Simplex

LINEAR PROGRAMMING – THE SIMPLEX METHOD (1) Problems involving both slack and surplus variables

A linear programming model has to be extended to comply with the requirements of the simplex procedure, that is, 1. All equations must be equalities. 2. All variables must be present in all equations.

(I) In the case of ‘ ≤ ’ constraints, slack variables are added to the actual variables to make the equation an equality.

(II) In the case of ‘ ≥ ’ constraints, surplus variables are used to make the equation an equality.

A surplus variable is the difference between the total value of the true (decision) variables and the number (usually, total resource available) on the right-hand side of the equation. Thus, a surplus variable will always have a negative value.

Consider the following linear programming problem: Minimise cost = 21 86 XX + Subject to

0,7501500

5000

21

2

1

21

≥≥≤=+

XXX

XXX

There are three constraints, consisting of one equality, one ‘smaller than or equal to’ and one ‘greater than or equal to’.

The model should now be modified as follows:

0,,,750

15005000

2121

22

11

21

≥=−=+=+

SSXXSX

SXXX

where 1S is a slack variable and 2S is a surplus variable.

2

The presence of a surplus variable causes a problem when drawing the first simplex tableau because of its negative value. In any maximisation problem, this tableau must satisfy the following requirements:

1. All the slack variables (and thus surplus variables as well) must form part of the

initial solution mix (basis). 2. The table must contain as many rows as there are constraints. 3. The elements in the columns of variables appearing in the basis must form a unit

vector. If 2S is included in the basis, the elements of the 2S will be 0, 0 and –1 and thus not a

unit vector. Furthermore, the values of the true variables being 0 in the initial solution, this would imply that 2S would assume a negative value in the initial solution (it can be easily checked by means of the third equality that, if 2X = 0, 2S = –750). This is definitely contrary to the non-negativity restriction – all variables must have a positive value. (2) Artificial variables

This problem is solved by adding an artificial variable (denoted by iA ) to the equation, that is, a variable that has a positive value. Artificial variables are also used in equations which are already equalities in order to comply with the requirements (1) and (2) above. Just remember that an artificial variable has no significance pertaining to the solution of the problem – it is used merely to find a solution mix in the first simplex tableau.

Although artificial variables will always form part of the initial solution mix, the

objective is to remove them as soon as possible by means of the simplex procedure. As long as an artificial variable still appears in the solution mix, the final solution has not yet been found.

As any other variable, artificial variables are included in the objective function but the

value attached to them creates another problem. In minimisation problems, since the objective is to find the lowest cost possible, a very large value is allocated to artificial variables. This value is denoted by M where M is a very large positive value.

The above linear programming model is thus rewritten as

Minimise cost = 212121 0086 MAMASSXX +++++ Subject to

0,,,,,75010101015000001015000010011

212121

212121

212121

212121

≥=++−++=+++++=+++++

AASSXXAASSXXAASSXXAASSXX

before using the simplex procedure.

3

Example 1 (slack variables only) [Maximisation] Linpro Limited, a manufacturing enterprise, produces two products, namely product A and product B. Both products are manufactured from the same raw material and processed by the same machine. The available machine capacity is 600 hours and 10 000 kilogram of raw materials are available. The production of one unit of product A utilises 30 minutes of machine time while one unit of product B takes one hour to produce. To produce one unit of product A, 12.5 kg of raw material is required whereas 10 kg is required for one unit of product B. The marginal income of product A, of which a maximum of 700 units can be sold, amounts to R4 per unit. Product B has a marginal income of R6 per unit. The company wants to determine what combination of product A and product B must be produced in order to obtain maximum profit. Formulate a linear programming model for the above problem and hence find the optimal solution by using the simplex method. Solution 1 We first identify our decision variables, objective function and thus write down the constraints. Let the decision variables 1X = “number of units of product A to be manufactured” and

