1. Number of Variables 2. Transportation Example 1

 On 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. 2

 A 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? 3

 V  Stock: 40  B A  Needs: 30 Needs: 25  W  Stock: 45 4 $5 $3 $9 $6

 The 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. 5

 V  Stock: 40  B A  Needs: 30 Needs: 25  W  Stock: 45 6 $5 $3 $9 $6 x y 25 - y 30 - x

 Warehouse W:  x + y < 45  Warehouse V:  (30 - x ) + (25 - y ) < 40  Nonnegative restrictions:  0 0 7

 x + y < 45  x + y > 15  x < 30  y < 25  0 < x  0 < y 8

 The cost of transporting the TVs is to be minimized.  [cost] = 9 x + 6 y + 5(30 - x ) + 3(25 - y )  [cost] = 4x+3y+225 9

10 Step y < 45 - x 2. y > 15 - x 3. x < 30, 0 < x 4. y < 25, 0 < y

11 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)

VertexCost = 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=

V Stock: 40 B A Needs: 30 Needs: 25 W Stock: $5 $3 $9 $

 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. 14