Urgenthomework Logo
UrgentHomeWork
+1-617-874-1011 (US)
help@urgenthomework.com
Live chat
Download as:
Rating : ⭐⭐⭐⭐⭐
Price: $10.99
Language:EN
Pages: 45
Words: 4796

Entering variable isx minimum ratio and its row index

Exercise 25:

8x1 − 4x2 − 6x3 − 8x4 + x5 + 3x6 + S2= 24

12x1 + 5x2 − 9x3 + 6x4 − 9x5 + 8x6 + S3= 360

R2(old) = 24 8 -4 -6 -8 1 3 0 0
R2(new)=R2(old)÷8 3 1 -0.5 -0.75 -1 0.125 0.375 0 0
R3(old) = 360 12 5 -9 6 -9 8 0 1
R2(new) = 3 1 -0.5 -0.75 -1 0.125 0.375 0 0
12×R2(new) = 36 12 -6 -9 -12 1.5 4.5 0 0
R3(new)=R3(old) - 12R2(new) 324 0 11 0 18 -10.5 3.5 0 1

Negative minimum Zj-Cj is -28M-4 and its column index is 4. So, the entering variable is x4.

Minimum ratio is 10.8 and its row index is 1. So, the leaving basis variable is A1.

∴ The pivot element is 10.

Entering =x4, Departing =A1, Key Element =10

R1(new) = R1(old)/ 10

R1(old) = 108 0 10 6 10 9.5 -2.5 0
R1(new)=R1(old)÷10 10.8 0 1 0.6 1 0.95 -0.25 0
R3(old) = 324 0 11 0 18 -10.5 3.5 1
R1(new) = 10.8 0 1 0.6 1 0.95 -0.25 0
18×R1(new) = 194.4 0 18 10.8 18 17.1 -4.5 0
R3(new)=R3(old) - 18R1(new) 129.6 0 -7 -10.8 0 -27.6 8 1
R3(old) = 129.6 0 -7 -10.8 0 -27.6 8
R3(new)=R3(old)÷8 16.2 0 -0.875 -1.35 0 -3.45 1

R1(new) = R1(old)* 0.25 R3(new)

R2(new) = R2(old) - 0.125 R3(new)

R2(old) = 13.8 1 0.5 -0.15 0 1.075 0.125
R3(new) = 16.2 0 -0.875 -1.35 0 -3.45 1
0.125×R3(new) = 2.025 0 -0.1094 -0.1687 0 -0.4312 0.125
R2(new)=R2(old) - 0.125R3(new) 11.775 1 0.6094 0.0187 0 1.5062 0
Iteration-5 Cj 1 2 3 3 2 1
B CB XB x1 x2 x3 x4 x5 x6 Min/Ratio
XB/x3
x4 3 14.166 -0.0581 0.7459 (0.2614) 1 0 0 14.166/0.2614=54.1905→
x5 2 7.8174 0.6639 0.4046 0.0124 0 1 0 7.8174/0.0124=628
x6 1 43.1701 2.2905 0.5207 -1.3071 0 0 1 ---
Z=101.3029 Zj 3.444 3.5674 -0.4979 3 2 1
Zj-Cj 2.444 1.5674 -3.4979↑ 0 0 0

Negative minimum Zj-Cj is -3.4979 and its column index is 3. So, the entering variable is x3.

Minimum ratio is 54.1905 and its row index is 1. So, the leaving basis variable is x4.
∴ The pivot element is 0.2614.
Entering =x3, Departing =x4, Key Element =0.2614

Exercise 26:

a)

Max Z = x1 + 2 x2 + 3 x3
subject to
and x1,x2,x3≥0;

