Practice: The Linear Relaxation and Its Bound

Recognition · Interpretation

An integer minimisation has relaxation optimum z LP ∗ = 250 . What does this establish about the integer optimum z IP ∗ ?

2 hints available, least help first.

Hint 1: Retrieval cue

Which feasible set is larger, and what does optimising over a larger set do to the value?

Hint 2: Concept cue

For a minimisation, does a larger menu give a smaller or larger best value?

Direct application · Evaluation · Explanation

Consider

max 4 x 1 + 3 x 2 subject to 3 x 1 + 4 x 2 ≤ 12 , 4 x 1 + x 2 ≤ 10 , x 1 , x 2 ≥ 0  and integer .

(a) Solve the relaxation and state the bound with its direction. (b) Round its solution and test the result. (c) Find the integer optimum and give the gap.

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

2 hints available, least help first.

Hint 1: Retrieval cue

Solve the two binding constraints simultaneously for the relaxation.

Hint 2: Strategy cue

Test both rounding directions, and check feasibility before comparing values.

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 relaxation. Drop integrality. The two constraints bind where 3 x 1 + 4 x 2 = 12 and 4 x 1 + x 2 = 10 . From the second, x 2 = 10 − 4 x 1 ; substituting, 3 x 1 + 40 − 16 x 1 = 12 , so − 13 x 1 = − 28 and x 1 = 28 / 13 ≈ 2.154 , giving x 2 = 10 − 112 / 13 = 18 / 13 ≈ 1.385 .

Value: 4 ( 28 / 13 ) + 3 ( 18 / 13 ) = ( 112 + 54 ) / 13 = 166 / 13 ≈ 12.77 .

This is a maximisation, so the relaxation gives an upper bound: no integer plan exceeds 12.77 . Since the objective coefficients are whole and the variables are integer, the value is a whole number, so no integer plan exceeds 12 .

(b) Rounding. ( 2.154 , 1.385 ) rounds to ( 2 , 1 ) . Check: 3 ( 2 ) + 4 ( 1 ) = 10 ≤ 12 and 4 ( 2 ) + 1 = 9 ≤ 10 . Feasible, with value 8 + 3 = 11 .

Rounding up on x 2 instead gives ( 2 , 2 ) : 6 + 8 = 14 > 12 . Infeasible.

(c) The integer optimum. Check the feasible whole-number points with high value. ( 0 , 3 ) : 12 ≤ 12 , 3 ≤ 10 , feasible, value 9 . ( 1 , 2 ) : 3 + 8 = 11 ≤ 12 , 4 + 2 = 6 ≤ 10 , feasible, value 10 . ( 2 , 1 ) : feasible as above, value 11 . ( 2 , 0 ) : value 8 . ( 1 , 1 ) : value 7 .

The integer optimum is 11 at ( 2 , 1 ) .

The gap. 12.77 − 11 = 1.77 , or against the tightened bound of 12 , a gap of 1 .

What to notice. Here rounding to nearest happened to land on the optimum, and that is luck, not method. The other rounding direction was infeasible, and nothing in the rounding step distinguishes the two cases. The bound is what actually certifies ( 2 , 1 ) : no integer plan can exceed 12 , and we have found 11 , so at most 1 remains.

A complete answer does each of these:

  • forms relaxation
  • states bound direction
  • rejects rounding
  • draws valid inferences

Comparison · Method selection

The linear relaxation of an integer program is feasible, with optimum 30 . What follows about the integer program?

2 hints available, least help first.

Hint 1: Retrieval cue

Infeasible relaxation proves infeasible integer program. Does the converse hold?

Hint 2: Concept cue

Can you write an interval containing no whole number?

Direct application · Construction

An integer program has x 1 , x 2 ∈ { 0 , 1 } , x 3 ≥ 0 integer, and constraints x 1 + x 2 + x 3 ≤ 5 , x ≥ 0 . What is its linear relaxation?

2 hints available, least help first.

Hint 1: Retrieval cue

What exactly does 'relaxation' drop?

Hint 2: Concept cue

A binary variable has two restrictions on it: bounds, and wholeness.

Error diagnosis · Explanation · Evaluation

An analyst reports on

max x 1 + 10 x 2 subject to 10 x 1 + 11 x 2 ≤ 21 , x 1 , x 2 ≥ 0  and integer :

The relaxation gives x 2 = 21 / 11 ≈ 1.909 , x 1 = 0 , value 19.09 . Rounding down to stay feasible gives ( 0 , 1 ) with value 10 . Rounding is conservative, so ( 0 , 1 ) is the integer optimum or very near it.

