chap 3 artificial variables(28.02.13)

7/27/2019 Chap 3 Artificial Variables(28.02.13)

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CHAPTER 3

Artificial Variables Techniques

3.1 INTRODUTION

LPP in which constraints may also have ≥ and = signs after ensuring that all 0 b i ≥ are considered in

this section. In such cases basis of matrix cannot be obtained as an identity matrix in the starting simplex

table, therefore we introduce a new type of variable called the artificial variable. These variables are

fictitious and cannot have any physical meaning. The artificial variable technique is a device to get the

starting basic feasible solution, so that simplex procedure may be adopted as usual until the optimal

solution is obtained. To solve such LPP there are two methods.

(i) The Big M Method or Method of Penalties.

(ii) The Two-phase Simplex Method.

3.2 THE BIG M METHOD

The following steps are involved in solving an LPP using the Big M method.

Step 1 Express the problem in the standard form.

Step 2 Add non-negative artificial variables to the left side of each of the equations corresponding to

constraints of the type ≥ or = However, addition of these artificial variable causes violation of thecorresponding constraints. Therefore, we would like to get rid of these variables and would not allow

them to appear in the final solution. This is achieved by assigning a very large penalty (-M for

maximization and + M for minimization) in the objective function.

Step 3 Solve the modified LPP by simplex method, until anyone of the three cases may arise.

BIG M METHOD CRIERION: OPTIMALITY

The optimal solution to the augmented problem to the original problem if there are noartificial variables with nonzero value in the optimal solution.

BIG M METHOD CRIERION: NO FEASIBLE SOLUTION

If any artificial variable is in the basis with non zero value at the optimal solution of the

augmented problem, then the original problem has no feasible solution. The solution satisfies

the constraints but does not optimize the objective function, since it contains a very large

penalty M and is called pseudo optimal solution.



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Note: While applying simplex method, whenever an artificial variable happens to leave the basis, we drop

that artificial variable and omit all the entries corresponding to its column from the simplex table.

Example 3.1 Use Big M method to solve the example 2.2 Turkey Feed Problem as given in Table 3.1.

TABLE 3.1 Turkey meal’s Data

Ingredient Compostitoin of each Pound Feed(Oz.) Minimum MonthlyRequirement Per

Turkey(Oz.)Brand Feed 1 Brand Feed 2

A

B

C

Cost per pound

5

4

1/2

Rs.2

10

3

0

Rs.3

90

48

3/2

21 x3x2zMinimize +=

subject to these constraints:

90x10x5 21 ≥+ ounces (ingredient A constraint)

48x3x4 21 ≥+ ounces (ingredient B constraint)

2

3x

2

11 ≥ ounces (ingredient C constraint)

0x1 ≥ , 0x 2 ≥

SOLUTION

Step 1. Express the problem in standard form

Slack variables s1, s2 and s3 are subtracted from the left-hand sides of the constraints to convert them to

BIG M METHOD CRIERION: DEGENERATE SOLUTION

If at least one artificial variable in the basis at zero level and the optimality condition is

satisfied then the current solution is an optimal basic feasible solution (though degenerated

solution).



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Eliminating 321 A,A,A from the first, second and third equations modified objective function can be

written as

M141MsMsMsx)3M13(x)2M10(zMinimize 32121 ++++−−−−=

Or M141MsMsMsx)3M13(x)2M10(z 32121 =−−−−+−+

Problem, now, has eight variables and three constraints. five of the variables have to be zeroised to get

initial basic feasible solution to the 'artificial system'. Putting,

3A,48A,90Aand,0sssxx 32132121 ======== we get

The starting feasible solution is .M141zand3A,48A,90A 321 ====

Note that we are starting with a very heavy cost (compare it with zero profit in maximization problem)

which we shall minimze during the solution procedure. Table 3.2 represents the problem and its solution.

.

TABLE 3.2 Initial simplex tableau

To improve this solution, we determine that x2 is the entering variable, because 10M-2 is the largest

positive entry in the bottom row.