2X = “number of units of product B to be manufactured” [Decision variables are normally those which appear in the objective function. It is always better to use similar symbols instead of the traditional x and y to avoid confusion. We can, for example, choose to denote our decision variables by the first letters of the names of the products. However, for the simplex method, it is advisable to use indexed variables.] The objective function may be formulated as follows (it can be obtained from the table below): Maximise Profit = 21 64 XX + A table may also reveal to be helpful in understanding the problem better and to write down the constraints easily as well. Consider the following:

Product A Product B Total available resource Machine capacity (hrs) 0.5 1 600 Raw material (kg) 12.5 10 10000 Demand (at most) 700 Marginal income R4 R6

The constraints follow:

0,70010000105.126005.0

21

1

21

21

≥≤≤+≤+

XXX

XXXX

4

Now, we must ‘prepare’ this linear system of inequalities before using the simplex method. We convert all inequalities to equalities by introducing slack variables. The new model is

Maximise Profit = 32121 00064 SSSXX ++++ subject to

0,70010000105.126005.0

21

31

221

121

≥=+=++=++

XXSX

SXXSXX

where 21 , SS and 3S are slack variables. Tableau of the initial solution

Solution mix Quantity 1X 2X 1S 2S 3S

1S 600 0.5 1 1 0 0

2S 10000 12.5 10 0 1 0

3S 700 1 0 0 0 1 The first simplex tableau

jC 4 6 0 0 0 Solution mix Quantity 1X 2X 1S 2S 3S

0 1S 600 0.5 1 1 0 0 0 2S 10000 12.5 10 0 1 0 0 3S 700 1 0 0 0 1 jZ 0 0 0 0 0 0

jj ZC − 4 6 0 0 0 [Explanation for notation jC (row) = coefficients of variables in the objective function

jC (column) = coefficients of solution mix variables in the objective function

jZ = sum of the products of the jC column elements and corresponding variable column elements. For example, the bold zero in the table is obtained as follows:

)00()100()10( ×+×+× ] Note that the shaded cell gives the optimal profit.

5

The simplex procedure can be summarised as follows:

1. Find the pivot column – the column that has the greatest positive entry in the jj ZC − row. If there are no positive entries in a tableau, it means that the optimal

solution has already been reached.


0 1S 600 0.5 1 1 0 0 0 2S 10000 12.5 10 0 1 0 0 3S 700 1 0 0 0 1 jZ 0 0 0 0 0 0

jj ZC − 4 6* 0 0 0 2. Find the pivot row. Divide each solution mix entry in the Quantity column by its

corresponding entry in the pivot column and record the result in the Ratio column. The smallest non-negative ratio determines the pivot row.

jC 4 6 0 0 0 Solution mix Quantity 1X 2X 1S 2S 3S Ratio

0 1S 600 0.5 1 1 0 0 600* 0 2S 10000 12.5 10 0 1 0 1000 0 3S 700 1 0 0 0 1 ∞ jZ 0 0 0 0 0 0

jj ZC − 4 6 0 0 0 3. Select the pivot element in order to start applying basic row operations. The pivot

element is found at the intersection of the pivot row and the pivot column. It indicates which variable will enter the solution mix and which one will leave. In the above table, the pivot element is 1 (indicated in bold). We deduce that 2X will enter the solution mix and that 1S will leave. Why 1S ? Simply because, it is the solution mix slack variable which is found in the first row.

Pivot row

6

The second simplex tableau

Now, we have to perform basic row operations in order to make the 2X column entries identical to those in the unit vector that was in the 1S column. In this context, we will consider only two (out of the possible three) basic row operations:

1. Multiply a row by a non-zero constant. 2. Add a multiple of a row to another row.


0 1S 600 0.5 1 1 0 0 600* 0 2S 10000 12.5 10 0 1 0 1000 0 3S 700 1 0 0 0 1 ∞ jZ 0 0 0 0 0 0

jj ZC − 4 6 0 0 0 For example, if we have to make 10 (indicated in bold above) become 0, we have to take 10 subtract 10 times1 (the pivot element). In so doing, all the entries of the second row must also subtract 10 times the entries in the first. Remember, we only consider the entries in the rows involving the solution mix variables – in this case, 2S and 3S . The last entry in the pivot column being a zero, we don’t need any further basic row operations – at least one good thing here! The entries of the jZ and jj ZC − rows need to be recalculated. The complete second simplex tableau looks like


6 2X 600 0.5 1 1 0 0 0 2S 4000 7.5 0 –10 1 0 0 3S 700 1 0 0 0 1 jZ 3600 3 6 6 0 0

jj ZC − 1 0 –6 0 0 This procedure is repeated until there are no more positive entries in the jj ZC − row, in which case we would have reached the optimal solution.

