Can Excel graph the following equation? If so, how would I go about
doing this?
z=((.03y+.15)x+150.75)x
Think how you'd go about doing it with a calculator, a pen and a sheet
of grid paper. You'd choose some values of x and y, calculate z, and
mark the calculated z positions on the grid paper. Then you'd join the
points up with lines.
For efficiency, you might do all the calculating first, writing the
numbers in a table, then use the table values to plot the points on the
grid paper.
Doing it with a spreadsheet is just the same process. Make the table
using the spreadsheet cells, then use the table to make the chart.
The formula for your z column in this case will be
= ( ( 0.03*Y1 + 0.15 ) * X1 + 150.75 ) * X1
where "X1" and "Y1" are cell references.
Basically, I have a different cost formula for each of 4 different
suppliers. All have a different twist, but all contain the same 2
variables (x and y) and result in a single value (z). I need to be
able to graph the equations so that I can visually present where the
costs intersect.
Four cost formulae means four z columns, which is simple enough. The
search for an intersect between two lines is trickier, because each of
your equations is one equation in three unknowns (x, y and z), making
them equations of *surfaces*, not lines. That means they won't have
intersection points, but intersection lines.
If you had only two different suppliers to compare, I'd suggest graphing
four lines of z against x for four values of y, for each of two
suppliers, and joining the four points of intersection to show the line.
But the intersection line will have variable y, making it hard to
interpret on a graph of z against x. Plus, you'll need six different
graphs to compare every pairwise combination of four suppliers.
Have you considered using algebra instead of graphing for the first
step? Then you can plot the six lines as x against y on a single graph,
neglecting z, which I assume you don't care so much about except that
it's what's equal at the crossover point. If you really care about z,
you can label the points, but then I'd use six graphs, not one, to avoid
cluttering the space up.