Math Is Fun Forum

the1

Guest

linear programming

Hi, I really need some help on this! I've been working on the problem for about a week and I am not 100% sure that I found the right answer! Here's what I need to do:
1. Max L(x1,x2,x3,x4)=32.5x1+60x2+50x3+225x4

6 * x1 + 12 * x2 + 9 * x3 + 30 * x4 <= 9000
15 * x1 + 12 * x2 + 9 * x3 + 144 * x4 <= 30000
4 * x1 + 4 * x2 + 16 * x3 + 120 * x4 <= 16000

x1>=0 x2>=10 x3>=30 x4>=0

I have to solve this, but as you see x2>=10 and x3>=30, so here's what I did:
Min L(x1,x2,x3,x4)=32.5x1+60x2+50x3+225x4

6 * x1 + 12 * x2 + 9 * x3 + 30 * x4 <= 9000
15 * x1 + 12 * x2 + 9 * x3 + 144 * x4 <= 30000
4 * x1 + 4 * x2 + 16 * x3 + 120 * x4 <= 16000
X5-X2<=-10 -X5+X2>=10
X6-X3<=-30 -X6+X3>=30

x1>=0 x2>=0 x3>=0 x4>=0 x5>=0 x6>=0
I solved it and it looks fine!
2. Now I need to write the normal form and solve it:
Min L(x1,x2,x3,x4)=32.5x1+60x2+50x3+225x4

6 * x1 + 12 * x2 + 9 * x3 + 30 * x4+x7 = 9000
15 * x1 + 12 * x2 + 9 * x3 + 144 * x4+x8 = 30000
4 * x1 + 4 * x2 + 16 * x3 + 120 * x4+x9 = 16000
-X5+X2=10
-X6+X3=30

x1>=0 x2>=0 x3>=0 x4>=0 x5>=0 x6>=0
I did it and it looks OK
3. From the normal form I have to write the dual form and solve it. here's what I did:
6*y1+15*y2+4*y3>=32,5
12*y1+12*y2+4*y3+y4>=60
9*y1+9*y2+16*y3+y5>=50
30*y1+144*y2+120*y3<=255
-y4<=0
-y5=0
y1=0
y2=0
y3=0
y4=0
y5=0

What I get as answer from Excel Solver is: Solver could not find feasible solution! Is it possible? Could someone find any errors in my efforts? Am I right at all? I am completely lost now! Thanks for the help!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: linear programming

Hi the1;

Welcome to the forum!

One question and bear with me on these. What were your two answers for 1)?

Your syntax might be a little bit off.

6*y1+15*y2+4*y3>=32,5

Is that a 32.5 on the end there?

Please show exactly what you put into your solver. Exactly what you were trying to solve.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

the1

Guest

Re: linear programming

Yes, I mean 32.5. The two answers for 1. are: x2=10 and x3=30

                                L(X1,X2)     Formula1
X1    X2    X3    X4    X5    X6            Min    2100
0    10    30    0    0    0               
C1    C2    C3    C4    C5    C6               
32,5    60    50    225    0    0               
Matrice                                Value    Formula
6    12    9    30    0    0        <=    9000    390
15    12    9    144    0    0        <=    30000    390
4    4    16    120    0    0        <=    16000    520
0    1    0    0    -1    0        >=    10    10
0    0    1    0    0    -1        >=    30    30

Here's how it looks Solver(Excel). The cell under Formula1 actually is the target cell. Under Formula cell I use sumproduct to calculate same for the cell under Formula1.  I solve the same way the others.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: linear programming

Okay to put that 32,5 in that cell like that? I do not use excel so...

May I have the full answers for both the max and the min for 1)?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

the1

Guest

Re: linear programming

That's the whole thing that I use and together with the solution!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: linear programming

What I mean is there are the variables x1,x2,x3,x4,x5,x6. What did you get for the maximum value and the minimum value and for all the variables for 1)

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

the1

Guest

Re: linear programming

when I solve it as Max x1=1000, x2=0, x3=0, x4=100 but I have for x2>=10 and x3>=30, so I need to do something and then I put some extra constraints and I solve it as MIN. Here are the answers:  x1=0, x2=10, x3=30,x4=0,x5=0, x6=0

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: linear programming

Hi;

Here are the answers:  x1=0, x2=10, x3=30,x4=0,x5=0, x6=0

That is not the whole thing. The max and the min are what is important.

I am getting:

x1 = 948
x2 = 10
x3 = 30
x4 = 97.4

Maximum = 54825

For the minimum:

x2 = 20
x3 = 60

The rest are 0, with minimum of 4200

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

the1

Guest

Re: linear programming

Thanks! I found my mistake! I've been solving this for days...I tried also substitution! And I just figured it out! I should not use any extra constraints! I just use x2>=10 x3=>30 for limitation! Thanks you were very helpful!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: linear programming

Hi the1;

I did not do a whole lot but I had fun talking to you. Hope to see you again.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.