Evaluate this.

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

2 hints available, least help first.

Hint 1: Retrieval cue

Try increasing x 1 from zero and check feasibility.

Hint 2: Concept cue

Compare the gap between the bound and the reported value. Is it small?

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.

The rounded point is feasible. ( 0 , 1 ) gives 0 + 11 = 11 ≤ 21 . Value 10 . So far the analyst is correct.

It is not optimal. Consider ( 1 , 1 ) : 10 + 11 = 21 ≤ 21 , feasible, with value 1 + 10 = 11 > 10 . The rounded plan is beaten.

Why rounding could never have found it. Rounding adjusts each coordinate to a neighbour of its fractional value. Here x 1 was 0 in the relaxation and is 1 at the optimum, rounding 0 gives 0 . The improvement required increasing a variable the relaxation had set to zero, which no rounding rule does.

Why 'conservative' is the wrong idea. Rounding down preserves feasibility for constraints of this form, so it avoids one failure mode. It says nothing about optimality, and 'conservative' quietly suggests the answer is safe in a sense it is not. It is safe to use and may be far from best.

What the bound actually tells us. The relaxation gives an upper bound of 19.09 , so no integer plan exceeds 19 . Against the analyst's ( 0 , 1 ) with value 10 that is a gap of 9 , enormous relative to the value, and a clear signal that searching further was worthwhile. The analyst had the number that showed their answer was questionable and read it as support instead.

The general point. The continuous and integer optima answer different questions, so there is no reason for them to be near each other. Here the relaxation puts everything into x 2 because its ratio is far better, while the integer optimum takes one unit of each. A structurally different plan, not a nearby one.

What should have been reported. The plan ( 1 , 1 ) with value 11 , together with the bound: no integer plan exceeds 19 , so at most 8 remains unclaimed, and whether to keep searching depends on what 8 is worth.

A complete answer does each of these:

  • forms relaxation
  • states bound direction
  • rejects rounding
  • draws valid inferences

Transfer · Interpretation · Evaluation

A solver working on an integer minimisation reports after an hour:

```
best bound 18240
incumbent 18795
gap 3.0%
```

A manager asks: what is the answer, is it optimal, and should we keep running?

Answer all three, and say what would change your advice.

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

2 hints available, least help first.

Hint 1: Retrieval cue

Which of the two numbers corresponds to a plan that exists?

Hint 2: Strategy cue

The gap converts a computing question into a question about value. What is the remaining amount worth?

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.

What the two numbers are. The incumbent, 18795 , is the best integer plan found so far. A real, implementable plan with that cost. The best bound, 18240 , comes from relaxations and says no integer plan costs less. Together:

18240 ≤ z IP ∗ ≤ 18795 .

What the answer is. The plan costing 18795 . That is a genuine answer available now, not a provisional one, and it can be implemented today.

Is it optimal? Unknown. It may be. The gap can persist because the bound is slack rather than because a better plan exists. What is certain is that it is within 555 of optimal, which is the 3 % the solver reports.

Should we keep running? That depends on what 555 is worth against the cost of computing, not on how long the solver has run. If these are pounds on a one-off schedule, an overnight run to chase at most 555 is likely poor value. If they are pounds per day on an annual contract, it is worth continuing. The bound has converted a computational question into a business one, which is its real utility.

What would change the advice.

  • A shrinking gap. If the gap has moved substantially in the last few minutes, progress is being made and continuing is more attractive.
  • A static gap. An hour with no movement suggests the remaining work is large, and the marginal value of more time is low.
  • The decision's own tolerance. If the plan must be acted on this afternoon, a 3 % guarantee is a good position and further computation is irrelevant.
  • Model quality. A large stubborn gap often indicates loose linking bounds. Tightening the formulation can close it far faster than more computing time, which is why 'improve the model' is sometimes the right answer to 'should we run longer'.

What not to say. That the answer is 18240 . Nothing establishes any plan attains the bound, and frequently none does. Reporting it as the cost would understate a real plan's price by 555 .

A complete answer does each of these:

  • forms relaxation
  • states bound direction
  • rejects rounding
  • draws valid inferences
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