Graphical Solution Method

1
Example 3.2

• Graphical Solution Method

2
Background Information

• To illustrate the graphical approach, we will use
  a slightly different scaled down version of
  Monets product mix problem.
• Now there are only two frame types,1 and 2, and
  only two scarce resources, labor hours and metal.
• The algebraic model is given below
• max 2.25x1 2.60x2 (profit objective)
• subject to 2x1 x2 ? 4000 (labor constraint)
• x1 2x2 ? 5000 (metal constraint)
• x1, x2 ? 0 (nonnegativity constraint)

3
Background Information -- continued

• The objective implies that each type 1 frame
  contributes a profit of 2.25, whereas each type
  2 frame contributes a profit of 2.60.
• The first constraint is a labor hour constraint.
  There are 4000 hours available. Each type 1 frame
  requires 2 labor hours, and each type 2 frame
  requires 1 labor hour.
• Similarly, the second constraint is a metal
  constraint. There are 5000 ounces of metal
  available. Each type 1 frame requires 1 ounce of
  metal and each type 2 frame requires 2 ounces of
  metal.
• Find the optimal product mix graphically.

4
Solution

• The idea is to graph the constraints on a
  two-dimensional graph to see which points (x1,
  x2) satisfy all of the constraints. This set of
  points is labeled the feasible region. Then we
  see which point in the feasible region provides
  the largest profit.
• The graphical solution appears on the next slide.

5
Graphical Solution
6
Solution -- continued

• To produce the graph, we first locate the lines
  where the constraints hold as equalities.
• For example, the line for labor is 2x1 x2
  4000. The easiest way to graph this is to find
  the two points where it crosses the axes.
• Joining the points (0,4000) and (2000,0), we get
  the line where the labor constraint is satisfied
  exactly, that is, as an equality.
• All points below and to the left of this line are
  also feasible there are these are the points
  where less than the maximum number of 4000 labor
  hours are used.

7
Solution -- continued

• We indicate the feasible side of the line by the
  short arrows pointing down to the left from the
  labor constraint line.
• Similarly the metal constraint line crosses the
  axes at the points (0,2500) and (5000,0), so we
  join these two points to find the line where all
  5000 ounces of metal are used.
• Finally the points on or below both of these
  lines constitute the feasible region. These are
  the point below the heavy lines.

8
Solution -- continued

• You can think of the feasible region as all
  points on or inside the figure formed by four
  points (0,0), (0,2500), (2000,0), and the point
  where labor hour and metal constraint lines
  intersect.
• The next step is to bring profit into the
  picture. We do this by constructing isoprofit
  lines that is lines where total profit is a
  constant. Any such line can be written as 2.25x1
  2.60x2 P where P is a constant profit level.
  Solving for x2, we can put this equation in
  slope-intercept form x2 P/2.60 (2.25/2.60)x1

9
Solution -- continued

• This shows that any isoprofit line has slope
  2.25/2.60, and it crosses the vertical axis at
  the value P/2.60. Three of these isoprofit lines
  appear in the chart as dotted lines.
• Therefore, to maximize profit, we want to move
  the dotted line up and to the right until it just
  barely touches the feasible region.
• Graphically, we can see that the last feasible
  point it will touch is the point indicated in the
  figure, where the labor hour and metal constraint
  lines cross.

10
Solution -- continued

• We can then solve two equations in two unknowns
  to find the coordinates of this point. They are
  x1 1000 and x2 2000, with a corresponding
  profit of P 7450.
• Note that if the slope of the isoprofit lines
  were much steeper, the the optimal point would be
  (2000,0). On the other hand,m if the slope were
  mush less steep, the optimal point would be
  (0,2500).These statements make intuitive sense.
• If the isoprofit lines are steep, this is because
  the unit profit from frame type 1 is large
  relative to the unit profit from frame type 2.

11
Solution -- continued

• The crucial point, however, is that only three
  points can be optimal (2000,0), (0, 2500), or
  (1000, 2000), the three corner points other
  than (0,0) in the feasible region.
• The best of these depends on the relative slopes
  of the constraint lines and isoprofit lines in
  the graph.