1x 2x s1 s2 s3 A1 A2 A3 b B.V. Ratio

5 10 -1 0 0 1 0 0 90 s1 90/5=19

4 3 0 -1 0 0 1 0 48 s2 48/4=12

1 0 0 0 -1 0 0 1 3 s3 3/1=3

10M-2 13M-3 -M -M -M 0 0 0 141M z

Let M =50 as default value, then we have

TABLE 3.3 First iteration simplex tableau

1x 2x s1 s2 s3 A1 A2 A3 b B.V. Ratio

5 -1 0 0 1 0 0 90 s1 90/10=9

4 3 0 -1 0 0 1 0 48 s2 48/3=16

1 0 0 0 -1 0 0 1 3 s3 --498 647 -50 -50 -50 0 0 0 7050 z

Entering Current z-value

Leaving variable10



5

TABLE 3.4 Second iteration simplex tableau

1x 2x

s1 s2 s3 A1 A2 A3 b B.V. Ratio

1/2 1 -1/10 0 0 1/10 0 0 9 s1 9/0.5=18

5/2 0 3/10 -1 0 -3/10 1 0 21 s2 42/5=8.4

0 0 0 -1 0 0 1 3 s3 3/1=3

174.5 0 14.7 -50 -50 -64.7 0 0 1227 z


TABLE 3.5 Third iteration simplex tableau

1x

2x

s1 s2 s3 A1 A2 A3 b B.V. Ratio

0 1 -1/10 0 1/2 1/10 0 -1/2 7.5 x2 7.5/0.5=15

0 0 3/10 -1 -3/10 1 -5/2 13.5 A1 27/5=5.4

1 0 0 0 -1 0 0 1 3 x1

0 0 14.7 -50 124 -64.7 0 -174.5 703.5 z


TABLE 3.6 Optimal simplex tableau

1x

2x

s1 s2 s3 A1 A2 A3 b B.V.

0 1 -0.16 0.20 0 0.16 -0.20 0 4.8 x2

0 0 0.12 -0.40 1 -0.12 0.40 -1 5.4 s3

1 0 0.12 -0.40 0 -0.12 0.40 0 8.4 x1

0 0 -0.24 -0.20 0 -49.76 -49.8 -50 31.20 z

Since all z coefficients are negative it becomes optimal solution with minimum cost Rs. 31.20.

Hence, the minimum cost solution is to purchase 8.4 pounds of brand 1 feed and 4.8 pounds of brand 2

feed per turkey per month.

EXAMPLE 3.2 Use Big M method to Maximize z = 3x1 + 2x2 Subject to the constraints

12x4x3

2xx2

21

21

≥+

≤+

0x,x 21 ≥

1 Leaving variable

5/2 Leaving variable



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SOLUTION

Step 1. Express the problem in standard form

Slack variables s1 and s2 are add and subtracted from the left-hand sides of the constraint 1 and 2

respectively to convert them to equations. These variable s2 is called negative slack variable or surplusvariable. Variable s1 represents excess of availability of 2 units on constraint 1, s2 represents excess of

requirement of 12 on constraint 2 . Since they represent 'free', the cost/profit coefficients associated with

them in the objective function are zeros. The problem, therefore, can be written as follows

Maximize Z = 3x1 + 2x2+0s1+0s2

12sx4x3

2sxx2

221

121

=−+

=++

0,s,s,x,x 2121 ≥

Step 2. Find initial basic feasible solution

Putting x1 = x2 = 0, we get s1=2, s2 =-12 as the first basic solution but it is not feasible as s2 have negative

values that do not satisfy the non-negativity restrictions. Therefore, we introduce artificial variables A1 in

the second constraint, which take the form

12Asx4x3

2sxx2

1221

121

=+−+

=++

0A,s,s,x,x 12121 ≥

Now artificial variables with values greater than zero violate the equality in constraints established in step

1. Therefore, A1 should not appear in the final solution. To achieve this, they are assigned a large unit

penalty (a large negative value, - M) in the objective function, which can be written as

12121 MAs0s0x2x3zMaximize −+++=

Subject to

12Asx4x3

2sxx2

1221

121

=+−+

=++

0A,s,s,x,x 12121 ≥

Eliminating 1A from the second equation, modified objective function can be written as

M12Mss0x)M42(x)M33(zMaximize 2121 −−++++=



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Let M =50 as default value, then, we have

600s5s0x202x153zMaximize 2121 −−++=

Problem, now, has five variables and two constraints. Three of the variables have to be zeroised to get

initial basic feasible solution to the 'artificial system'. Putting 0sxx 221 === , we get

The starting feasible solution is 600zand12A,2s 11 −=== .

Note that we are starting with a very heavy negative profit which we shall maximize during the solution

procedure. Table 3.7 represents the problem and its solution.


To improve this solution, we determine that x2 is the entering variable, because -202 is the smallest entry

in the bottom row.

1x 2x s2 s1 A1 b B.V. Ratio

2 0 1 0 2 s1 2/1=2

3 4 -1 0 1 12 A1 12/4=3

-153 -202 50 0 0 -600 z

. Entering Current z-value


1x 2x s2 s1 A1 b B.V. Ratio

2 1 0 1 0 2 x2 2/1=2

-5 0 -1 -4 1 12 A1 12/4=3

251 0 50 202 0 -196 z

.