Pivot row

7

The third simplex tableau


6 2X 600 0.5 1 1 0 0 1200

0 2S 4000 7.5 0 –10 1 0 533* 0 3S 700 1 0 0 0 1 700 jZ 3600 3 6 6 0 0

jj ZC − 1 0 –6 0 0 The largest positive entry in the last row is 1 so that the pivot column is the 1X column. The ratios of Quantity: 1X are recalculated and it is found that 533 is the smallest non-negative value, meaning that the pivot row is the 2S row. Thus, the pivot element is 7.5 (indicated in bold) so that 2S has to leave the solution mix for 1X to enter. Logically, we don’t expect any other tableaus to come up since both decision variables have entered the solution mix! Using basic row operations, we first divide the second row entries by 7.5 to obtain a 1 in its place before converting the 0.5 and 1 in the same column to zeros.


6 2X 600 0.5 1 1 0 0 0 2S 533 1 0 –1.33 0.13 0 0 3S 700 1 0 0 0 1 jZ 3600 3 6 6 0 0

jj ZC − 1 0 –6 0 0 Basic row operations: ‘First row minus half second row’ and ‘third row minus second row’.


6 2X 333 0 1 1.67 –0.067 0 4 1X 533 1 0 –1.33 0.13 0 0 3S 167 0 0 1.33 –0.13 1 jZ 4130 4 6 4.67 0.13 0

jj ZC − 0 0 –4.67 –0.13 0

Pivot row

pivot column

8

As predicted, this is the final simplex tableau since there are no more positive entries in the last row. We have therefore reached the optimal solution! Remember that the slack variable

3S is not a decision variable. Interpretation of the final simplex tableau

1. The final (optimal) solution is 5331 =X and 3332 =X so that the optimal profit is R4130. In fact, this solution mix corresponds to a corner-point (vertex) on graph. Note that each number in the columns of the variables occurring in the solution mix represents a unit vector.

2. The entry at the intersection of the 1S column and the jj ZC − row, –4.67,

represents the amount by which the total profit will decrease for every unit of 1S added to the solution mix, that is, R4.67. If the available machine hours are reduced by one hour, 2X will be reduced by 1.67 units in the solution mix whereas 1X will be increased by 1.33 units.

Decrease in profit = R6 × 1.67 = R10.00 Increase in profit = R4 × 1.33 = R 5.33 Net decrease in profit = R 4.67

3. The entry at the intersection of the 1S column and the jZ row, 4.67, represents the

amount by which the total profit will increase if one additional machine hour is made available for the production of product A and production B (that is, 601 machine hours). The shadow price with respect of one machine hour is thus R4.67.

4. The entry at the intersection of the 2S column and the jZ row, 0.13, is the shadow

price with respect to one kilogram of raw material. If the available raw material is increased by one kilogram, the total profit will increase by 13 cents.

5. The entry at the intersection of the 3S column and the jZ row, 0, is the shadow

price with respect to the demand for product A. Since the current solution mix does not fully satisfy the demand for product A, ( )67.13 =S , an increase in the demand for this product will have no influence on the total profit.

