# linear programing graphical solution problem

• Aug 30th 2008, 06:46 AM
jack steven
linear programing graphical solution problem
hey guys...
i have a question and not able to find the equations to solve the problem... please help... its urgent....

Solve the problem graphically to find the maximum value of the profit and the optimal values of the decision variables.
For one of its clients, Propaganda Ltd. must run a series of advertisements on television for one week, for which there are two tariffs: off-peak and peak (prime-time). There is a budget of £100,000 available for the advertising.

Off-peak advertising costs £750 per minute and generates a revenue of £1,200 per minute.
Prime-time advertising costs £5,000 per minute and generates a revenue of £7,000 per minute.

The rules of the broadcaster mean that the company must buy at least one hour of off-peak advertising.

What mix of daily prime-time and off-peak advertising should Propaganda Ltd. buy in order to maximise profit, and what is the maximum profit? (NB: Profit = Revenue – Cost)

Solve the problem graphically to find the maximum value of the profit and the optimal values of the decision variables.
• Aug 30th 2008, 10:04 AM
TKHunny
These really are not as hard as they look. I do admit their appearance is daunting. There is an awful lot to say, but I believe firmly that the hardest part is simply keeping everything organized.

Let's just get some equations on paper. Before we can do Equations, we need definitions.

Let's say
OP is the number of Off-Peak Advertising minutes and
PT is the number of Prime-Time Advertising minutes.

This gives right away, a formula for the profit so generated:

OP(1200-750) + PT(7000-5000) = 450*OP + 2000*FP = TotalProfit

It should be obvious that the more minutes of FP we can get, the better off we'll be.

We have only one constraint: 100,000

This gives the formula

OP*750 + PT*5000 = 100000 <== You need to stare at this until you see where it came from. This is the line we cannot cross.

Just a little more housekeeping. I hope it's obvious that OP = 0 and PT = 0 are also lines we cannot cross. we cannot purchase -3 minutes of advertising!

After all that, we need only choose which variable, OP or PT will be on the horizontal axis. The other will be on the vertical axis. We are graphing, so you knew we would need coordiante axes, right?

I picked OP as the vertical axis. It does not matter. I just picked one.

Solving everything for OP, then, gives:

Constraint:

OP*750 + PT*5000 = 100000 ==> OP = (-20/3)*PT + (40/3)

Profit Equation:

450*OP + 2000*FP = TotalProfit ==> OP = (-40/9)*PT + (TotalProfit/450)

That's it. You should be able to graph the constraint equation. I put it in Slope-intercept form for you.

The trickier one is the Profit Equation, since it needs the Total Profit and that's what we are trying to find. Not to worry! We can just look at both solutions and see which one we like.

"Both"? That's right. Only intersections matter. The intersections in this problem are on the coordinate axes.

One intersection point is at (PT,OP) = (0,40/3). We knew this from the y-intercept when we graphed it. The other intersection is at (PT,OP) = (2,0). Let's just make the profit equation run through each of those two points.

(PT,OP) = (2,0)
0 = (-40/9)*2 + (TotalProfit/450) ==> TotalProfit = 4000
This gives the equation OP = (-40/9)*PT + (4000/450)

(PT,OP) = (0,40/3)
40/3 = (-40/9)*0 + (TotalProfit/450) ==> TotalProfit = 6000
This gives the equation OP = (-40/9)*PT + (6000/450)

Graph those two and, again, stare at it until it soaks in. Notice how the SLOPE of these two lines is the same. Think about what happens as you move this line up or down. You should be able to think through why 6000 is the maximum solution.

Note: More constraints lead to more of these important intersections. It isn't any harder, mathematically, but there is more going on and more stuff that needs to be tracked and kept organized.
• Aug 30th 2008, 02:29 PM
jack steven