# Linear profit and plotting graphs with results.

#### nihil404

##### New member
I am not sure how to interpret this problem and where to start to get my values to plot my graphs and get my x, y and etc values as there is too much going on. Can someone shed some light of how to do this problem?

The problem is:

You are to help a popular online VOD service maximise their profits by advising the company’s sale team on set ‘sales targets’ for “HD single screen” subscriptions and “SD single screen” subscriptions given that it is 2 times more profitable to sell SD single screen subscriptions than HD single screen ones. It must also be noted that the company must honour its existing HD single screen (5) and SD single screen (10) subscribers. Being an online company profits increase linearly in relation to the number of servers running. In this instance we will only look for the costings of running a single server. That being said, due to hardware limitations, a server can only stream 200 IO streams, where a HD stream consumes 4 IO streams and a SD stream only consumes 1 IO stream. In terms of bandwidth cost, we have a simple contract that provides the first £100 worth (per month) of bandwidth free, after which the cost for bandwidth significantly increases such that users do not exceed this free £100 limit. It is to be noted that on average the bandwidth consumption (per month) of a typical HD subscription never exceeds £1.50 with a similar case for SD subscription being £0.75. Finally - in addition to the above, it must also be noted that this company is a daughter company from which we lease all our content from our parent company. The costs of the leases are £1 (per month) for HD subscription and £0.75 (per month) for SD subscription. Our aim is not to exceed £80 of leasing cost.

I am not sure if is this how I solve this:

How do I find that the HD subscriptions is twice as profitable? And what the HD and SD existing customers have to do with the problem?

IO streams = 200

HD stream = 4 = x

SD stream = 1 = y

4x + 1y = 200

=y=200-4x

X intercepts (50,0) Y intercepts (0,200)

Bandwidth cost = 100

1.50 HD x 0.75 SD Y

1.5x + 0.75y = 100

y=2(3-x+200)/3 X intercepts (200/3) Y intercepts (0, 400/3)
Which I believe is related to the profitability X being twice as profitable than Y.

Lease cost =80

HD = 1 x

SD = 0.75y

1x + 0.75y = 80

y=4(-x+80)/3

X intercepts (80,0) Y intercepts (0,320/3)

Am I doing the correct calculations?

Many thanks in advance for any help.

#### ksdhart2

##### Senior Member
I really don't like this problem. There's so much going on here and it's tough to parse out what's important. It almost seems like the type of exercise that's maybe meant to prepare students for "the real world," in which they may have to deal with such a "word vomit" from their boss. I'll step through it one part at a time and give my interpretation of what it means.

You are to help a popular online VOD service maximise their profits by advising the company’s sale team on set ‘sales targets’ for “HD single screen” subscriptions and “SD single screen” subscriptions given that it is 2 times more profitable to sell SD single screen subscriptions than HD single screen ones.
The problem starts off fine, informing you that your task is to maximize profits. We're told that each SD subscription is twice as profitable as an HD subscription. So if we let $$x$$ represent the number of HD subscriptions and $$y$$ represent the number of SD subscriptions (which it seems like you did in your work, though you also state "HD stream =4 = x" which confuses the matter even more, by implying that $$x$$ simultaneously stands for the number of HD subscriptions and is always equal to 4) then the core task at hand is to maximize $$x + 2y$$, subject to some constraints...

It must also be noted that the company must honour its existing HD single screen (5) and SD single screen (10) subscribers.
This is where the problem first starts to go "off the rails," but what I get out of this is it gives our first two constraints of $$x \ge 5$$ and $$y \ge 10$$

Being an online company profits increase linearly in relation to the number of servers running. In this instance we will only look for the costings of running a single server. That being said, due to hardware limitations, a server can only stream 200 IO streams, where a HD stream consumes 4 IO streams and a SD stream only consumes 1 IO stream.
Okay, so there's a bit of fluff here, but the core aspect I pull out of this passage is establishing another constraint. It makes sense that the company would want to use all of it's available streams, so we have the constraint $$4x + y = 200$$, from which you can obtain $$y = 200 - 4x$$.

In terms of bandwidth cost, we have a simple contract that provides the first £100 worth (per month) of bandwidth free, after which the cost for bandwidth significantly increases such that users do not exceed this free £100 limit. It is to be noted that on average the bandwidth consumption (per month) of a typical HD subscription never exceeds £1.50 with a similar case for SD subscription being £0.75.
This passage is the most confusing. Are we to just assume that users never exceed £100 in bandwidth? What happens if they do? Further, who is the "user" in this context? Is it the people who are subscribing to this company's service? Or is it the company itself?

If we assume that "the user" is the company whose profits we're trying to maximize and also assume that £100 is a hard and fast limit, it seems to give another constraint of $$1.5x + 0.75y \le 100$$

Finally - in addition to the above, it must also be noted that this company is a daughter company from which we lease all our content from our parent company. The costs of the leases are £1 (per month) for HD subscription and £0.75 (per month) for SD subscription. Our aim is not to exceed £80 of leasing cost.
In terms of substantive information, this is nearly identical to the previous passage and gives us our final constraint of $$x + 0.75y \le 80$$.

So, basically, what the problem really boils down to, after stripping away all of the extraneous and confusing information, is to maximize $$x + 2y$$, subject to the constraints:

$$4x + y = 200$$
$$1.5x + 0.75y \le 100$$
$$x + 0.75y \le 80$$
$$x \ge 5$$
$$y \ge 10$$

That hopefully shouldn't be too bad.

• nihil404

#### nihil404

##### New member
This is pretty much what I did as well after watching a couple of video tutorials:
To rephrase the question: given that x is the number of HD subscriptions sold and y is the number of SD subscriptions sold, maximize x+2y, considering the following;

x>=5, y>=10 (company honouring contract)
4x+y=200 (IO server streaming)
x+0.75y<80 (leasing cost)
1.5x+0.75y<100 (bandwidth)

Now, how do I maximize the x+2y and apply into the results??

#### ksdhart2

##### Senior Member
There's a couple of different methods you can use, but I prefer an algebraic one. You've correctly identified that $$4x + y = 200 \implies y = 200 - 4x$$, so we can freely substitute that into the expression you want to maximize, giving $$x + 2y = x + 2(200 - 4x) = \text{???}$$.

From there, think about what happens to $$x + 2(200 - 4x)$$ as $$x$$ increases and what happens as $$x$$ decreases. That should give you a good intuition about what the desired strategy for maximization is, and solving the inequalities for $$x$$ will allow you to hone in on a specific numeric answer.

#### Subhotosh Khan

##### Super Moderator
Staff member
This is pretty much what I did as well after watching a couple of video tutorials:
To rephrase the question: given that x is the number of HD subscriptions sold and y is the number of SD subscriptions sold, maximize x+2y, considering the following;

x>=5, y>=10 (company honouring contract)
4x+y=200 (IO server streaming)
x+0.75y<80 (leasing cost)
1.5x+0.75y<100 (bandwidth)

Now, how do I maximize the x+2y and apply into the results??
When I did this type of problem (many moons ago) - I graphed the inequalities on a "graph paper" and established "critical points" (generally the intersection of the domains established by the inequalities) and calculates the objective function (the function to be optimized) at those "critical" points.