9

Example 2 (slack and artificial variables) [Minimisation] A producer of fertiliser receives an order to supply exactly 5000 kg of fertiliser, consisting of a special mixture of phosphate and ammonium nitrate. The mixture must not contain more than 1500 kg of phosphate and not less than 750 kg of ammonium nitrate. The cost per kilogram of phosphate and ammonium nitrate are R6 and R8 respectively. The producer would like to determine the ratio in which raw materials must be combined so as to limit costs to a minimum. Formulate a linear programming model for the above problem and hence find the optimal solution by using the simplex method. Solution 2 We first identify our decision variables, objective function and thus write down the constraints. Let the decision variables 1X = “number of kg of phosphate” and

2X = “number of kg of ammonium nitrate”

The objective function may be formulated as follows: Minimise Cost = 21 86 XX +

The constraints are

0,70015005000

21

2

1

21

≥≥≤≤+

XXX

XXX

[Note that this the very same example on artificial variables mentioned above.]

The above linear programming model is thus modified as

Minimise cost = 212121 0086 MAMASSXX +++++ Subject to

0,,,,,75010101015000001015000010011

212121

212121

212121

212121

≥=++−++=+++++=+++++

AASSXXAASSXXAASSXXAASSXX

before using the simplex procedure.

10

Tableau of the initial solution

Solution mix Quantity 1X 2X 1S 2S 1A 2A

1A 5000 1 1 0 0 1 0

1S 1500 1 0 1 0 0 0

2A 750 0 1 0 –1 0 1 The first simplex tableau

jC 6 8 0 0 M M Solution mix Quantity 1X 2X 1S 2S 1A 2A

M 1A 5000 1 1 0 0 1 0 0 1S 1500 1 0 1 0 0 0 M 2A 750 0 1 0 –1 0 1 jZ 5750M M 2M 0 –M M M

jj ZC − 6–M 8–2M 0 M 0 0 This time, to identify the pivot column, we have to find the largest negative entry in the jj ZC − row (since this is a minimisation problem). This is the 2X column because, remember, M is a large positive number so that 8–2M is more negative than 6–M.


M 1A 5000 1 1 0 0 1 0 0 1S 1500 1 0 1 0 0 0 M 2A 750 0 1 0 –1 0 1 jZ 5750M M 2M 0 –M M M

jj ZC − 6–M 8–2M* 0 M 0 0

Pivot column

11

To identify the pivot row, the ratios Quantity: 2X have to be calculated and the Ratio column created. Again, the smallest non-negative entry determines the pivot row.

jC 6 8 0 0 M M Solution mix Quantity 1X 2X 1S 2S 1A 2A Ratio

M 1A 5000 1 1 0 0 1 0 5000 0 1S 1500 1 0 1 0 0 0 ∞ M 2A 750 0 1 0 –1 0 1 750* jZ 5750M M 2M 0 –M M M

jj ZC − 6–M 8–2M 0 M 0 0 The pivot element is therefore 1 (indicated in bold above), that is, 2X enters and 2A leaves the solution mix. By means of basic row operations, we make the 2X column become a unit vector and calculate the values of jZ and jj ZC − accordingly to obtain the following tableau: The second simplex tableau


M 1A 4250 1 0 0 1 1 –1 0 1S 1500 1 0 1 0 0 0 8 2X 750 0 1 0 –1 0 1 jZ 4250M + 6000 M 8 0 M–8 M –M+8

jj ZC − 6–M 0 0 –M+8 0 2M–8 The pivot column, row (the ratios are recalculated) and element are again identified.

jC 6 8 0 0 M M

Solution mix Quantity 1X 2X 1S 2S 1A 2A Ratio

M 1A 4250 1 0 0 1 1 –1 4250 0 1S 1500 1 0 1 0 0 0 1500* 8 2X 750 0 1 0 –1 0 1 ∞ jZ 4250M + 6000 M 8 0 M–8 M –M+8

jj ZC − 6–M* 0 0 –M+8 0 2M–8

Pivot row

12

After applying the relevant basic row operations, we obtain The third simplex tableau


M 1A 2750 0 0 –1 1 1 –1 6 1X 1500 1 0 1 0 0 0 8 2X 750 0 1 0 –1 0 1 jZ 2750M + 15000 6 8 6–M M–8 M –M+8

jj ZC − 0 0 M–6 –M+8 0 2M–8 The procedure is repeated since there is still an artificial variable in the solution mix (that is, a negative entry in the last row, –M+8; the pivot column, row (the ratios are recalculated) and element are again identified.

