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solniwko [45]
3 years ago
6

Which of the following is TRUE about the​ northwest-corner rule? A. All allocations are based on lowest cost B. All demand and s

upply constraints are​ satisfied, but cost is ignored C. All demand and supply constraints are satisfied while minimizing cost D. All demand and supply constraints are ignored while minimizing cost

Mathematics
1 answer:
Paul [167]3 years ago
3 0

Answer:

B. All demand and supply constraints are​ satisfied, but cost is ignored

Step-by-step explanation:

The​ northwest-corner rule is a method to compute the initial feasible solution of a transportation problem.

In this method you start to supply the demand with the northwest-corner option without taking into account the related cost. You assign the whole  production of the first source to first demand, if that is more than necessary, then you assign the excess to the second demand, and so on; but if is not enough, then you assign the whole production of the next source to unsatisfied demand. Repeat this procedure until all demands are satisfied.

In the picture attached, an example is given. The table shows the unit transportation cost and in the circles the feasible solution obtained with northwest-corner rule.

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Answer:

  • <em>convert </em>the mixed fractions to improper fractions (where the numerator is greater than or equal to the denominator): multiply the whole number part by the fraction's denominator, add that to the numerator, write the result on top of the denominator.
  • if the denominators are not the same, work out the common denominator and <em>rewrite </em>the fractions with the same denominators
  • subtract by subtracting the numerators and writing the result over the denominator
  • convert back to mixed fractions by dividing the numerator by the denominator, write down the whole number answer, write down the remainder above the denominator.

Example

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convert to improper fractions:

\dfrac{3 \times 3+2}{3}-\dfrac{1 \times 5+4}{5}=\dfrac{11}{3}-\dfrac{9}{5}

common denominator = 3 × 5 = 15, so:

\dfrac{11}{3}-\dfrac{9}{5}=\dfrac{11\times 5}{3\times 5}-\dfrac{9\times 3}{5\times 3}=\dfrac{55}{15}-\dfrac{27}{15}

subtract:

\dfrac{55}{15}-\dfrac{27}{15}=\dfrac{55-27}{15}=\dfrac{28}{15}

convert back to mixed fractions:

28 \div 15=1 \textsf{ remainder }13=1 \frac{13}{15}

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