3.3 Linear Programming III1. Number of Variables2. Transportation Example

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Number of VariablesOn the surface some problems may appear to have more than two variables. However, sometimes they can be translated into mathematical language so that only two variables are required.

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Example TransportationA TV dealer has stores in city A and B and warehouses in cities W and V. The cost of shipping a TV from W to A is $6, from V to A is $3, from W to B is $9 and from V to B is $5. Store in A orders 25 TV sets and store in B orders 30 sets. The W warehouse has a stock of 45 sets and V warehouse has 40. What is the most economical way to supply the two stores the requested TV sets?

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Step 1AV

Stock: 40

B ANeeds: 30 Needs: 25

WStock: 45

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$5 $3

$9 $6

Step 1BThe number of variables can be reduced by observing that what is not shipped from the warehouse in W must be shipped from the warehouse in V.Let x be the number of TVs shipped from the W warehouse to the store in B and y be the number of TVs shipped from W to A. Then 30 - x is going from V to B and 25 - y from V to A.

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Step 1A&BV

Stock: 40

B ANeeds: 30 Needs: 25

WStock: 45

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$5 $3

$9$6

x y

25 - y30 - x

Step 1CWarehouse W:

x + y < 45Warehouse V:

(30 - x) + (25 - y) < 40Nonnegative restrictions:

0 < x and 30 - x > 00 < y and 25 - y > 0

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Step 1C - Simplifiedx + y < 45

x + y > 15x < 30y < 250 < x0 < y

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Step 1DThe cost of transporting the TVs is to be minimized.

[cost] = 9x + 6y + 5(30 - x) + 3(25 - y)[cost] = 4x+3y+225

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Step 2

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2 3

3 4

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•1. y < 45 - x•2. y > 15 - x•3. x < 30, 0 < x•4. y < 25, 0 < y

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Step 3

•y < 45 - x•y > 15 - x•x < 30 •0 < x•y < 25 •0 < y (0,15)

(0,25)

(30,15)

(30,0)(15,0)

(20,25)

Step 4Vertex Cost = 4x+3y+225(0,15) Cost= 270(0,25) Cost= 300(20,25) Cost= 380(30,15) Cost= 390(30,0) Cost= 345(15,0) Cost= 285

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•V•Stock: 40

• B A•Needs: 30 Needs: 25

•W•Stock: 45

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$5 $3

$9$6

0 15

1030

Summary Section 3.3 Sometimes it is necessary to use algebra to reduce the number of variables. Once the number of variables is reduced to two, the steps for solving a linear programming problem are followed.

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