APM261: Quiz 5. Advanced Simplex Methods

Quiz 5. Advanced Simplex Methods

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Each question has only one correct answer. The results of the quiz do not affect the final marks.

1: An artificial variable ...

Represent difference between the right-hand side and the left-hand side of a technological constraint
Is added to a technologinal constraint for finding an initial basic feasible solution
Equals zero identically
Is a dual variable associated to a constraint

2: The following tableau is final for Phase I of the two-phase method:

	`x₁`	`x₂`	`y₁`	`y₂`
`x₁`	1	-1	0	-3	3
`y₁`	0	2	1	4	0
`z'`	0	-3	0	-5	0

Here x₁ and x₂ are decision variables, y₁ and y₂ are artificial variables, and

z' = y₁ + y₂

The original objective function in minimization problem is

z = 2 x₁ - 3 x₂

What is the next step of the two-phase method?

Terminate the two-phase method: the problem has no feasible solutions
Delete the columns for x₂ and y₂ and go to Phase II
Delete the columns for y₁ and y₂ and go to Phase II
Delete the column for y₂ and go to Phase II

3: Find minimum of

z = 3 x₁ + 4 x₂ - 5 x₃ - 6 x₄

subject to the constraints:

x₁ + 2 x₂ + 2 x₃ = 4
x₁ + 4 x₂ + 3 x₄ = 6

where all variables x₁, x₂, x₃, and x₄ are non-negative. Suggest an suitable initial solution to start the simplex method.

Select x₃ and x₄ as basic variables and write the system as the simplex tableau
Select x₁ and x₂ as basic variables and write the system as the simplex tableau
Select x₃ and x₄ as basic variables, eliminate x₃ and x₄ from the objective function, and write the system as the simplex tableau
Add artificial variables y₁ and y₂ to the constraints and start the simplex method with the objective function z

4: The objective function for the minimization problem is

z = x₁ - x₂

Find the missing entry A in the simplex tableau:

	`x₁`	`x₂`	`x₃`	`x₄`
`x₁`	1	0	`A`	2	1
`x₂`	0	1	-1	1	3
`z`	0	0	-1	1	-2

A = -2
A = -1
A = 1
A = 2

5: The objective function for the minimization problem is

z = 4 x₁ + 5 x₂ - x₄ - 3 x₅

Find the missing entry B in the simplex tableau:

	`x₁`	`x₂`	`x₃`	`x₄`	`x₅`
`x₂`	3	1	2	4	0	1
`x₅`	-1	0	3	2	1	3
`z`	`B`	0	-1	-15	0	-4

B = 12
B = 14
B = 18
B = 22

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