-->Phase-1<--

Iteration-1 Cj 0 0 0 0 0 0 0 0 0 0 0 -1 -1 -1 -1
B CB XB x1 x2 x3 S1 S2 S3 S4 S5 S6 S7 S8 A1 A2 A3 A4 Min Ratio
XB/x2
A1 -1 3 -3 (15) -3 -1 0 0 0 0 0 0 0 1 0 0 0 3/15=0.2→
S2 0 60 6 3 6 0 1 0 0 0 0 0 0 0 0 0 0 60/3=20
S3 0 21 -6 6 3 0 0 1 0 0 0 0 0 0 0 0 0 21/6=3.5
A2 -1 21 9 5 -1 0 0 0 -1 0 0 0 0 0 1 0 0 21/5=4.2
A3 -1 3 -3 5 2 0 0 0 0 -1 0 0 0 0 0 1 0 3/5=0.6
S6 0 30 6 8 -4 0 0 0 0 0 1 0 0 0 0 0 0 30/8=3.75
S7 0 12 0 8 -4 0 0 0 0 0 0 1 0 0 0 0 0 12/8=1.5
A4 -1 12 3 0 3 0 0 0 0 0 0 0 -1 0 0 0 1 ---
Z=-39 Zj -6 -25 -1 1 0 0 1 1 0 0 1 -1 -1 -1 -1
Zj-Cj -6 -25↑ -1 1 0 0 1 1 0 0 1 0 0 0 0
Iteration-3 Cj 0 0 0 0 0 0 0 0 0 0 0 -1 -1
B CB XB x1 x2 x3 S1 S2 S3 S4 S5 S6 S7 S8 A3 A4 Min Ratio
XB/x3
x2 0 0.6 0 1 -0.2 -0.06 0 0 -0.02 0 0 0 0 0 0 ---
S2 0 46.2 0 0 6.6 -0.02 1 0 0.66 0 0 0 0 0 0 46.2/6.6=7
S3 0 29.4 0 0 4.2 0.56 0 1 -0.48 0 0 0 0 0 0 29.4/4.2=7
x1 0 2 1 0 0 0.0333 0 0 -0.1 0 0 0 0 0 0 ---
A3 -1 6 0 0 (3) 0.4 0 0 -0.2 -1 0 0 0 1 0 6/3=2→
S6 0 13.2 0 0 -2.4 0.28 0 0 0.76 0 1 0 0 0 0 ---
S7 0 7.2 0 0 -2.4 0.48 0 0 0.16 0 0 1 0 0 0 ---
A4 -1 6 0 0 3 -0.1 0 0 0.3 0 0 0 -1 0 1 6/3=2
Z=-12 Zj 0 0 -6 -0.3 0 0 -0.1 1 0 0 1 -1 -1
Zj-Cj 0 0 -6↑ -0.3 0 0 -0.1 1 0 0 1 0 0

Negative minimum Zj-Cj is -1 and its column index is 8. So, the entering variable is S5.

Minimum ratio is 0 and its row index is 8. So, the leaving basis variable is A4.

∴ The pivot element is 1.

Entering =S5, Departing =A4, Key Element =1

R8(new)=R8(old)

R1(new)=R1(old) + 0.0667R8(new)

R2(new)=R2(old) - 2.2R8(new)

R3(new)=R3(old) - 1.4R8(new)

R4(new)=R4(old)

R5(new)=R5(old) + 0.3333R8(new)

R6(new)=R6(old) + 0.8R8(new)

R7(new)=R7(old) + 0.8R8(new)

Iteration-5 Cj 0 0 0 0 0 0 0 0 0 0 0
B CB XB x1 x2 x3 S1 S2 S3 S4 S5 S6 S7 S8 Min Ratio
x2 0 1 0 1 0 -0.0667 0 0 0 0 0 0 -0.0667
S2 0 33 0 0 0 0.2 1 0 0 0 0 0 2.2
S3 0 21 0 0 0 0.7 0 1 -0.9 0 0 0 1.4
x1 0 2 1 0 0 0.0333 0 0 -0.1 0 0 0 0
x3 0 2 0 0 1 -0.0333 0 0 0.1 0 0 0 -0.3333
S6 0 18 0 0 0 0.2 0 0 1 0 1 0 -0.8
S7 0 12 0 0 0 0.4 0 0 0.4 0 0 1 -0.8
S5 0 0 0 0 0 -0.5 0 0 0.5 1 0 0 -1
Z=0 Zj 0 0 0 0 0 0 0 0 0 0 0
Zj-Cj 0 0 0 0 0 0 0 0 0 0 0

