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Operations Management and Strategic Management · Optimum Allocation of Resources - LPP

Special Cases in Graphical Solution of LPP

Updated 10 October 2026 · Fact-checked

Special cases in the graphical method are outcomes that do not give one neat optimal corner. Multiple optima mean the objective line is parallel to a binding constraint. Unbounded means the feasible region lets Z improve forever. Infeasible means no point satisfies all constraints. A redundant constraint does not change the feasible region.

Understand Special Cases in Graphical Solution

In the graphical method you draw every constraint, shade the feasible region, and test the corner points. Most questions give one best corner. Four special cases break this pattern, and the examiner uses them to test whether you understand what the graph is showing.

Multiple optimal solutions. The objective function line is parallel to one of the constraint lines that forms an edge of the feasible region. When you slide the objective line to its best position, it lies along that whole edge. Two corner points then give the same best value of Z. Every point on the line segment between them also gives that value. So there are infinitely many optimal solutions.

Unbounded solution. The feasible region is open in the direction in which Z improves. You can keep moving along the region and Z keeps growing (maximisation) or falling (minimisation). No finite optimum exists. In real problems this signals a missing constraint, because no firm has unlimited resources or demand.

Infeasible solution. The constraints contradict each other. There is no point that satisfies all of them together, so the feasible region is empty. An unbounded region is still a region. An infeasible problem has none.

Redundant constraint. A constraint that lies completely outside, or only touches, the feasible region formed by the other constraints. Removing it does not change the region or the answer. It is a valid constraint but never binding.

Key rules to remember

Multiple optima test
Slope of Z line = slope of a binding constraint edge, and both end corners give equal Z
General solution on the edge between corners A and B: λA + (1 − λ)B, for 0 ≤ λ ≤ 1. Z is the same everywhere on it.
Unbounded test
Feasible region open in the direction of improving Z
Maximisation is unbounded only if Z can rise without limit. An open region alone is not enough. A minimisation problem on an open region may still have a finite optimum. Minimisation can be unbounded only if Z can fall without limit, which needs some negative cost coefficient.
Infeasible test
Feasible region = ∅ (no common point)
Typical sign: conflicting constraints such as x + y ≤ 4 and x + y ≥ 6.
Redundant constraint test
Every corner point of the region satisfies the constraint without it being binding
Deleting it leaves the feasible region and the optimum unchanged.

How to solve Special Cases in Graphical Solution questions

Use the same graphical routine for every question. The special case shows up in steps 3 to 5.

  1. 1Convert each inequality to an equation, find two points (usually the intercepts) and draw each line. Use x ≥ 0 and y ≥ 0 unless told otherwise.
  2. 2Shade the side that satisfies each inequality. Test the origin if the line does not pass through it.
  3. 3Identify the common shaded region. If there is none, write: the problem is infeasible. Stop here.
  4. 4If the region exists but extends without limit in the direction in which Z improves, state that the solution is unbounded. Show a ray along which Z grows.
  5. 5If the region is closed, list every corner point and compute Z at each. Compare the values.
  6. 6If two adjacent corners give the same best Z, state that there are multiple optimal solutions. Write both corners and say that every point on the segment joining them is also optimal.
  7. 7Check each constraint against the final region. A constraint that touches no edge of the region is redundant. Name it.
  8. 8Write the conclusion in words, with the optimal values of x, y and Z where they exist.

Quickest way: Corner-table check with a slope comparison

When to use it: Use when you have limited time and the question has two variables and a closed or nearly closed region.

  1. Find the corner points by solving pairs of binding constraints. Do not plot every point.
  2. Compute Z at each corner and put the values in a small table.
  3. If two corners tie for the best value, confirm that the Z slope matches the edge slope. Then declare multiple optima.
  4. To test a doubtful constraint, substitute the corner points into it. If every corner satisfies it with room to spare, it is redundant.
  5. For unbounded or infeasible cases, ask two questions. Can any point satisfy all constraints? If yes, can I move to infinity and still improve Z? Answer from those.

Common mistakes in Special Cases in Graphical Solution

  • Calling a problem unbounded when it is infeasible, or the reverse.

    Both cases mean there is no single finite answer, so students mix them up.

    Fix: Unbounded has a feasible region that never ends. Infeasible has no feasible region at all. Always draw first and check whether any shaded area is common.

  • Giving only two points as the answer when there are multiple optima.

    Students stop after finding two corners with equal Z.

    Fix: State that infinitely many solutions exist on the line segment joining the two corners. Give the general form λA + (1 − λ)B with 0 ≤ λ ≤ 1.

  • Saying any open region is unbounded.

    Students look only at the shape of the region and forget the objective.

    Fix: Check the direction of improvement of Z. A minimisation problem on a region open to the upper right usually has a finite answer.

