Homework 1b

[Sidra]

Any insight on how folks plotted the points, 2-5 on the graph for
portion f of the homework?

Thanks,

Sidra

[Sudha Venkatesh]

Iam still working on the graph. What was optimal solution you derived for 1 -b?

[Subramaniam Narayanan]

Here is my graph... The optimal solution was 3,250.

[Sudha Venkatesh]

Optimal solution at x1= 130; x4=10,x2=x3=0

max e= 3250.

any other answers?

[Elliott Le]

I have the same plot. My answers below:

1. X3=260,  X4=140,  e=0
2. X2=86.67,  X4=-33.33

3. X2=70,  X3=50,  e=1750
4. X1=130,  X4=10,  e=3250

5. X1=140,  X3=-20

6. X1=100,  X2=20,  e=2700

Elliott Le

--
Elliott Le

[Darren Ng]

[Darren Ng]

I got the same answers as Elliot.  The optimal answer should be the same as the optimal solution to HW 1a (#4 below)

On Fri, Sep 30, 2011 at 8:52 PM, Elliott Le <ellio...@gmail.com> wrote:

[Rachna Patel]

question:

3. X2=70,  X3=50,  e=1750

Shouldn't e=70*15=1050?

[Rachna Patel]

one more question:

6. X1=100,  X2=20,  e=2700

isn't it (25*100)+ (20*15) = e = 2800?

[Elliott Le]

Nice catch.

Elliott Le

--
Elliott Le

[Rachna Patel]

I'm on the second problem. Is this what you guys got?

a.

Cj  25  15   0   0
Cb X1  X2 X3  X4  Bi
0   2     3    1    0     260 ---- 260/2=130 ***
0   1     2    0    1     140 ---- 140/1 = 140
Zj   0   0     0     0   
c-z 25 15   0    0
      ***

e=0

Pivot column = x1, pivot row = bi=260

c.
next iteration - x2, x3 = 0

After this im getting a bit lost....anyone know how to proceed from here?

[Rachna Patel]

anyone get this for the second part

               Cj  25  15   0   0

basis      Cb X1  X2 X3  X4  Bi
  x4           0   0   .5   -.5    1     140
   x1          25  1  1.5   .5    0     130
              Zj   25   37.5 12.5 0   
              c-z  0    -22.5 -12.5 0

 
      e=25*130=3250

[Sidra Haider]

I got something slightly similar but also a little different:

               Cj  25     15      0     0 

basis      Cb X1      X2    X3    X4  Bi

              25   1       1.5    0.5    0    130
               0    0       0.5   -0.5    1    10
              Zj   2.5     37.5  12.5  0    
              c-z  22.5  22.5   12.5  0

 
      e=25*130=3250

--
Sidra Haider
sidra....@gmail.com

[Almitra Sharma]

I am getting the following matrix which is similar to Sidra's and
Rachna's answer. But my last row is negative.

Cj 25 15 0 0
basis Cb X1 X2 X3 X4 Bi

X1 25 1 1.5 0.5 0 130
X4 0 0 0.5 -0.5 1 10
Zj 25 37.5 12.5 0

c-z 0 -22.5 -12.5 0

e=25*130+10*0=3250
--Almitra

[Gabriela Tchaga]

I got the same thing as almitra.

gabi

--
Gabriela G. Tchaga
GGTc...@gmail.com
510-386-5619

[Victoria Cheng]

I got the same as almitra!

On Sun, Oct 2, 2011 at 7:14 PM, Almitra Sharma <karnik....@gmail.com> wrote:

[Gabriela Tchaga]

Here is what I got for 1b:

1. x3=260 x4=140 e= 0

2.x2=86.67 x4= -33.33 not feasible

3. x2=70 x3= 50 e= 1050

4. x1=130 x4=10 e= 3250

5. x1=140 x3=-20 not feasible

6. x1=100 x2= 20 e=2800

good job everyone.

gabi

[Elliott Le]

How do I reference the pivot column/row? Is it X1/b1?

Elliott Le

--
Elliott Le

[Carey Deangelis]

On number 3 and 5 for problem set 1b, how do you not get the same answer as 2 and 4?  For example both 2 and 3 the graph would look like:

X2          X4(or X3)       Bi
3           0                  260
2           0                  140

Am I missing something?

Since both solutions are using X2 wouldn't both solutions use the coefficients 3 and 2?

thanks,

Carey

[Rachna Patel]

Hi Elliott,

I referenced it by pivot column = x1, pivot row = bi=260

On Mon, Oct 3, 2011 at 8:58 AM, Elliott Le <ellio...@gmail.com> wrote:

[Elliott Le]

Thanks Rachna.

Elliott Le

--
Elliott Le

[Sudha Venkatesh]

shouldn't we mention pivot row as X3 to indicate that the one that  leaves the tableau?

On Mon, Oct 3, 2011 at 9:50 AM, Rachna Patel <rachna....@gmail.com> wrote:

[Gabriel Bowers]

This is the answer that I am getting as well.

I think the last question part #d wants us to recognize that all columns show "c - z <= 0".  This indicates an optimal solution.  If any of these were greater than 0, then we would need to perform another pivot operation.

Does this jive with everyone else?

Gabriel

On Sun, Oct 2, 2011 at 7:14 PM, Almitra Sharma <karnik....@gmail.com> wrote:

[Manuel Severino]

I agree. Negative disposal values take us out of the feasible region so we need to stop.


Manny

[Gabriela Tchaga]

What did you guys write down for the last question on the second homework. Im kinda confused as to what he wants us to write about regarding the methods- not in his notes that I saw of. Thanks everyone

 

p.s. I got the same answers as Sidra on the second question.

 

gabi

[Rachna Patel]

someone in the class asked the prof about the second question and im almost positive he confirmed that there needed to be negative values on the second part.