jC 6 8 0 0 M M Solution mix Quantity 1X 2X 1S 2S 1A 2A Ratio

M 1A 2750 0 0 –1 1 1 –1 2750* 6 1X 1500 1 0 1 0 0 0 ∞ 8 2X 750 0 1 0 –1 0 1 –750 jZ 2750M + 15000 6 8 6–M M–8 M –M+8

jj ZC − 0 0 M–6 –M+8* 0 2M–8 The fourth (and final) simplex tableau


0 2S 2750 0 0 –1 1 1 –1 6 1X 1500 1 0 1 0 0 0 8 2X 3500 0 1 –1 0 1 0 jZ 37000 6 8 –2 0 8 0

jj ZC − 0 0 2 0 M–8 M

13

Interpretation of the final tableau In order to minimise cost, the mix must consist of 1500 kg phosphate ( 1X ) and 3500 kg of ammonium nitrate ( 2X ). In this mix, the minimum requirement set for ammonium nitrate will be exceeded by 2750 kg ( 2S ). The minimum total cost amounts to R37000.

14

(3) Extraordinary situations The following circumstances may also arise when the simplex method is applied.

CASES HOW TO RECOGNISE THEM IN A SIMPLEX TABLEAU

Multiple optimal solutions

If the index of a variable not contained in the solution mix is zero in the final tableau, it means that there is more than one optimal solution. Mathematically, the inclusion of this variable in the solution mix will not change the value of the objective function.

Degeneracy One or more constraints are redundant – as a result of this, two equally small non-negative ratios may occur in the table when the pivot row is to be identified. If the incorrect one is selected, the only problem is that an extra iteration will be carried out.

Infeasibility This case is easily recognised if there is still an artificial variable in the solution mix whilst all the indices indicate that an optimal solution has been found.

Unbounded solutions When identifying the pivot row, all the ratios will be either negative or ∞ . It means that one or more variables can increase indefinitely without the constraints being exceeded in a maximisation problem.

15

(4) Key concepts Artificial variable A variable that is only used in equations and ‘ ≥ ’ constraints to find a

feasible initial solution.

Index row jj ZC − This represents the net amount by which the objective function will increase (if positive) or decrease (if negative) if one additional unit of a variable is added to the solution mix.

Lower limit for shadow prices

The level to which a limited resource can be reduced with a constant reduction in the profit for each unit by which the resource is reduced.

Pivot column The column with the largest positive (negative) entry in the jj ZC − row in the case of a maximisation (minimisation) problem.

Pivot row The row with the smallest non-negative ratio between the quantity column and the number in the pivot column. The ratio represents the maximum quantity of the pivot variable that can be included in the next solution mix. The variable in the pivot row is replaced by the pivot variable in the next simplex tableau.

Pivot variable

The variable that will enter the solution mix.

Principal row The new row in the next simplex tableau that replaces the pivot row in the previous tableau. The value of the principal row is determined by dividing the pivot row by the pivot number.

Shadow prices The values of the slack variables in the jZ row. A shadow price is the value of one additional unit of resource that is already being fully utilised.

Slack variable A variable added to a ‘ ≤ ’ constraint to change it to an equality. It represents the unused portion of the resource.

Solution mix A combination of variables in the objective function representing a possible solution to the linear programming problem.

Surplus variable A variable added to a ‘ ≥ ’ constraint to change it to an equality. It represents the quantity by which the utilisation of a resource exceeds the prescribed minimum.

Upper bound for shadow prices

The level to which a limited resource can be reduced while the profit decreases by a constant amount with the removal of each unit of the resource.

jZ row It represents the total value of the solution mix in the quantity column. It further shows the amount by which the total value will be reduced if one additional unit of a slack variable is added to the solution mix.

16

Exercises Questions for discussion

1. Briefly explain the purpose of the simplex method as well as the procedure that is followed.

2. Enumerate the points of similarity between the simplex method and the graphical

method. Under what circumstances will preference be given to each method?

