Practice: The Revised Simplex Method

Recognition · Comparison

The same program is solved twice from the same starting basis, once by full tableau and once by revised simplex, using the most-negative entering rule in both. What can be said about the two runs?

2 hints available, least help first.

Hint 1: Retrieval cue

Which of the six steps of an iteration differ between the two methods: the decisions, or the way the quantities are obtained?

Hint 2: Concept cue

Ask what the tableau's bottom row holds, and what the revised method computes when it prices a column.

Direct application

In a revised simplex iteration the multipliers are p T = ( − 5 / 6 , 0 ) . A nonbasic variable x 2 has cost c 2 = − 4 and column A 2 = ( 4 , 2 ) T .

What is the reduced cost c ¯ 2 ? Give your answer as a decimal.

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

The reduced cost is c j − p T A j . Compute the dot product before subtracting.

Hint 2: Concept cue

( − 5 / 6 ) ( 4 ) + ( 0 ) ( 2 ) = − 10 / 3 . Now subtract that from − 4 .

Direct application

In a revised simplex iteration the basis inverse is

A B − 1 = ( 1 − 1 0 1 / 2 ) ,

and the entering variable x 2 has column A 2 = ( 3 , 1 ) T .

Compute the direction u = A B − 1 A 2 . What is its first component u 1 ?

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

The first component of a matrix-vector product is the first row dotted with the vector.

Hint 2: Concept cue

Row one of A B − 1 is ( 1 , − 1 ) . Take its dot product with ( 3 , 1 ) .

Construction · Direct application · Explanation

Consider

min − 4 x 1 − 3 x 2 subject to 2 x 1 + 3 x 2 + x 3 = 12 , 2 x 1 + x 2 + x 4 = 8 , x ≥ 0 ,

with the basis B = { 1 , 3 } , where x 1 occupies the position corresponding to the second constraint row.

(a) Write A B , compute A B − 1 , and show the check that confirms it.

(b) Compute the basic values x B and state the full point and its objective value. Confirm it is feasible.

(c) Form the multipliers p T , then price both nonbasic columns. State which variable enters and why.

(d) Compute the direction u for the entering column only, apply the ratio test, and give the leaving variable, the step, and the new basis.

(e) List every quantity your iteration computed, and say which of them a full tableau would have carried instead. Name one quantity the tableau would have updated that you never needed.

Write your answer, then compare it with the worked solution.

3 hints available, least help first.

Hint 1: Retrieval cue

The basis order matters: c B must list the basic costs in the same order as the columns of A B .

Hint 2: Concept cue

Form p T once, then each reduced cost is a single dot product c j − p T A j .

Hint 3: Strategy cue

Before computing any direction, decide which column enters. Only that column needs u .

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) The basis matrix and its inverse. With x 3 in row 1 and x 1 in row 2, the basis columns in that order are A 3 = ( 1 , 0 ) T and A 1 = ( 2 , 2 ) T :

A B = ( 1 2 0 2 ) , det A B = 2 , A B − 1 = ( 1 − 1 0 1 / 2 ) .

Check: A B A B − 1 = ( 1 2 0 2 ) ( 1 − 1 0 1 / 2 ) = ( 1 0 0 1 ) ✓ (b) Basic values. x B = A B − 1 b = ( 1 − 1 0 1 / 2 ) ( 12 8 ) = ( 4 4 ) , so x 3 = 4 and x 1 = 4 . The full point is x = ( 4 , 0 , 4 , 0 ) , with objective − 4 ( 4 ) − 3 ( 0 ) = − 16 . Both basic values are nonnegative, so the basic solution is feasible. Checking the constraints directly: 2 ( 4 ) + 0 + 4 = 12 ✓ and 2 ( 4 ) + 0 + 0 = 8 ✓. (c) Multipliers and pricing. The basic costs in basis order are c B = ( c 3 , c 1 ) T = ( 0 , − 4 ) T , so

p T = c B T A B − 1 = ( 0 , − 4 ) ( 1 − 1 0 1 / 2 ) = ( 0 , − 2 ) .

Pricing the nonbasic columns x 2 and x 4 :

c ¯ 2 = − 3 − ( 0 , − 2 ) ⋅ ( 3 , 1 ) T = − 3 + 2 = − 1 ,
c ¯ 4 = 0 − ( 0 , − 2 ) ⋅ ( 0 , 1 ) T = 2 .

c ¯ 2 = − 1 < 0 , so x 2 enters: each unit of x 2 lowers the objective by 1 once the basic variables adjust. c ¯ 4 = 2 is positive, so returning x 4 to the basis would raise the objective. (d) Direction, ratio test, update. For the entering column only:

u = A B − 1 A 2 = ( 1 − 1 0 1 / 2 ) ( 3 1 ) = ( 2 1 / 2 ) .

Both components are strictly positive, so both rows are eligible. The ratios are x B ( 1 ) / u 1 = 4 / 2 = 2 and x B ( 2 ) / u 2 = 4 / ( 1 / 2 ) = 8 . The minimum is 2 at row 1, so x 3 leaves, θ = 2 , and the new basis is B ′ = { 2 , 1 } with x 2 now in row 1. The new point: x 2 = θ = 2 , and the remaining basic variable x 1 = 4 − 2 ( 1 / 2 ) = 3 . So x = ( 3 , 2 , 0 , 0 ) with objective − 4 ( 3 ) − 3 ( 2 ) = − 18 , down from − 16 by θ c ¯ 2 = 2 ( − 1 ) = − 2 ✓ (e) What was computed, and what a tableau would have carried. Computed this iteration: A B − 1 (4 entries), x B (2), p T (2), two reduced costs (2 scalars), and one direction u (2). Fourteen numbers. A full tableau would have carried the whole 3 × 5 array, all four variable columns plus the right-hand side, across two constraint rows and the objective row, and updated every entry at the pivot. A quantity never needed: the direction for column 4, A B − 1 A 4 . The tableau maintains that column at every iteration; here x 4 was priced as a single scalar, found positive, and dismissed, so its direction was never formed. On a program with hundreds of columns this is where the saving lies: most columns are priced and discarded, and only the entering one has a direction computed.

A complete answer does each of these:

  • forms multipliers
  • prices a column
  • computes direction on demand
  • ratio test and update
  • accounts for storage
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