Since all Z coefficients and an artificial variable appears in the objective row at positive level, the

solution of given LPP does not possess any feasible solution. .

EXAMPLE 3.3 Solve the LPPMinimize z = 4x1 +x2

Subject to

3x2x

6x3x4

3xx3

21

21

21

≤+

≥+

=+

0x,x 21 ≥

1 Leaving variable



8

SOLUTION

Since the objective function is minimization, we covert it into maximization using

Min z = Max (-z)

Subject to

3x2x

6x3x43xx3

21

21

21

≤+

≥+

=+

0x,x 21 ≥

Convert the given LPP into standard form by adding artificial variables A1, A2, surplus variable s1 and

slack variable S2 to get the initial basic feasible solution.

212121 MAMAs0s0xx4zMaximize −−++−−=

Subject to

3Sx2x

6ASx3x4

3Axx3

221

2121

121

=++

=+−+

=++

0x,x 21 ≥

Modified objective function:

M9Ms)xM4()x4M7(zMaximize 121 −−−+−=

The starting feasible solution is A1 = 3, A2 = 6, S2 = 3.

Let M =50

TABLE 3.9 Initial iteration simplex tableau

1x 2x s1 A2 A1 s2 b Basic Varible

3 1 0 1 0 0 3 A2

4 3 -1 0 1 0 6 A1

1 2 0 0 0 1 3 s2

-346 -199 50 0 0 0 -450 z

To improve this solution, we determine that x2 is the entering variable, because -346 is the smallest entry

in the objective row.


1x 2x s1 A2 A1 s2 b B.V. Ratio

1 0 1 0 0 3 A2 3/1=23 Leaving variable



9

4 3 -1 0 1 0 6 A1 6/4=1.5

1 2 0 0 0 1 3 s2 3/1=3

-346 -199 50 0 0 0 -450 z




1 .33 0 .33 0 0 1 x1 1/.33

0 -1 -1.33 1 0 2 A1 2/1.67

0 1.67 0 -0.33 0 1 2 s2 2/1.67

0 -83.67 50 115.3 0 0 -104 z

Entering Current z-valueTABLE 3.12 Third iteration simplex tableau


1 0 0.2 .33 0 0 0.6 x1 1/.2

0 1 -0.6 -1.33 1 0 1.2 A1

0 0 1 -0.33 0 1 0 s1 2/1

0 0 -0.2 48.4 50.2 0 -3.6 z


TABLE 3.13 Optimal simplex tableau

1x 2x s1 A2 A1 s2 b B.V.

1 0 0 .40 0 -0.2 0.6 x1

0 1 0 -0.2 0 0.6 1.2 x2

0 0 1 1 -1 1 0 s1

0 0 0 48.6 50 0.2 -3.6 z

since all coefficients of z are positive solution is optimum and is given by x1=0.6=3/5, x2 =1.2=6/5,

and Max z = -3.6 = -18/5

Min z= Max (-z)= 18/5.

1.67 Leaving variable

1 Leaving variable



10

EXAMPLE 3.4

Consider the following linear programming model and solve it using Big M method.

321 x15x18x12zMinimize ++=

Subject to

0x,x,x

96x12x6x364x6x8x4

321

321

321

≥

≥++

≥++

SOLUTION

The canonical form of the given problem is shown below:

2121321 MAMAs0s0x15x18x12zMinimize ++++++=

Subject to

0A,A,s,s,x,x,x96Asx12x6x3

64Asx6x8x4

2121321

22321

11321

≥

=+−++

=+−++

The starting feasible solution is A1 = 64, A2 =96.

Let M =50,

Then we have modified objective function:

8000s50s50x885x682x338

M160s50s50x)M1815(x)M1418(x)M712(zMinimize

21321

21321

+++−−−=

+++−+−+−=

Subject to

0A,A,s,s,x,x,x

96Asx12x6x3

64Asx6x8x4

2121321

22321

11321

≥

=+−++

=+−++

TABLE 3.14 Initial simplex tableau

1x 2x x3 s1 s2 A1 A2 b B.V.

4 8 6 -1 0 1 0 64 A1

3 6 12 0 -1 0 1 96 A2

338 682 885 -50 -50 0 0 8000 z

To improve this solution, we determine that x3 is the entering variable, because 885 is the largest entry

in the objective row.



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1x 2x x3 s1 s2 A1 A2 b B.V.