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=2,x2=1,x3=2

Max Z=0

Iteration-2 Cj 1 2 3 0 0 0 0 0 0 0 0
B CB XB x1 x2 x3 S1 S2 S3 S4 S5 S6 S7 S8 Min Ratio
XB/S4
x2 2 2 0 1 0 -0.0333 0 0.0476 -0.0429 0 0 0 0 ---
S2 0 0 0 0 0 -0.9 1 -1.5714 (1.4143) 0 0 0 0 0/1.4143=0→
S8 0 15 0 0 0 0.5 0 0.7143 -0.6429 0 0 0 1 ---
x1 1 2 1 0 0 0.0333 0 0 -0.1 0 0 0 0 ---
x3 3 7 0 0 1 0.1333 0 0.2381 -0.1143 0 0 0 0 ---
S6 0 30 0 0 0 0.6 0 0.5714 0.4857 0 1 0 0 30/0.4857=61.7647
S7 0 24 0 0 0 0.8 0 0.5714 -0.1143 0 0 1 0 ---
S5 0 15 0 0 0 0 0 0.7143 -0.1429 1 0 0 0 ---
Z=27 Zj 1 2 3 0.3667 0 0.8095 -0.5286 0 0 0 0
Zj-Cj 0 0 0 0.3667 0 0.8095 -0.5286↑ 0 0 0 0

Since all Zj-Cj≥0

Hence, optimal solution is arrived with value of variables as :
x1=2,x2=2,x3=7

Max Z=27

b)

Max Z = x1 + 2 x2 + 3 x3
subject to
and x1,x2,x3≥0;
Iteration-1 Cj 0 0 0 0 0 0 0 0 0 0 0 0 -1 -1 -1 -1
B CB XB x1 x2 x3 S1 S2 S3 S4 S5 S6 S7 S8 S9 A1 A2 A3 A4 Min/Ratio
XB/x2
A1 -1 3 -3 (15) -3 -1 0 0 0 0 0 0 0 0 1 0 0 0 3/15=0.2→
S2 0 60 6 3 6 0 1 0 0 0 0 0 0 0 0 0 0 0 60/3=20
S3 0 21 -6 6 3 0 0 1 0 0 0 0 0 0 0 0 0 0 21/6=3.5
A2 -1 21 9 5 -1 0 0 0 -1 0 0 0 0 0 0 1 0 0 21/5=4.2
A3 -1 3 -3 5 2 0 0 0 0 -1 0 0 0 0 0 0 1 0 3/5=0.6
S6 0 30 6 8 -4 0 0 0 0 0 1 0 0 0 0 0 0 0 30/8=3.75
S7 0 12 0 8 -4 0 0 0 0 0 0 1 0 0 0 0 0 0 12/8=1.5
A4 -1 12 3 0 3 0 0 0 0 0 0 0 -1 0 0 0 0 1 ---
S9 0 1 0 0 2 0 0 0 0 0 0 0 0 1 0 0 0 0 ---
Z=-39 Zj -6 -25 -1 1 0 0 1 1 0 0 1 0 -1 -1 -1 -1
Zj-Cj -6 -25↑ -1 1 0 0 1 1 0 0 1 0 0 0 0 0
Iteration-3 Cj 0 0 0 0 0 0 0 0 0 0 0 0 -1 -1
B CB XB x1 x2 x3 S1 S2 S3 S4 S5 S6 S7 S8 S9 A3 A4 Min Ratio
XB/x3
x2 0 0.6 0 1 -0.2 -0.06 0 0 -0.02 0 0 0 0 0 0 0 ---
S2 0 46.2 0 0 6.6 -0.02 1 0 0.66 0 0 0 0 0 0 0 46.2/6.6=7
S3 0 29.4 0 0 4.2 0.56 0 1 -0.48 0 0 0 0 0 0 0 29.4/4.2=7
x1 0 2 1 0 0 0.0333 0 0 -0.1 0 0 0 0 0 0 0 ---
A3 -1 6 0 0 3 0.4 0 0 -0.2 -1 0 0 0 0 1 0 6/3=2
S6 0 13.2 0 0 -2.4 0.28 0 0 0.76 0 1 0 0 0 0 0 ---
S7 0 7.2 0 0 -2.4 0.48 0 0 0.16 0 0 1 0 0 0 0 ---
A4 -1 6 0 0 3 -0.1 0 0 0.3 0 0 0 -1 0 0 1 6/3=2
S9 0 1 0 0 (2) 0 0 0 0 0 0 0 0 1 0 0 1/2=0.5→
Z=-12 Zj 0 0 -6 -0.3 0 0 -0.1 1 0 0 1 0 -1 -1
Zj-Cj 0 0 -6↑ -0.3 0 0 -0.1 1 0 0 1 0 0 0

Negative minimum Zj-Cj is -6 and its column index is 3. So, the entering variable is x3.

