ISO profit lines

[Maddy Fouts]

hey has anyone started the hwk yet??

[Darren Ng]

Yup. For the isoprofit lines, i just followed the examples in ch 2 ( I believe this is correct, I don't have it in front of me). Let me know if you have other questions.

[maddy...@gmail.com]

It seems simple but I  guess where I'm get confused is how he is picking the values. ( Ex 12,000) and then how he then solved .

My book hasn't arrived yet :(

Sent from my iPhone

[Rachna Patel]

The three corner pts I got were (0,70) (130,0), (100,20). The optimal
solution pt is (130,0).

For iso profit lines I used the following
25x1 + 15x2 = 3250
25x1 + 15x2 = 2250
25x1 + 15x2 = 1250

Anyone else get this or am way off?

--
Sent from my mobile device

[Sudha Venkatesh]

To my understanding we can assign any 'arbitary' number to the objective function to find a slope that falls in the feasible region .Then move the slope up till we reach the  point that yields maximum profit or move down to find minimum cost.

[Horsea08]

I just completed the question and got the same corner points
(including 0,0)

My optimum solution was the intersection of the two lines at ~100,20.
I also followed the book and calculated it mathematically to get a max
profit of 2,800. Is that accurate or is Rachna's 130,0 with a profit
of 3,250 accurate?

Thoughts?

On Sep 24, 9:21 am, Sudha Venkatesh <sudha.venkat...@gmail.com> wrote:
> To my understanding we can assign any 'arbitary' number to the objective
> function to find a slope that falls in the feasible region .Then move the
> slope up till we reach the  point that yields maximum profit or move down to
> find minimum cost.
>
>
>
>
>
>
>
> On Sat, Sep 24, 2011 at 8:46 AM, <maddy.fo...@gmail.com> wrote:
> >  It seems simple but I  guess where I'm get confused is how he is picking
> > the values. ( Ex 12,000) and then how he then solved .
> > My book hasn't arrived yet :(
>
> > Sent from my iPhone
>

> > On Sep 24, 2011, at 12:58 AM, Darren Ng <darren...@gmail.com> wrote:
>
> >   Yup. For the isoprofit lines, i just followed the examples in ch 2 ( I
> > believe this is correct, I don't have it in front of me). Let me know if you
> > have other questions.

[Sidra]

I'm a little stuck on #2a, do let me know what solution and how others
got it. Thanks!

[Subramaniam Narayanan]

Here are my solutions. I double checked these using Excel solver as well...

#1a
Objective Fn, Maximize Profit, e = 25X1 + 15X2 subject to the following constraints:
2X1 + 3X2 <= 260
1X1 + 2X2 <= 140
X1, X2 >= 0
Feasible Region is bounded by (0,0) (0,70), (100, 20), (130, 0)
And the optimal solution is X1 = 130, X2 = 0 and the profit at optimal solution is $3,250

#2a
Objective Fn, Minimize cost of advertising, e = 1,000X1 + 4,000X2 subject to the following constraints:
2X1 + 1X2 >= 50
1X1 + 3X2 >= 30
X1, X2 >= 0
Feasible Region is bounded by (0,50), (24, 2), (30, 0) and it extends to the right from these points.
And the optimal solution is X1 = 30, X2 = 0 and the cost of advertising at optimal solution is $30,000

[Sudha Venkatesh]

I too arrived at the optimal solution (for #1 a ) at the corner point (130,0).

[Rachna Patel]

Is the isocost line for #2 the line that goes through (0,50) and (25,0)?

On 9/24/11, Sudha Venkatesh <sudha.v...@gmail.com> wrote:
> I too arrived at the optimal solution (for #1 a ) at the corner point
> (130,0).
>
> On Sat, Sep 24, 2011 at 11:42 AM, Subramaniam Narayanan <
> ssnara...@gmail.com> wrote:
>
>> Here are my solutions. I double checked these using Excel solver as
>> well...

>> *
>>
>> #1a*

>> Objective Fn, Maximize Profit, e = 25X1 + 15X2 subject to the following
>> constraints:
>> 2X1 + 3X2 <= 260
>> 1X1 + 2X2 <= 140
>> X1, X2 >= 0
>> Feasible Region is bounded by (0,0) (0,70), (100, 20), (130, 0)
>> And the optimal solution is X1 = 130, X2 = 0 and the profit at optimal
>> solution is $3,250
>>

>> *#2a*

--

[Darren Ng]

I got a max of 2800, and intersection of 100, 20 as well.

