Lesson 3D.5.4
3D.5.4 Two-stage Simplex and big-M methods Quiz: Pearson Edexcel Further Maths, Unit 39
20 questions
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Lesson 3D.5.4, Two-stage Simplex and big-M methods: 20 multiple choice questions for the Pearson Edexcel Further Maths (9FM0), Unit 39: Linear programming, written with Revision Ninja.
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The 20 questions
-
In the constraint x + y - s + t = 5, what is the purpose of the artificial variable t?
- Objective maximiser
- Slack variable
- Surplus measure
- Starting basic variable
-
When writing a greater-than-or-equal constraint in standard form, what sign does the surplus variable have?
- Divided
- Added
- Multiplied
- Subtracted
-
In the big-M method for a maximisation problem, what coefficient does an artificial variable receive?
- Zero
- One
- Positive M
- Negative M
-
What quantity is minimised in stage one of the two-stage Simplex method?
- Decision variables
- Slack variables
- Objective function
- Artificial variables
-
At the end of stage one, the minimum sum of artificial variables is zero. What does this indicate?
- Optimal solution reached
- No solution exists
- Problem is unbounded
- Feasible solution found
-
Why does the standard Simplex method fail for a greater-than-or-equal-to constraint with a positive right-hand side?
- Origin is infeasible
- Objective is unbounded
- Origin is optimal
- Slack is negative
-
In the big-M method, what is the purpose of the parameter M?
- Very large penalty
- Maximum objective bound
- Minimum constraint value
- Multiplier for slack
-
Minimise C = 2x + 3y subject to x + y ≥ 4 and x + 3y ≥ 6 with x, y ≥ 0. What is the minimum value of C?
- 12
- 9
- 10
- 8
-
For minimise C = 2x + 3y subject to x + y ≥ 4 and x + 3y ≥ 6 with x, y ≥ 0, at which point is the minimum attained?
- (3, 1)
- (6, 0)
- (4, 0)
- (0, 4)
-
What is the maximum of P = x + 2y subject to x + y ≤ 5 and x + y ≥ 1 with x, y ≥ 0?
- 5
- 10
- 12
- 8
-
What is the maximum of P = x + y subject to x + 2y ≥ 4 and 3x + y ≤ 9 with x, y ≥ 0?
- 9
- 3.4
- 6
- 4
-
A maximisation problem has two ≥ constraints and one ≤ constraint. How many artificial variables are needed?
- 0
- 1
- 2
- 3
-
For x + y ≥ 4 and x + y ≤ 3 with x, y ≥ 0, what is the minimum possible sum of the artificial variables in stage one?
- 4
- 3
- 0
- 1
-
Write 2x + 3y ≥ 6 in standard form with surplus variable s and artificial variable t.
- 2x + 3y - s + t = 6
- 2x + 3y + s - t = 6
- 2x + 3y + s + t = 6
- 2x + 3y - s - t = 6
-
Maximise P = 2x + y subject to x + y ≥ 1 and x + y ≤ 4 with x, y ≥ 0. What is the maximum?
- 4
- 6
- 2
- 8
-
In the big-M method, what is the purpose of choosing M as a very large positive number?
- Eliminate slack variables
- Increase objective value
- Maximise profit function
- Penalise artificial variables
-
What does a positive artificial variable remaining in the optimal big-M tableau indicate?
- Alternative optima
- Degenerate solution
- Unbounded solution
- No feasible solution
-
After artificial variables reach zero in stage one, what happens to their columns in stage two?
- Inverted in tableau
- Dropped completely
- Doubled in size
- Set as basic
-
What error occurs if a student incorrectly uses positive M instead of negative M for a maximisation problem?
- Causes immediate cycling
- Rewards artificial variables
- Makes problem unbounded
- Forces zero objective
-
What does substituting the origin into a greater-than-or-equal-to constraint with a positive RHS yield?
- Undefined value
- True statement
- Zero remainder
- False statement
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