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Answer: TRUE
Diff: 2 Page Ref: Ch 2 review
Main Heading: Formulation and Computer Solution Key words: formulation, standard form
5) Fractional relationships between variables are not permitted in the standard form of a linear program.
Answer: TRUE
Diff: 2 Page Ref: Ch 2 review
Main Heading: Formulation and Computer Solution Key words: formulation, standard form
1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
6) A constraint for a linear programming problem can never have a zero as its right-hand-side value.
Answer: FALSE
Diff: 2 Page Ref: Ch 2 review
Main Heading: Formulation and Computer Solution Key words: formulation, standard form
7) The right hand side of constraints cannot be negative. Answer: FALSE
Diff: 2 Page Ref: Ch 2 review
Main Heading: Formulation and Computer Solution Key words: formulation
8) A systematic approach to model formulation is to first define decision variables. Answer: TRUE
Diff: 1 Page Ref: Ch 2 …show more content…
Answer: X12 + X22 + X32 ≤ 400
Diff: 2 Page Ref: 127-131
Main Heading: A Transportation Example
Key words: transportation problem, supply constraint
9
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
50) Small motors for garden equipment is produced at 4 manufacturing facilities and needs to be shipped to 3 plants that produce different garden items (lawn mowers, rototillers, leaf blowers). The company wants to minimize the cost of transporting items between the facilities, taking into account the demand at the 3 different plants, and the supply at each manufacturing site. The table below shows the cost to ship one unit between each manufacturing facility and each plant, as well as the demand at each plant and the supply at each manufacturing facility.
Write the formulation for this problem.
Answer: MIN Z = 4x1A + 4.5x1B + 3.2x1C + 3.5x2A + 3x2B +4x2C + 4x3A + 3.5x3B + 4.25x3C
s.t.
x1A + x1B +x1C = 200 x2A + x2B +x2C = 200 x3A + x3B +x3C = 300 x1A + x2A +x3A = 250 x1B + x2B +x3B = 150 x1C + x2C +x3C = 200
Diff: 2 Page Ref: 127-131
Main Heading: A Transportation Example
Key words: computer solution,
Diff: 2 Page Ref: Ch 2 review
Main Heading: Formulation and Computer Solution Key words: formulation, standard form
5) Fractional relationships between variables are not permitted in the standard form of a linear program.
Answer: TRUE
Diff: 2 Page Ref: Ch 2 review
Main Heading: Formulation and Computer Solution Key words: formulation, standard form
1
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
6) A constraint for a linear programming problem can never have a zero as its right-hand-side value.
Answer: FALSE
Diff: 2 Page Ref: Ch 2 review
Main Heading: Formulation and Computer Solution Key words: formulation, standard form
7) The right hand side of constraints cannot be negative. Answer: FALSE
Diff: 2 Page Ref: Ch 2 review
Main Heading: Formulation and Computer Solution Key words: formulation
8) A systematic approach to model formulation is to first define decision variables. Answer: TRUE
Diff: 1 Page Ref: Ch 2 …show more content…
Answer: X12 + X22 + X32 ≤ 400
Diff: 2 Page Ref: 127-131
Main Heading: A Transportation Example
Key words: transportation problem, supply constraint
9
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
50) Small motors for garden equipment is produced at 4 manufacturing facilities and needs to be shipped to 3 plants that produce different garden items (lawn mowers, rototillers, leaf blowers). The company wants to minimize the cost of transporting items between the facilities, taking into account the demand at the 3 different plants, and the supply at each manufacturing site. The table below shows the cost to ship one unit between each manufacturing facility and each plant, as well as the demand at each plant and the supply at each manufacturing facility.
Write the formulation for this problem.
Answer: MIN Z = 4x1A + 4.5x1B + 3.2x1C + 3.5x2A + 3x2B +4x2C + 4x3A + 3.5x3B + 4.25x3C
s.t.
x1A + x1B +x1C = 200 x2A + x2B +x2C = 200 x3A + x3B +x3C = 300 x1A + x2A +x3A = 250 x1B + x2B +x3B = 150 x1C + x2C +x3C = 200
Diff: 2 Page Ref: 127-131
Main Heading: A Transportation Example
Key words: computer solution,