 0), (0, 5), (5, 0), (2.5, 2.5)
Step 5: Substitute the coordinates of the vertices into the objective function to determine the maximum and minimum values of Z.
Maximum Z: Z = 3(2.5) + 2(2.5) = 12.5
Minimum Z: Z = 3(0) + 2(0) = 0
Question:
Maximize Z=3x+4y subject to the constraints: x+y≤4,x≥0,y≥0
Answer:
Step 1: Write the objective function in standard form: Maximize Z = 3x + 4y
Step 2: Write the constraints in standard form: x + y ≤ 4 x ≥ 0 y ≥ 0
Step 3: Draw a graph of the constraints.
Step 4: Find the feasible region and identify the corner points.
The feasible region is the area bounded by the lines x + y = 4, x = 0, and y = 0. The corner points are (0,0), (0,4), and (4,0).
Step 5: Evaluate the objective function at each corner point and identify the maximum value.
At (0,0): Z = 0 At (0,4): Z = 16 At (4,0): Z = 12
The maximum value of Z is 16, which occurs at (0,4).