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**austin81****Member**- Registered: 2005-03-21
- Posts: 39

Please LI'll like to know the meaninings of these terms as applied in Linear Programming:

Objective function, Contraints, feasible solution, feasible region, optimal solution, raetest and Least values.

Thanks

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**austin81****Member**- Registered: 2005-03-21
- Posts: 39

Thanks ery much> I just discovered there's more in Linear Programming than I thought of.

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**336337****Guest**

I can't put this problem into a model. I hope that there is someone who has some time for me.

Company X produces the product A. Before A is finished, there're 3 processus. Proces 1, from the raw material to an intermediate product B. Proces 2, from intermdiate B to intermediate product C. And proces 3, from intermediate C to the final product A.

The maximum demand for product A in the next 4 weeks is:

week 1: 375

week 2: 570

week 3: 750

week 4: 480

Product A is sold at 350/unit

Company X can employ their labourers on a weekly contract at 250/week. Each labourer works 40 hours.

It takes 0.5 hour for a labourer to complete process 1 for 1 unit, 1 hour to complete process 2 for 1 unit, and 0.25 hour to complete process 3.

The company can employ maximum 12 people. And a labourer needs to do the same process a whole week, so it can't work on process one on monday and the next day continue with process 2.

The intermediate and final products can aslo be stored, the cost for storage is 40 a week for product A, 15 a week for product B and 20 a weel for product C. At the end of week 4 all storage must be sold.

I need to maximise the profit but I can't find it. It would be great if someone could help me.

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

I'm not impressed that the laborers can't switch places each day because constraints

employers put like that simply induce things like carpal tunnel and other

repetitive injuries. Variety is a good thing.

**igloo** **myrtilles** **fourmis**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

The question isn't so hard but there are a lot of things to consider.

IPBLE: Increasing Performance By Lowering Expectations.

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**mathsyperson****Moderator**- Registered: 2005-06-22
- Posts: 4,900

I love your festive avatar, krazzi!

This question looks like it needs decision maths, which I don't do. I have a textbook lying around though, so if no one else manages to do this then I'll teach myself how to do this type of problem and then solve yours for you.

Why did the vector cross the road?

It wanted to be normal.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

I am seeing my avatar whitout any hat!

I can't understand the problem at all. But i'll try:

So if x is number of workers that are working only I process, y-second, z-third, then

x+y+z<=12

For 4 weeks thet must be producted equal mass from all the products,so

4(40.0,5.x)=4(40.1.y)=4(40.0,25.z)

∴

x=2z

y=4z

But x,y and z are natural, so 7z<=12; and z should be 1.

Hm, interesting. Here we don't have to use operational theory.

Is my answer correct?

{2,4,1}

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

No ,sorry. I haven't saw that we can sell every week.

IPBLE: Increasing Performance By Lowering Expectations.

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**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Cost of raw material ordering quantities is missing from the given data.

Assume if we maximize # units sold, we maximize profit, since we aren't

given raw material price or other overhead costs. Storage costs is not

specified if it is per unit or for all of each type. Also I am assuming

that no B's or C's are made, so you have to wait 1.5 hours before the

last step can be started in week 1, so this affects week 1 a little.

The results are guesswork below;I have no equations or methods:

WEEK 1:

making B 4men 160/.5=320units

making C 7men 312/1=312units

making A 1men 38.5/.25=154units

week 1 sold 154, 8 B's left, 158 C's left

WEEK 2:

making B 4men 320made + 8 units leftover = 328 units

making C 6men 240made + 158 units leftover = 398 units

making A 2men 320 units

week 2 sold 320, 88 B's left, 78 C's left

WEEK 3:

making B 4men 320made + 88 units leftover = 408 units

making C 6men 240made + 78 units leftover = 320 units

making A 2men 320 units

week 3 sold 320, 168 B's left, 0 C's left

WEEK 4: (ten workers on last week)

making B 1man 80made + 168 leftover = 248 units

making C 7men 248made with some wasted time

making A 2men 248made with much wasted time

week 4 sold 248, no product left over.

This analysis was just guesswork, so it's better than nothing.

Hope it helps.

**igloo** **myrtilles** **fourmis**

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