Minimum ratio is 0.5 and its row index is 9. So, the leaving basis variable is S9.

∴ The pivot element is 2.

Entering =x3, Departing =S9, Key Element =2

R9(new)=R9(old)÷2

R1(new)=R1(old) + 0.2R9(new)

R2(new)=R2(old) - 6.6R9(new)

R3(new)=R3(old) - 4.2R9(new)

R4(new)=R4(old)

R5(new)=R5(old) - 3R9(new)

R6(new)=R6(old) + 2.4R9(new)

R7(new)=R7(old) + 2.4R9(new)

R8(new)=R8(old) - 3R9(new)

Negative minimum Zj-Cj is -0.3 and its column index is 4. So, the entering variable is S1.

Minimum ratio is 11.25 and its row index is 5. So, the leaving basis variable is A3.

∴ The pivot element is 0.4.

Entering =S1, Departing =A3, Key Element =0.4

R5(new)=R5(old)÷0.4

R1(new)=R1(old) + 0.06R5(new)

R2(new)=R2(old) + 0.02R5(new)

R3(new)=R3(old) - 0.56R5(new)

R4(new)=R4(old) - 0.0333R5(new)

R6(new)=R6(old) - 0.28R5(new)

R7(new)=R7(old) - 0.48R5(new)

R8(new)=R8(old) + 0.1R5(new)

R9(new)=R9(old)

 

Iteration-5 Cj 0 0 0 0 0 0 0 0 0 0 0 0 -1
B CB XB x1 x2 x3 S1 S2 S3 S4 S5 S6 S7 S8 S9 A4 Min Ratio
XB/S4
x2 0 1.375 0 1 0 0 0 0 -0.05 -0.15 0 0 0 -0.125 0 ---
S2 0 43.125 0 0 0 0 1 0 0.65 -0.05 0 0 0 -3.375 0 43.125/0.65=66.3462
S3 0 21 0 0 0 0 0 1 -0.2 1.4 0 0 0 0 0 ---
x1 0 1.625 1 0 0 0 0 0 -0.0833 0.0833 0 0 0 0.125 0 ---
S1 0 11.25 0 0 0 1 0 0 -0.5 -2.5 0 0 0 -3.75 0 ---
S6 0 11.25 0 0 0 0 0 0 0.9 0.7 1 0 0 2.25 0 11.25/0.9=12.5
S7 0 3 0 0 0 0 0 0 (0.4) 1.2 0 1 0 3 0 3/0.4=7.5→
A4 -1 5.625 0 0 0 0 0 0 0.25 -0.25 0 0 -1 -1.875 1 5.625/0.25=22.5
x3 0 0.5 0 0 1 0 0 0 0 0 0 0 0 0.5 0 ---
Z=-5.625 Zj 0 0 0 0 0 0 -0.25 0.25 0 0 1 1.875 -1
Zj-Cj 0 0 0 0 0 0 -0.25↑ 0.25 0 0 1 1.875 0
Max Z = x1 + 2 x2 + 3 x3
subject to
and x1,x2,x3≥0;

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

1. As the constraint-1 is of type '≥' we should subtract surplus variable S1 and add artificial variable A1

2. As the constraint-2 is of type '≤' we should add slack variable S2

3. As the constraint-3 is of type '≤' we should add slack variable S3

4. As the constraint-4 is of type '≥' we should subtract surplus variable S4 and add artificial variable A2

After introducing slack, surplus, artificial variables

Iteration-1 Cj 0 0 0 0 0 0 0 -1 -1
B CB XB x1 x2 x3 S1 S2 S3 S4 A1 A2 Min/Ratio
XB/x2
A1 -1 3 -3 (15) -3 -1 0 0 0 1 0 3/15=0.2→
S2 0 60 6 3 6 0 1 0 0 0 0 60/3=20
S3 0 21 -6 6 3 0 0 1 0 0 0 21/6=3.5
A2 -1 21 9 5 -1 0 0 0 -1 0 1 21/5=4.2
Z=-24 Zj -6 -20 4 1 0 0 1 -1 -1
Zj-Cj -6 -20↑ 4 1 0 0 1 0 0

Negative minimum Zj-Cj is -20 and its column index is 2. So, the entering variable is x2.

Minimum ratio is 0.2 and its row index is 1. So, the leaving basis variable is A1.

∴ The pivot element is 15.