4 8 6 -1 0 1 0 64 A1

3 6 0 -1 0 1 96 A2

338 682 885 -50 -50 0 0 8000 z


TABLE 3.16 Second Iteration Simplex tableau

1x 2x x3 s1 s2 A1 A2 b B.V.

2.5 0 -1 0.5 1 -0.5 16 A1

0.25 0.5 1 0 -0.08 0 0.08 8 x3

116.75 239.5 0 -50 23.75 0 -73.75 920 z


TABLE 3.17 Optimal Simplex tableau

1x 2x x3 s1 s2 A1 A2 b B.V.

2.5 1 0 -1 0.5 1 -0.5 3.2 x2

0.25 0 1 0 -0.08 0 0.08 6.4 x3

-3 0 0 -2.1 -0.20 -47.9 -49.8 153.6 z

optimum z-value

since all coefficients of Z are negative solution is optimum and is given by x1=0, x2 =3.2=6/5, x3 = 6.4

and Min z = 153.6

EXERCISES 3.2

Use Big M method to solve the following LPP.

1. Minimize z = 12x1 +20x2

Subject to

0x,x

120x12x7

100x8x6

21

21

21

≥

≥+

≥+

[Ans: x1 =15, x2= 4 and Min z =205]

2. 321 x3xx2zMaximize ++= ,

Subject to

Leaving variable

5

12

Leaving variable



12

0x,x,x

12x4x3x2

5x2xx

321

321

321

≥

=++

≤++

[Ans: x1 =3, x2=2, x3,=0 and Max z = 8]

3. 321 x1x3x4zMinimize += ,Subject to

0x,x,x

8xx2x3

12x4x2x

321

321

321

≥

≥++

≥++

[Ans. x1 =0, x2=10/3,x3 =4/3 and Min z=34/3]

4. 321 xx4x2zMaximize ++=

Subject to

0x,x,x

1x2x2x

2x2xx25xx2x

321

321

321

321

≥

≥++−

=+−

≤−−

[Ans: Unbounded solution]

5. 321 x5x3xzMaximize ++= ,

Subject to

0x,x,x

4xx2x

33x9x10x33

15x5x10x3

321

321

321

321

≥

≥++

≤+−

≤++

[Ans: No solution]

3.3 THE TWO PHASE METHOD

In the preceeding section we observed that it was frequently necessary to add artificial variables to the

constraints to obtain an initial basic feasible solution to an L.P. problem. If the problem is to be solved,

the artificial variables must be driven to zero. The two-phase method is another method to handle these

artificial variables. Here the L.P. problem is solved in two phases.

PHASE I

In this phase we find an initial basic feasible solution to the original problem. For this all artificial

variables are to be driven to zero. To do this an artificial (Auxilary) objective function (r) is created which



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is the sum of all the artificial variables. This new objective function is then minimized, subject to the

constraints of the given (original) problem, using the simplex method. At the end of phase I, two cases

arise:

If the minimum value of r = 0 and one or more artificial variables appear in the basis at zero level, then a

feasible solution to the original problem is obtained. However, we must take care of this artificial variable

and see that it never becomes positive during phase II computations. Zero cost coefficient is assigned to

this artificial variable and it is retained in the initial table of phase II. If this variable remains in the basis

at zero level in all phase II computations, there is no problem. However, the problem arises if it becomes

positive in some iteration. In such a case, a slightly different approach is adopted in selecting the outgoing

variable. The lowest non-negative replacement ratio criterion is not adopted to find the outgoing variable.

Artificial variable (or one of the artificial variables if there are more than one) is selected as the outgoing

variable. The simplex method can then be applied as usual to obtain the optimal basic feasible solution to

the given L.P. problem.

PHASE II

Use the optimum basic feasible solution of phase I as a starting solution for the original LPP. Assign the

actual costs to the variable in the objective function and a zero cost to every artificial variable in the basis

at zero level. Delete the artificial variable column from the table which is eliminated from the basis in

phase I. Apply simplex method to the modified simplex table obtained at the end of phase I till an

optimum basic feasible is obtained or till there is an indication of unbounded solution.

REMARKS

1. In phase I, the iterations are stopped as soon as the value of the new (artificial) objective function

becomes zero because this is its minimum value. There is no need to continue till the optimality is

TWO PHASE METHOD CRIERION: NO FEASIBLE SOLUTIONIf the minimum value of r > 0, and at least one artificial variable appears in the basis at a

positive level, then the given problem has no feasible solution and the procedure terminates.

TWO PHASE METHOD CRIERION: OPTIMALITY

If the minimum value of r = 0, and no artificial variable appears in the basis, then a basic

feasible solution to the given problem is obtained. The artificial column (s) are deleted for

phase II computations.