3. Define the following terms and indicate under what circumstances each is used:

(a) slack variable (b) surplus variable (c) artificial variable

4. How will you go about selecting the following after the first simplex tableau has been

completed?

(a) the pivot column (b) the pivot row (c) the pivot element

5. Explain fully why the row with the smallest ratio is identified as the pivot row. 6. Describe the lost important differences between maximisation and minimisation

problems in the application of the simplex method.

7. Why should an artificial variable appear in the optimal solution mix and what should the decision-maker do if this happens?

8. Explain what additional information is made available by the final simplex tableau.

17

Linear programming problem

Toskry Limited manufactures three types of videocassette tapes:

The L-500 with a recording time of two hours The L-750 with a recording time of three hours The P-2000 with a recording time of eight hours

The production manager consults you to help him determine the number of units of each type

of cassette tape to be manufactured per week, given an available production capacity of 450 hours per week. The cassette tapes are packed in containers of 100 cassettes each, which must be considered as a unit for production purposes.

The production manager has the following information available: L-500 L-750 P-2000 Production cost per unit (rands) 30 45 70 Production time per unit (mins) 15 30 60 The following information is provided by the marketing manager:

The demand for the L-500 is virtually indefinite and it is dispatched to the trade as it becomes available, without taking permanent orders.

There is a permanent order of 200 units of the L-750 per week. However, if more is produced, it could be sold without difficulty.

The P-2000 is specially manufactured for Video Scene Limited, which has placed an order of 150 units per week.

The profit margin of the company is the same for all three types of tapes. Required

(a) Formulate a linear programming model. (b) Construct an extended model that can be solved by means of the simplex method. (c) Find the optimal solution.

Solution to the linear programming problem

Minimise Cost = 321 704530 XXX ++ Subject to

0,,1502004505.025.0

321

3

2

321

≥=≥≤++

XXXX

XXXX

18

Minimise Cost = 2121321 00704530 MAMASSXXX ++++++ Subject to

0,,,,,,1502004505.025.0

2121321

23

122

1321

≥=+=+−=+++

AASSXXXAX

ASXSXXX

Tableau of the initial solution

Solution mix Quantity 1X 2X 3X 1S 2S 1A 2A

1S 450 0.25 0.5 1 1 0 0 0

1A 200 0 1 0 0 –1 1 0

2A 150 0 0 1 0 0 0 1 The first simplex tableau

jC 30 45 70 0 0 M M Solution mix Quantity 1X 2X 3X 1S 2S 1A 2A Ratio

0 1S 450 0.25 0.5 1 1 0 0 0 900 M 1A 200 0 1 0 0 –1 1 0 200* M 2A 150 0 0 1 0 0 0 1 ∞ jZ 350M 0 M M 0 –M M M

jj ZC − 30 45–M* 70–M 0 M 0 0 The second simplex tableau

jC 30 45 70 0 0 M M Solution mix Quantity 1X 2X 3X 1S 2S 1A 2A Ratio

0 1S 350 0.25 0 1 1 0.5 –0.5 0 350 45 2X 200 0 1 0 0 –1 1 0 ∞ M 2A 150 0 0 1 0 0 0 1 150* jZ 9000+150M 0 45 M 0 –45 45 M

jj ZC − 30 0 70–M* 0 45 M–45 0

19

The third and final simplex tableau

jC 30 45 70 0 0 M M Solution mix Quantity 1X 2X 3X 1S 2S 1A 2A

0 1S 200 0.25 0 0 1 0.5 –0.5 –1 45 2X 200 0 1 0 0 –1 1 0 70 3X 150 0 0 1 0 0 0 1

jZ 19500 0 45 70 0 –45 45 70

jj ZC − 30 0 0 0 45 M–45 M–70 Optimal solution: 2002 =X , 1503 =X Minimum profit = R 19 500

Courtesy Mr. Rajesh Gunesh

(E-mail: [email protected])

Documents

Simplex