  • Treating a redundant constraint as an error and dropping it before drawing.

    Students feel a constraint that does nothing must be a misprint.

    Fix: Draw all constraints. Show that the line lies outside or does not bind the region. Then state that it is redundant and does not affect the optimum.

  • Declaring multiple optima without comparing Z values at the corners.

    Students notice that lines look parallel and skip the calculation.

    Fix: Parallel slopes are only a sign. Confirm by computing Z at both end corners and showing they are equal.

Worked examples

Example 1

Maximise Z = 4x + 2y subject to 2x + y ≤ 10; x ≤ 4; y ≤ 8; x + y ≤ 12; x, y ≥ 0. Find the optimal solution and comment on any special feature.

Show the solution
  1. Draw the lines 2x + y = 10, x = 4, y = 8 and x + y = 12 in the first quadrant.
  2. Corner points of the feasible region: O(0, 0); A(4, 0); B(4, 2), from x = 4 and 2x + y = 10; C(1, 8), from y = 8 and 2x + y = 10; D(0, 8).
  3. Compute Z: at O = 0; at A = 16; at B = 4(4) + 2(2) = 20; at C = 4(1) + 2(8) = 20; at D = 16.
  4. Maximum Z = 20 at both B and C. Note that Z = 2(2x + y), so the Z line has the same slope as the constraint 2x + y = 10. The edge BC lies on that constraint.
  5. Every point on BC is optimal: (x, y) = λ(4, 2) + (1 − λ)(1, 8), 0 ≤ λ ≤ 1.
  6. Check x + y ≤ 12: the largest value of x + y at the corners is 9 (at C), which is less than 12. The line never touches the region, so this constraint is redundant.

Answer: Maximum Z = 20. There are multiple optimal solutions: every point on the segment joining (4, 2) and (1, 8). The constraint x + y ≤ 12 is redundant.

Example 2

(a) Maximise Z = 3x + 2y subject to x − y ≤ 2; x ≥ 1; x, y ≥ 0. (b) Maximise Z = 2x + y subject to x + y ≤ 4; x + y ≥ 6; x, y ≥ 0. Identify the nature of each solution.

Show the solution
  1. Part (a): x − y ≤ 2 means y ≥ x − 2. With x ≥ 1 and y ≥ 0, the feasible region is the area to the right of x = 1 and above both y = 0 and y = x − 2. It is open to the right and upward.
  2. Take the points (t, t) for t ≥ 1. They satisfy x − y = 0 ≤ 2, x ≥ 1 and y ≥ 0, so they are feasible.
  3. At (t, t), Z = 3t + 2t = 5t. As t increases, Z increases without limit. So no finite maximum exists.
  4. Part (b): the first constraint needs x + y ≤ 4. The second needs x + y ≥ 6. No pair of values can be both at most 4 and at least 6.
  5. So there is no common feasible region, and the problem has no feasible solution.

Answer: (a) Unbounded solution: Z can be increased indefinitely. (b) Infeasible: the constraints conflict and the feasible region is empty.

Exam tips

  • When a question says 'comment on the solution' or 'what do you observe', expect a special case. Compute Z at all corners before commenting.
  • Write the reason in one line: 'Z line is parallel to constraint 1', 'region is open and Z increases', or 'no common region'. This earns the step marks.
  • In MCQs, read the words carefully. 'No feasible region' means infeasible. 'Z can be made arbitrarily large' means unbounded. 'More than one optimal point' means multiple optima.
  • Draw a neat, labelled graph with shaded region and corner coordinates. A rough sketch with the right label still earns marks, but an unlabelled one does not.
  • Do not say a redundant constraint is wrong. Say it does not affect the feasible region or the optimum.

Practice questions from Optimum Allocation of Resources - LPP

Special Cases in Graphical Solution: frequently asked questions

What is the difference between unbounded and infeasible solution in LPP?

In an unbounded problem, a feasible region exists but it extends so that Z can improve forever. In an infeasible problem, the constraints contradict each other and there is no feasible region. Draw the graph first to tell them apart.

How do you know if an LPP has multiple optimal solutions?

Two adjacent corner points give the same best value of Z. This happens when the objective function line is parallel to a binding constraint. Every point on the segment between those corners is then also optimal.

What is a redundant constraint in linear programming?

It is a constraint that does not form any boundary of the feasible region. Removing it leaves the region and the optimum unchanged. You can confirm it by checking that all corner points satisfy it without it being binding.

Can a minimisation problem be unbounded?

Yes, if Z can fall without limit over an unbounded feasible region, which needs some negative cost coefficient. With non-negative variables and non-negative cost coefficients, Z cannot fall below zero, so the problem is not unbounded.