That's the answer I was hoping you'd get. Well done!

Bob

ps. Is that all your questions sorted out?

]]>Actually your first answer is so good that it can be solved by inspection. No calculation at all. I am not sure that it can be done for any other problem but it works fine here.

I got your answer. That always makes me feel good!

The Russians say,"measure the cloth seven times before cutting it once." When I get the same answer he does, I recheck seven times.

]]>thanks for all of your help. I got the correct answer to be 5x + 5y + 1800

It is correct. I double checked.

]]>My post 8 had an error as I muddled up which one was x. But I revised my calculation and arrived at one of the answers you have here. So apply the method and it should work for you too.

I've got to log out now as there's a pile of washing up to do. I'll check back when I've finished.

Bob

]]>this problem is giving me a headache. My options are as follows:

A. -15x + 15y + 180

B. 45x + 65y

C. 105x + 105y

D. 5x + 5y + 1800

Please help

]]>Why are you trying to jam this into two variables?

Because I had my variables muddled I couldn't see how to write 'z' and 'w' in terms of x and y. So I introduced extras. But look on the bright side. I got your answer. That always makes me feel good!

Bob

]]>Bob

]]>Post #7 and post #10 are now solved by the simplex method.

If you are determined to represent this in just x and y then 0 and 0 are the obvious answers. You would naturally want to ship from the cheapest locations.

]]>Back to the drawing board.

Bob

]]>hi mom,

That example is what I call a Linear Programming Problem. I'm OK with them.

The car transport problem is a lot more complicated. Here's what I have done. (note: it may all be rubbish as I've never done one of these before.)

Let z be the number of cars transported from Balt. To Phil. and w the number from NY. to Phil.

As we want to minimise the cost, there's no point transporting more cars than necessary so

x + y = 20

z + w = 15Then the cost equation is

60x + 65y + 45z + 40 w = 60x + 60y + 5y + 40z + 40 w + 5z = 60(x+y) + 40(z+w) + 5y + 5z = 60 x 20 + 40 x 15 + 5(y+z)

So my only chance to reduce the cost is to make y+z as small as possible.

Just looking to see if I can make it zero.

Bob

Isn't x Balt. to Phil. already?

]]>bph >= 0

np >= 0

nph >= 0

bp >= 0

0 <= bp + bph <= 30

0 <= np + nph <= 18

c < 1800,

np + bp = 20

bph + nph =15

60 bp + 45 bph + 65 np + 40 nph = cost

cost = 1800

]]>cost = 20 x 60 + 15 x 40 = 1200 + 600 = 1800.

Bob

]]>That example is what I call a Linear Programming Problem. I'm OK with them.

The car transport problem is a lot more complicated. Here's what I have done. (note: it may all be rubbish as I've never done one of these before.)

Let z be the number of cars transported from Balt. To Phil. and w the number from NY. to Phil.

As we want to minimise the cost, there's no point transporting more cars than necessary so

x + y = 20

z + w = 15

Then the cost equation is

60x + 65y + 45z + 40 w = 60x + 60y + 5y + 40z + 40 w + 5z = 60(x+y) + 40(z+w) + 5y + 5z = 60 x 20 + 40 x 15 + 5(y+z)

So my only chance to reduce the cost is to make y+z as small as possible.

Just looking to see if I can make it zero.

Bob

]]>That is what I thought at first too. But that has an obvious answer of zero for both.

The lowest cost is I believe,1800 and has x and y as both 0.

]]>