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reached if this value becomes zero earlier than that.

2. Note that the new objective function is always of minimization type regardless of whether the original

problem is of maximization or minimization type.

EXAMPLE 3.5

Consider the example 3.4, the following linear programming model and solve it using the two-phase

method.


Subject to

0x,x,x

96x12x6x3

64x6x8x4

321

321

321

≥

≥++

≥++

SOLUTIONThe canonical form of the given problem is shown below:


Subject to

0A,A,s,s,x,x,x

96Asx12x6x3

64Asx6x8x4

2121321

22321

11321

≥

=+−++

=+−++

Phase 1

The model for phase 1 with its revised objective function is shown below. The corresponding initial table

is presented in Table 3.18. Further iterations towards optimality are shown in Tables 3.19 and 3.21.

Auxilary objective function:

21321

21

ssx18x14x7160

AAr Minimize

++−−−=

+=

subject to

0A,A,s,s,x,x,x

96Asx12x6x3

64Asx6x8x4

2121321

22321

11321

≥

=+−++

=+−++

TABLE 3.18 Initial Table of Phase 1 (Example 3.5)

1x 2x x3 s1 s2 A1 A2 b B.V.

4 8 6 -1 0 1 0 64 A1

3 6 12 0 -1 0 1 96 A2

7 14 18 -1 -1 0 0 160 r



15


1x 2x x3 s1 s2 A1 A2 b B.V.

4 8 6 -1 0 1 0 64 A1

3 6 0 -1 0 1 96 x3

7 14 18 -1 -1 0 0 160 r

Entering Current r-value


1x 2x x3 s1 s2 A1 A2 b B.V.

2.5 0 -1 0.5 1 0.5 64 A1

0.25 0.5 1 0 -0.093 0 -0.093 96 A2

2.5 5 0 -1 0.5 0 1.5 16 r


TABLE 3.21 Optimal simplex tableau of pahase 1

1x 2x x3 s1 s2 A1 A2 b B.V.

0.5 1 0 -0.2 0.1 -0.1 -0.1 3.2 x2

0 0 1 0.1 -0.13 -0.093 0.13 6.4 x3

0 0 0 0 0 0 0 0 r

The set of basic variables in the optimal table of phase 1 does not contain artificial variables. So, the

given problem has a feasible solution.

Phase 2

The only one iteration of phase 2 is shown in Table 3.22. One can verify that Table 3.23 gives the optimal

solution. The solution in Table 3.23 is optimal and feasible. The optimal results are presented below.

by x1=0, x2 =3.2=6/5, x3 = 6.4 and min z=153.6

TABLE 3.22 Initial Table of Phase 2

1

x 2

x x3 s1 s2 b B.V.

0.5 1 0 -0.2 0.1 3.2 x2

0 0 1 0.1 -0.13 6.4 x3

-12 -15 -18 0 2.1 0 z

TABLE 3.23 Optimal Table of Phase 2

Leaving variable12

5 Leaving variable



16

1x 2x x3 s1 s2 b B.V.

0.5 1 0 -0.2 0.1 3.2 x2

0 0 1 0.1 -0.13 6.4 x3

3 0 0 0 2.1 153.6 z

The optimal results are presented by x1=0, x2 =3.2=6/5, x3 = 6.4 and min z=153.6

EXAMPLE 3.6

Use the two-phase simplex method to maximize

321 x3x4x5zMaximize +−=

Subject to

0x,x,x

50x6x3x8

76x10x5x620x6xx2

321

321

321

321

≥

≤+−

≤++

=−+

SOLUTION


321 x3x4x5zMaximize +−=

Subject to

0A,s,s,x,x,x

50sx6x3x8

76sx10x5x6

20Ax6xx2

121321

2321

1321

1321

≥

=++−

=+++

=+−+

Phase 1




321

1

x6xx220

Ar Minimize

+−−=

=

Subject to



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0A,s,s,x,x,x

50sx6x3x8

76sx10x5x6

20Ax6xx2

121321

2321

1321

1321

≥

=++−

=+++

=+−+

TABLE 3.24 Initial Table (Example 3.6)

1x 2x x3 A1 s1 s2 b B.V.

2 1 -6 1 0 0 20 A1

6 5 10 0 1 0 76 s1

8 -3 6 0 0 1 50 s2

2 1 -6 0 0 0 20 r

TABLE 3.25 First Iterationl simplex tableu

1x 2x x3 A1 s1 s2 b B.V.