Entering =x2, Departing =A1, Key Element =15

R1(new)=R1(old)÷15

R2(new)=R2(old) - 3R1(new)

R3(new)=R3(old) - 6R1(new)

R4(new)=R4(old) - 5R1(new)

Iteration-3 Cj 0 0 0 0 0 0 0
B CB XB x1 x2 x3 S1 S2 S3 S4 Min Ratio
x2 0 0.6 0 1 -0.2 -0.06 0 0 -0.02
S2 0 46.2 0 0 6.6 -0.02 1 0 0.66
S3 0 29.4 0 0 4.2 0.56 0 1 -0.48
x1 0 2 1 0 0 0.0333 0 0 -0.1
Z=0 Zj 0 0 0 0 0 0 0
Zj-Cj 0 0 0 0 0 0 0

Negative minimum Zj-Cj is -3.4 and its column index is 3. So, the entering variable is x3.

Minimum ratio is 7 and its row index is 3. So, the leaving basis variable is S3.

 ∴ The pivot element is 4.2.

Entering =x3, Departing =S3, Key Element =4.2

R3(new)=R3(old)÷4.2

R1(new)=R1(old) + 0.2R3(new)

R2(new)=R2(old) - 6.6R3(new)

R4(new)=R4(old)

Iteration-2 Cj 1 2 3 0 0 0 0
B CB XB x1 x2 x3 S1 S2 S3 S4 Min Ratio
XB/S4
x2 2 2 0 1 0 -0.0333 0 0.0476 -0.0429 ---
S2 0 0 0 0 0 -0.9 1 -1.5714 (1.4143) 0/1.4143=0→
x3 3 7 0 0 1 0.1333 0 0.2381 -0.1143 ---
x1 1 2 1 0 0 0.0333 0 0 -0.1 ---
Z=27 Zj 1 2 3 0.3667 0 0.8095 -0.5286
Zj-Cj 0 0 0 0.3667 0 0.8095 -0.5286↑

Negative minimum Zj-Cj is -0.5286 and its column index is 7. So, the entering variable is S4.

Minimum ratio is 0 and its row index is 2. So, the leaving basis variable is S2.

∴ The pivot element is 1.4143.

Entering =S4, Departing =S2, Key Element =1.4143

R2(new)=R2(old)÷1.4143

R1(new)=R1(old) + 0.0429R2(new)

R3(new)=R3(old) + 0.1143R2(new)

R4(new)=R4(old) + 0.1R2(new)

-6X1 + 6X2 + 3X3 ≤ 21,

9X1 + 5X2 - X3 ≥ 21,

X1, X2, X3>0

Exercise 28: LP Transportation problem

17X41+14X42+17X43+10X44+16X45+18X46

Subject to

X11+X21+X31+X41=13

X12+X22+X32+X42=12

Number of basic variables = m + n – 1 = 4+ 6 – 1 = 9.

The total transportation cost is calculated by multiplying each xij in an occupied cell with the corresponding cij and adding as follows:

Exercise 29: ManualSimplex LP implementation

Origin:

x12+x14-z=0

x23+x43-x36=0

Destination:

x25<=3

x23<=5

Send 2 unit from 1 to 2

Send 2 unit from 2 to 5

The maximum flow is 3 unit.

Chapter 12:

Exercise 30—W = 15 and

  1. Initialize maxProfit = 0;

    • Dequeue the Node u and and enqueue next level node I,e. node 1.

    • Profit of node 1 = 64, update maxProfit if profit of node is more.

  • So, Enqueue Node 2 to Queue.

  • Again, Repeat till Queue is not empty.

  1. Print the MaxProfit = 154.

Item 1 fits give profit of 64 and weight of 4.

64 + 90 + 65 = 219.

{}
0 0
219

90 + 91+ 24 = 205, for remove node 1 and enqueue 2

{1}
64 4
219
{1}
0 0
205
{1,2}
64 4
219
{1,2, 3}
0 0
154

Exercise 31: Java code to implement the solution

Output:

Q32)

Bad1 traversal is better score than Bad2. As we can see the output screen for both bad1 and bad2. Bad1 is far better than bad2

Exercise 33

56.29+8.54+31.40+26.93+22.14+53.85+25.00+16.03+13.60+14.00+50.70=318.48

Urgenthomework Logo
  Urgent HomeWork
Quick Links
Quick Links
Follow Us
Copyright © 2009-2023 UrgentHomework.com, All right reserved.