[Maddy Fouts]

Thanks for all the help team.... can someone that thinks they have the correct answers scan their papers so we can all compare and ask questions. 

I guess i'm still confused about the iso profit lines

--

Madelaine G. Fouts
Class of 2011
Leavey School of Business

Santa Clara University

[Subramaniam Narayanan]

Attaching the files as separate msgs...

On Sat, Sep 24, 2011 at 11:14 PM, Subramaniam Narayanan <ssnara...@gmail.com> wrote:

Hope these graphs clear things up...
For #1a, the slope of the isoprofit line is > than the slope of both the constraints. Hence the optimal maximizing
solution will be @(130,0).

[Subramaniam Narayanan]

Attaching the files as separate msgs...

[Gabriela Tchaga]

Hi everyone,

sorry it took me a while to get back to you with the homework solutions I got. I was at a work convention and just got to the homework today. here are my answers:

1a

optimum solution: 130, 0  3250

iso profit lines

25x1 + 15x2= 3000

25x1 + 15x2 = 2000

25x1 + 15x2 = 1000

feasible region: 

130,0

0,0

0,70

100,20

2a:

optimum solution:

30,0 30,000

iso cost line:-- still working on this one

feasible region:

30,0

24,2

0,50

hope this helps. Thanks everyone for your emails! see you in class tomorrow!

gabi tchaga

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

[Rachna Patel]

Hey guys I think we have different answers for the iso profit lines. I'm still confused on this. Can anyone explain this? I believe one iso profit line does have to go through the optimal solution so one should be

25x1+15x2=3250?

[Gabriel Bowers]

Just finished the HMWK. 

 

My understanding of the isocost or isoprofit lines are that we can find them by identifying the indices in the feasible region.

For instance in #1A, the two indices would be: (100,20) and (130,0)

Two isoprofit lines generated:

A) For (100,20) the profit is $2800.  The isoprofit line would be "2800 = 25*X1 + 15*X2"

B) For (130,0) the profit is $3250.  The isprofit line would be "3250 = 25*X1 + 15*X2"

 

After drawing the isoprofit #A ($2800) and #B ($3250) it shows that the optimal solution lies in a small triangle at the bottom right of the feasible region.

 

Gabriel

[Carey Deangelis]

I am still confused on how to find the slope of the isoprofit line.  How did you guys determine that?

Thanks,

Carey

[Gabriel Bowers]

If you have an ISOPROFIT equation of '"3250 = 25*X1 + 15*X2"

Then the intercepts would be:

@ X1 = 0; X2 = 3250/15 = ~215

@ X2 = 0, X1 = 3250/25 = 130

For the X2 intercept of ~215, I did not draw the line up to 215, I just showed it going up in that direction.

 

If X1 is your Y-axis, then you could say that it is 3250 = 25y + 15x.  Then the linear equation is "y = (15x - 3250)*(1/25)".  The slope is then -3250/25.

 

Gabriel

[Victoria Cheng]

I think we can all have different iso profit lines since we put in arbitrary numbers as the profit into e = 25X1 + 15X2 but when we set the optimal profit (3250) into the equation, that line will cross the optimal point. Although it says isoprofits can help you find the optimal solution, for me, I had to find the optimal solution and then find the isoprofit line.  I drew the line by setting the two intercepts like Gabriel.  For the slope, I set X2 as y-axis and got 25x+15y = 3250 and got a slope of (-25/15)...

[Gabriel Bowers]

I think you're right regarding the slope.  This takes me back to grade 11 math...  :-)

 

Y = 3250/15 - X*(25/15).

 

Gabriel

[Victoria Cheng]

haha i know! do you know if it matters if we set x1 as y or x2 as y? i think the slopes would be different...

[Almitra]

I think Victoria is right.. I too solved the problems using arbitrary values for profit.

[Manuel Severino]

I agree. I solved it in Excel first (kinda defeats the purpose) and then plotted a few isoprofit lines.

Is anyone else bothered with the optimal solutions involving selling 0 purses and buying only magazine subscriptions? I know the constraints made it such that way, but both problems were a bit silly ...


Manny

[Maddy Fouts]

can some one please post 

 for #2a  i'm a little confused