Daa unit iv - problems

UNIT IV - ITERATIVE IMPROVEMENT
SOLVED PROBLEMS
1. Solve the given linear equations by simplex method.
Maximize p = 3x + 4y subject to x+3y ≤ 30; method
Step 1: Insert slack variables S1 and S2.
x + 3y + S1 = 30
2x + y + S2 = 20
Step 2: Rewrite the objective function
x + 3y + S1 = 30
2x + y + S2 = 20
-3x - 4y + P = 0
Step 3: Form the initial simplex tableau
1 2x y S S P
1 3 1 0 0 30
2 1 0 1 0 20
3 4 0 0 1 0
 
   
   
Step4: Find the pivot element
1 2x y S S P
1 11 0 0 10
3 3
5 10 1 0 10
3 3
5 40 0 1 40
3 3
 
 
 
 
   
Repeat the above step till there is no negative value is last row.
1 2x y S S P
2 10 1 0 8
5 5
311 0 0 6
5 5
30 0 1 1 50
5
 
 
 
 
 
  
P reaches maximum value of 50 at x = 6 and y = 8.

2. Solve the given linear programming problem.
Maximize Z = -18x1 + 4x2 + 5x3
Subject to x1 ≥ 0, x2 ≥ 0, x3 ≥ 0
8x1 – 3x2 + 3x3  21
3x1 + 2x2 + x3 = 6
-2x1 + 4x2 + 3x3 ≥ 15
Step 1: Insert stack variable
8x1 – 3x2 + 3x3 + S1 = 21
3x1 + 2x2 + x3 = 6
-2x1 + 4x2 + 3x3 + S2 = 15
Step2: Rewrite the objective function
8x1 – 3x2 + 3x3 + S1 = 21
3x1 + 2x2 + x3 = 6
-2x1 + 4x2 + 3x3 + S2 = 15
Z + 181 – 4x2 – 5x3 = 15
Step 3: form the simplex tableau
1 2 3 1 2x x x S S Z
8 3 3 1 0 0 21
3 2 1 0 0 0 6
2 4 3 0 1 0 15
18 4 5 0 0 1 0
 
 
 
 
 
  
Step 4: select the pivot element and solve it iteratively.
The maximum value for Z = 30 at x1 = 0, x2 = 0 and x3 = 6.

3. Solve the following transportation problem.
fi – flow in ith
iteration
Iteration 1
S  4  3  t
f1 = min {2, 1, 1}
f1 = 1
Iteration 2
S  1  2  t
f2 = min {4, 4, 3}
f2 = 3
Iteration 3
S  4  t
f3 = min {1, 3}
f3 = 1
0/11/1
4
4
1
0/1
3
32
S
3 t
4
1 2
12
4
4
1
1
3
32
S
3 t
4
1 2

Iteration 4
S  1  3  4  t
f4 = min {1, 2, 1, 2}
f4 = 1
The maximum flow value = f1 + f2 + f3 + f4
= 1 + 3 + 1 + 1
= 6
0/10/2
0/4
1/3
1
0/1
2/1
0/32
S
3 t
4
1 2

4. Find the maximum flow of the network
O – source T – sink
Iteration 1
O  B  D  T
f1 = min {5, 6, 5}
f1 = 5
Iteration 2
O C  T
f2 = min {4, 5}
f2 = 4
Iteration 3
Redirect 1 unit of flow from path 0  B  D  T to path 0  B  C  T,
then freed capacity of arc D  T could be used.
O  A  D  T
 O  A  D  B  C  T
f3 = min {4, 4, 5, 4, 1}
f3 = 1
 Total flow = f1 + f2 + f3
= 4 + 5 + 1
= 10
44
4 4
5
6
5
50
B D
C
A
T

5. Find the maximal matching for the following graph: A  {1, 2, 4}, B  {1}, C 
{2, 3}, D  {4, 5}, E3.
Solution
Choose the path with unique matching
This is the maximal matching
A 1
B 2
C 3
D 4
E 5
A 1
B 2
C 3
D 4
E 5

6. Find the maximum matching for the following instance. Tcff  {Britta, Annie,
Super model} Vaugh  {Britta, Annie} Troy  {Annie, Lauch lady, Sherley}
Pierce  {Launch Lady} Abed  {Annie, Launch Lady }
The maximum matching
Teff Britta
Vaugh
n
Annie
Troy Launch lady
Pierce Shirley
Abed Super model
Teff Britta
Vaugh
n
Annie
Troy Launch lady
Pierce Shirley
Abed Super model

7. Find a good matching for marriage for the following people.
Men’s preference
Women’s preference
Finding the matching .
Solution: Men’s preference
Women’s preference
The matching M={(1, 4), (2, 3), (3, 2), (4, 1)}
1:
2:
3:
4:
2
4
1
2
1
3
4
1
1
1
1
1
1
1
4
1
3
4
2
2
3
3
1:
2:
3:
4:
2
3
2
4
4
1
3
1
1
1
1
1
1
1
1
4
1
3
3
2
4
2
1
2
3
4
2
4
1
2
1
3
4
1
1
1
1
1
1
1
4
1
3
4
3
2
2
3
1
2
3
4
2
3
2
4
4
1
3
1
1
1
1
1
1
1
1
4
1
3
3
2
4
2

8. Find the static matching for the following instance.
Men’ preference Women’s preference
If A  , B  β, C  , D  
A and β forms a blocking pair
A  , B  , C  β, D  
This is the stable matching.
1 2 3 4

β


A
C
C
B
B
A
B
A
D
D
D
C
C
B
A
D
1 2 3 4
A
B
C
D

β
β

β










β

Daa unit iv - problems

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