2 1 -6 1 0 0 20 A1

6 5 10 0 1 0 76 s1

-3 6 0 0 1 50 s2

2 1 -6 0 0 0 20 r



1x 2x x3 A1 s1 s2 b B.V.

0 1.75 -7.5 1 0 -0.25 7.5 A1

0 7.25 5.5 0 1 -0.75 38.5 s1

1 -0.38 0.75 0 0 0.13 6.25 x1

0 1.75 -7.5 0 0 -0.25 7.5 r



1x 2x

x3 A1 s1 s2 b B.V.

0 1 -4.29 0.57 0 -0.14 4.29 x2

0 0 35.57 -4.14 1 0.29 7.43 s1

1 0 -0.86 0.21 0 0.07 7.86 x1

0 0 0 -1 0 0 0 r

8

Leaving variable




18

Optimum r-value

Phase 2

The only one iteration of phase 2 is shown in Table 3.28. One can verify that Table 3.29 gives the

optimal solution. The solution in Table 3.29 is optimal and feasible.


1x 2x x3 s1 s2 b B.V.

0 1 -4.29 0 -0.14 4.29 x2

0 0 35.57 1 0.29 7.43 s1

1 0 -0.86 0 0.07 7.86 x1

-5 4 -3 0 0 0 z


1x 2x x3 s1 s2 b B.V.

0 1 -4.29 0 -0.14 4.29 x2

0 0 35.57 1 0.29 7.43 s1

1 0 -0.86 0 0.07 7.86 x1

0 0 9.86 0 0.93 22.14 z

Optimum z-value

The optimal results are presented by x1=7.86, x2 =-4.29, x3 = 0 and max z=22.14.

EXAMPLE 3.7

Use two-phase simplex method to solve the following LPP.


Subject to

0x,x,x

4xxx

10x2x3

8xx2x4

321

321

21

321

≥

≥++

≤+

≤++

SOLUTION



subject to



19

0A,s,s,s,x,x,x

4Asxxx

10sx2x3

8sxx2x4

1321321

13321

221

1321

≥

=+−++

=++

=+++

Phase 1




321

1

xxx4

Ar Minimize

−−−=

=

subject to

0A,s,s,s,x,x,x

4Asxxx

10sx2x3

8sxx2x4

1321321

13321

221

1321

≥

=+−++

=++

=+++


1x 2x x3 s3 s1 s2 A1 b B.V.

4 2 1 0 1 0 0 8 s1

3 2 0 0 0 1 0 10 s1

1 1 1 -1 0 0 1 4 A1

1 1 1 -1 0 0 0 4 r

TABLE 3.31 First iteration simplex tableu

1x 2x x3 s3 s1 s2 A1 b B.V.

2 1 0 1 0 0 8 s1

3 2 0 0 0 1 0 10 s2

1 1 1 -1 0 0 1 4 A1

0 1 1 -1 0 0 0 4 r


4 Leaving variable



20

TABLE 3.32 Second iteration simplex tableu

1x 2x x3 s3 s1 s2 A1 b B.V.

1 0.5 0.25 0 0.25 0 0 2 x1

0 0.5 0.75 0 -0.75 1 0 4 s2

0 0.5 -1 -0.25 0 1 2 A1

0 0.5 0.75 -1 -0.25 0 0 2 r



1x 2x x3 s3 s1 s2 A1 b B.V.

1 0.33 0 0.33 0.33 0 -0.33 1.33 x1

0 1 0 -1 -1 1 1 6 s2

0 0.67 1 -1.33 -0.25 0 1.33 2.67 x3

0 0 0 0 -0.33 0 -1 0 r

Optimum r-value

Phase 2

The only one iteration of phase 2 is shown in Table 3.34. One can verify that Table 3.35, itself gives the

optimal solution. The solution in Table 3.35 is optimal and feasible.


1x 2x x3 s3 s1 s2 b B.V.

1 0.33 0 0.33 0.33 0 1.33 x1

0 1 0 -1 -1 1 6 s2

0 0.67 1 -1.33 -0.25 0 2.67 x3

-8 -4 -2 0 -0.33 0 0 z

TABLE 3.35 Optmal Table of Phase 2

1x 2x x3 s3 s1 s2 b B.V.




21

1 0.33 0 0.33 0.33 0 1.33 x1

0 1 0 -1 -1 1 6 s2

0 0.67 1 -1.33 -0.25 0 2.67 x3

0 0 0 0 2 0 16 z

The optimal results are presented by x1=1.33, x2 = 0, x3 = 2.67 and max z = 16.

