# Real World Algebra Problem!

• Dec 17th 2011, 09:42 AM
Phugoid
Real World Algebra Problem!
Hi there.

Bit of a real-world problem here.

My current electricity supplier charges me 195p per week as a fixed-cost service charge. They charge me 15.91p per kWh at high-rate hours and 8.75p per kWh at low-rate hours.

Or you could say that my weekly cost is:

$\displaystyle \displaystyle \text{COST}_{\text{Supplier1}} = 15.91 \times \text{USEAGE}_{\text{LR}} + 8.75 \times \text{USEAGE}_{\text{HR}} + 195$

A potential electricity supplier would charge me 175p per week as a fixed-cost service charge. They would charge me 11.654p per kWh at all times, with no low rate hours.

Or you could say:

$\displaystyle \displaystyle \text{COST}_{\text{Supplier2}} = 11.654\times \text{USEAGE}_{\text{LR}} + 11.654\times \text{USEAGE}_{\text{HR}} + 175$

So my problem is... which supplier would give me the best deal?

Now, I don't know precisely how much energy I use in the low rate hours and how much I use in the high rate hours, but there's only really 4 things that we use specifically in the low rate hours (washing machine, dishwasher, tumbledryer, heating, etc) under supplier 1, and the rest (laptops, lights, TV, phone chagers, hairdryer, hair straighteners, cooker, fridge, freezer, water heating, alarm clock) is in the high rate hours, so I have something of an inkling which suggests that most of hour useage is in the high-rate hours.

However, what I'd like to know is precisely which ratios of $\displaystyle \frac{\text{USEAGE}_{\text{LR}}} {\text{USEAGE}_{\text{HR}} }$ give $\displaystyle \text{COST}_{\text{Supplier1}} \geq \text{COST}_{\text{Supplier2}}$, so I can more accurately estimate whether I'm better off switching or staying. :D

Any ideas?
• Dec 17th 2011, 01:31 PM
ILikeSerena
Re: Real World Algebra Problem!
Hi Phugoid! :)

Your ratio would not work very well since you have fixed offsets.
Unless of course your usage is so high that the offsets become negligible.

What I'd do is to calculate the difference in cost of the suppliers COST1 - COST2.
If it is positive supplier 2 is cheaper.

The resulting equation is:
COST1 - COST2 = 20 - 2.904 H + 4.256 L

The result comes out like this Wolfram plot.
The break even points are the points on the line in this plot.

If you really want a ratio, we would need to neglect the 20 term and set the difference in cost to zero.
So:
0 = - 2.904 H + 4.256 L
meaning:
L/H = 2.904 / 4.256
• Dec 17th 2011, 07:46 PM
Wilmer
Re: Real World Algebra Problem!
Let h = total hours; assume 75% = high

[1] 195 + 15.91(.75h) + 8.75(.25h) : [2] 175 + 11.654h

[1] 195 + 14.12h : [2] 175 + 11.654h

2nd proposal clearly much cheaper.