Lesson 4D.1.1-4D.1.2
4D.1.1-4D.1.2 Initial solutions and the stepping-stone method Quiz: Pearson Edexcel Further Maths, Unit 40
20 questions
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Lesson 4D.1.1-4D.1.2, Initial solutions and the stepping-stone method: 20 multiple choice questions for the Pearson Edexcel Further Maths (9FM0), Unit 40: Transportation problems, written with Revision Ninja.
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The 20 questions
-
What is the north-west corner method used to find in a transportation problem?
- Initial feasible solution
- Shadow cost values
- Improvement indices
- Optimal total cost
-
Which cell is allocated first when using the north-west corner method?
- Lowest cost cell
- Highest cost cell
- Top-left cell
- Bottom-right cell
-
In the north-west corner method, what limits the allocation quantity in a selected cell?
- Number of rows
- Row index value
- Remaining supply or demand
- Total matrix cost
-
How many occupied cells indicate a degenerate solution in an m by n transportation problem?
- Equal to m times n
- Exactly m + n - 1
- Greater than m + n - 1
- Fewer than m + n - 1
-
How many occupied cells does a non-degenerate basic solution of a 3 by 4 transportation problem need?
- 5
- 7
- 12
- 6
-
In the stepping-stone method, what does the improvement index of an unoccupied cell measure?
- Maximum supply limit
- Minimum demand bound
- Cost change per unit
- Total routing flow
-
What starting value is conventionally assigned to row variable R1 in shadow cost calculations?
- One
- Total supply
- Zero
- Minus one
-
The north-west corner solution for supplies A = 20, B = 30, demands P = 10, Q = 25, R = 15, with costs A: 4, 6, 8 and B: 5, 3, 7, allocates AP = 10, AQ = 10, BQ = 15 and BR = 15. What is its total cost?
- 260
- 230
- 240
- 250
-
A two by three transportation table has four occupied cells. Is it degenerate?
- Degenerate solution
- Unbounded solution
- Infeasible solution
- Not degenerate
-
For the same north-west corner solution with R_A = 0, what is the shadow cost R_B for row B?
- 5
- 3
- -5
- -3
-
In the same solution, what is the improvement index of unoccupied cell A-R (cost 8)?
- -2
- 0
- -10
- 2
-
Which cell should enter the solution next in that solution, if the aim is to reduce cost?
- Row B, column Q
- Row A, column R
- Row A, column P
- Row B, column P
-
Using the stepping-stone loop for cell A-R, what is the largest amount θ that can be moved around the loop?
- 15
- 20
- 10
- 5
-
After one improvement from A-R with θ = 10, which cell leaves the solution?
- Row A, column P
- Row A, column Q
- Row B, column Q
- Row B, column R
-
What is the total cost after the improvement from A-R with θ = 10?
- 230
- 260
- 240
- 220
-
A three by three transportation table has only four occupied cells. What must be added?
- Negative cost
- Dummy destination
- Zero-valued allocation
- Artificial variable
-
Why must the stepping-stone loop alternate between plus and minus cells?
- Maximise total flow
- Keep totals unchanged
- Satisfy cell demands
- Minimise unit cost
-
Which unoccupied cell enters the solution in a minimising transportation problem?
- Most negative index
- Largest positive index
- Zero improvement index
- Smallest cost coefficient
-
After the single improvement (AP = 10, AR = 10, BQ = 25, BR = 5), what is the improvement index of cell A-Q?
- 0
- -2
- 4
- 2
-
What indicates that a transportation solution is optimal for minimisation?
- All indices zero
- Non-negative indices
- Negative allocation values
- All costs positive
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