EXAMPLE 3.8 Use two-phase simplex method to solve

Maximize Z = 5x1 + 3x2

Subject to

0x,x

6x4x

1xx2

21

21

21

≥

≥+

≤+

SOLUTION


Maximize Z = 5x1 + 3x2

subject to

0x,x

6ASx4x

1Sxx2

21

1221

121

≥

=+−+

=++

Phase 1




221

1

sx4x6

Ar Minimize

+−−=

=

Subject to

0x,x

6ASx4x

1Sxx2

21

1221

121

≥

=+−+

=++

Initial basic feasible solution is given by S1 =1, A1 = 6.



22


1x 2x s2 s1 A1 b B.V.

2 1 0 1 0 1 s1

1 4 -1 0 1 6 A1

1 4 -1 0 0 6 r


1x 2x s2 s1 A1 b B.V.

2 1 0 1 0 1 s1

1 4 -1 0 1 6 A1

1 4 -1 0 0 6 r


1x 2x s2 s1 A1 b B.V.

2 1 0 1 0 2 x2

-7 0 -1 -4 1 1 A1

-7 0 -1 -4 0 2 r

Since all coefficients of objective row are negative as shown in Table 3.38, z is an optimum feasible

solution to the auxiliary LPP is obtained. But as Min r > 0, and an artificial variable A, is in the basis at a

positive level, the original LPP does not possess any feasible solution.

EXERCISES

Use two-phase method to solve the following LPP.

1. 321 xxx2zMaximize ++=

subject to

0x,x,x

4x5x3x2

1x4x6x3

8x3x6x4

321

321

321

321

≥

≥−+

≤−−

≤++

[Ans: x1=9/7, x2 =10/21, x3=0 and max z= 64/21]



23

2. 321 x5x3x2zMaximize ++=

subject to

0x,x,x

4xx2x

33x9x10x33

15x5x10x3

321

321

321

321

≥

≥++

≤+−

≤++

[Ans: Infeasible Solution.]

3. A company possesses two manufacturing plants, each of which can produce three products X, Y, Z

from common raw material. However, the proportions in which the products are produced are different in

each plant and so are the plants operating cost per hour. Data on production per hour and costs are given

in Table 3.38 together with current orders in hand for each product.

TABLE 3.39

Product Operating cost per hour (Rs)X Y Z

Plant I 2 4 3 9

Plant II 4 3 2 10

Order on hand 50 24 60

Use simplex method to find the number of production hours needed to fulfil the orders on hand on a

minimum cost.

[Ans: Min z =195, x1 = 35/2, x2 =15/4]

4. Reduce the following LPP to the canonical form, so as to minimize z:

321 x3xx2zMinimize ++=

subject to

0x,x,x

3x5x2

7x4x2x3

5x2x5

321

21

321

21

≥

≤+

≥++

≤+−

[Ans: Min z =4.88, x1 = 1.5, x2 = 0, x3 = 0.63]



24


subject to

0x,x,x

4xx2x

33x9x10x33

15x5x10x3

321

321

321

321

≥

≥++

≤+−

≤++

[Ans: No feasible solution]


subject to

0x,x,x

24x3x4x

16x2xx2

10x5xx

321

321

321

321

≥

≤++

≥++

=++

[Ans: Min z = 39, x1 = 5, x2 = 4, x3 =1]

7. Use the two-phase method to

21 xxzMinimize +=

subject to

0x,x

7x7x

4xx2

21

21

21

≥

≥+

≥+

[Ans: Min z = 31/13, x1 = 21/13, x2 = 10/13]

8. Solve the following linear programming problem, using the two phases of the simplex method

21 xx2zMinimize +=

subject to

0x,x,x,x

1xxx

8x3x10x5

4321

421

321

≥

=++

=−+

[Ans: Min z = 4/5, x1 = 0, x2 = 4/5, x3 =0, x4 = 1/5]



26

14. Using any appropriate method, solve the following problems. If it is not possible to arrive at an

optimal solution, state clearly, the reasons for that and the characteristics that you see in your finalsolution.

