BrainDen.com - Brain Teasers
• 0

PowerCo

Question

PowerCo has three plants that must provide enough electricity to four cities. Shipping to each city from the various plants costs varying amounts. Moreover each plant has a set supply of energy that it can send to the various cities. The specifics of this information are outlined in the chart below:

From City 1 City 2 city 3 city 4 Supply

Plant 1 \$8 \$6 \$10 \$9 35

Plant 2 \$9 \$12 \$13 \$7 50

Plant 3 \$14 \$9 \$16 \$5 40

Demand 45 20 30 30

How can PowerCo best minimize its costs?

• 1
• 1

Recommended Posts

• 0

Plant 1: 10 to City 2 @ \$6 and 25 to City 3 @ \$10

Plant 2: 45 to City 1 @ \$9 and 5 to City 3 @ \$13

Plant 3: 10 to City 2 @ \$9 and 30 to City 4 @ \$5

Total Cost = \$1020

As par for the course, this could probably have be made a lot more elegantly, but it seems to get the job done.

Fair warning: this took about 20 seconds to spit out an answer. Feel free to uncomment the cout after the first for loop if you are really into countdowns.

```int a, b, c, d, e, f, g, h, i, j, k, l, a1, b1, c1, d1, e1, f1, g1, h1, i1, j5, k1, l1, money=10000;

int main()
{
for(a=35; a>=0; a--)
{
//cout<<a<<endl;
for(b=((35-a)>20 ? 20 : (35-a)); b>=0; b--)
{
for(c=((35-(a+>30 ? 30 : (35-(a+); c>=0; c--)
{
d=35-(a+b+c);

for(e=45-a; e>=0; e--)
{
for(f=((50-e)>20 ? 20 : (50-e)); f>=0; f--)
{
for(g=((50-(e+f))>30 ? 30 : (50-(e+f))); g>=0; g--)
{
h=50-(e+f+g);

for(i=((a+e)<5 ? 0 : 45-(a+e)); i>=0; i--)
{
if(a+e+i==45)
{
for(j=((40-i)>20 ? 20 : (40-i)); j>=0; j--)
{
if(b+f+j==20)
{
for(k=((40-(i+j))>30 ? 30 : (40-(i+j))); k>=0; k--)
{
if(c+g+k==30)
{
l=40-(i+j+k);

if(d+h+l==30 and (8*a+6*b+10*c+9*d+9*e+12*f+13*g+7*h+14*i+9*j+16*k+5*l)<=money)
{
money=(8*a+6*b+10*c+9*d+9*e+12*f+13*g+7*h+14*i+9*j+16*k+5*l);
a1=a;b1=b;c1=c;d1=d;
e1=e;f1=f;g1=g;h1=h;
i1=i;j5=j;k1=k;l1=l;
}
}
}
}
}
}
}
}
}
}
}
}
}

cout<<a1<<"  "<<b1<<"  "<<c1<<"  "<<d1<<endl<<e1<<"  "<<f1<<"  "<<g1<<"  "<<h1<<endl<<i1<<"  "<<j5<<"  "<<k1<<"  "<<l1<<endl<<"= \$"<<money<<endl;
return 0;
}

```

Share on other sites

• 0

plant 1 should send 30 to city 3 @ \$10 and it's remaining 5 to city 2 @ \$6 = \$330

plant 2 should send 45 to city 1 @ \$9 and its remaining 5 to city 4 @ \$7 = \$440
plant 3 should send 25 to city 4 @ \$5 and 15 to city 2 @ \$9 = \$260
for a total cost of \$930 I think that's the best they can do.
Share on other sites

• 0

plant 1 should send 30 to city 3 @ \$10 and it's remaining 5 to city 2 @ \$6 = \$330

plant 2 should send 45 to city 1 @ \$9 and its remaining 5 to city 4 @ \$7 = \$440

plant 3 should send 25 to city 4 @ \$5 and 15 to city 2 @ \$9 = \$260

for a total cost of \$930 I think that's the best they can do.

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.