(i) (ii)

0x,x

40x2

10xx

tosubject

xx2zMax

21

1

21

21

≥

≤

≤−

+=

0x,x,x

12xx4x3

2xx2

tosubject

xx2x3zMax

321

321

21

321

≥

≥++

≤+

++=

(iii)

0x,x,x

40x3xx2

10x5xx

tosubject

x3xx2zMax

321

321

321

321

≥

≤+−

≤+−

+−=

15. A company manufactures three types of leather belts, namely A, B and C. The unit profits from then

three varieties are Rs. 10, Rs. 5 and Rs. 7 respectively. Leather is sufficient for only 800 belts per day (all

types together). Belt A requires thrice the time of belt B and belt C requires twice the tim of B. If all the

belts of type B are produced, a maximum of 1,000 belts per day can be produced. Belt A requires a fancy

buckle and 150 such buckles per day are available. There are sufficient number buckles for the other

varieties. Determine how many belts of each type be produced to maximize the total profit.16. A factory works 8 hours a day, producing three products viz. A, B and C. Each of these products is

processed in three different operations viz. 1, 2 and 3. The processing times in minutes for each of these

products in each of the operations are given in Table 3.40 along with utilisation of the processes and the

cost and price in rupees for each of these three products which have unlimited demand.

TABLE 3.40

Processing time (rains.) in operation Cost/unit Price/unit

Product 1 2 3 (Rs.) (Rs.)

A 4 3 1 10 16

B 2 1 4 8 12

C 3 4 5 5 10

Utilisation 80% 70% 90%

(i) Determine the optimal product mix using the simplex method.

(ii) Give the interpretations for the values obtained in the final simplex table.



27

17. Three types of cutting tools are produced in a factory using a lathe, a grinder and a polisher. The

duration in hours required to produce one batch of tools on each of these machines are given in the

following table 3.41 along with costs, selling prices of each batch of tools and the minimum number of

hours available on each machine per week.TABLE 3.41

Tool

type

Processing time (hrs.) per batch on a Cost the

(Rs.) per batch

Selling prices

(Rs.) per batch Lathe Grinder Polisher

A 7 2 5 100 145

B 3 3 8 65 100

C 4 4 2 80 120

per week 50 46 80

(i) Determine the optimum production schedule and maximum profit per we7k.

(ii) The factory wants to double the capacity of grinder at an additional c6st of Rs. 50 per week. Should it

go for it ? Substantiate your result.

18. A manufacturer produces three products A, B and C. Each product requires processing on two

machines I and Il. The time required to produce one unit of each product on a machine is

TABLE 3.42

Product Time to produce one unit (hrs.)

Machine I Machine II

A 0.5 0.6

B 0.7 0.8

C 0.9 1.05

There are 850 hours available on each machine. The operating cost is Rs. 5/hr. for machine I and Rs. 4/hr.

for machine II. The market requirements are at least 90 units of A, at least 80 units of B and at least 60

units of C. The manufacturer wishes to meet the requirements at minimum cost. Solve the problem by the

simplex method.

321

321

x)7.8x7.6x9.4

x)4x05.15x9.0(x)4x7.05x7.0(x)4x6.05x5.0(zMinimize

++=

+++++=



28

Subject to

0x,x,x

60x

80x

90x

850x05.1x8.0x6.0

850x9.0x7.0x5.0

321

3

2

1

321

321

≥

≥

≥

≥

≤++

≤++

SELF TEST

1.A basic feasible solution is a solution to an LP problem that corresponds to a corner point of the

feasible region.

a. True

b. False

2. In preparing a ~! constraint for an initial simplex tableau,

we would

a. add a slack variable.

b. add a surplus variable.

c. subtract an artificial variable.

d. subtract a surplus variable and add an artificial variable.

3. In the initial simplex tableau, the solution mix variables

can be

a. only slack variables.

b. slack and surplus variables.

c. artificial and surplus variables.

d. slack and artificial variables.

4. Even if an LP problem involves many variables, an optimal solution will always be found at a corner

point of the n-dimensional polyhedron forming the feasible region.

a. True

b. False

5. Which of the following in a simplex tableau indicates that

an optimal solution for a maximization problem has been found?

a. all the Cj — Zj values are negative or zero

b. all the C. — Z. values are positive or zero

c. all the substitution rates in the pivot column are nega-

tive or zero



29

d. there are no more slack variables in the solution mix

6. To formulate a problem for solution by the simplex

method, we must add slack variables to

a. all inequality constraints.

b. only equality constraints.c. only "greater than" constraints.

d. only "less than" constraints.

7. If in the optimal tableau of an LP problem an artificial

variable is present in the solution mix, this implies

a. infeasibility.

b. unboundedness.

c. degeneracy.

d. alternate optimal solutions.

8. If in the final optimal simplex tableau the C. — Z. value for

a nonbasic variable is zero, this implies

a. feasibility.

b. unboundedness.

c. degeneracy.

d. alternate optimal solutions.

9. In a simplex tableau, all of the substitution rates in the

pivot column are negative. This indicates

a. there is no feasible solution to this problem.

b. the solution is unbounded.

chap 3 artificial variables(28.02.13)

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