Question

The simplex algorithm is to continue in this manner, always performing basis exchanges which improve the objective function, until no more exchanges are possible. We conclude with an example: Buzz Buzz Buzz Coffee has on hand 1 kg of coffee grounds, 1 gallon of milk and 10 cups of sugar. They can use these to make espressos, containing 8 grams of grounds and no milk or sugar; lattes, containing 15 grams of grounds, 0.0625 gallons of milk and 0.125 cups of sugar; or caf´e cubano, containing 7.5 grams of grounds, no milk and 0.125 cups of sugar. They will be able to sell all they produce, which they will sell at prices of $2 for espressos, $4 for lattes and $5 for a caf´e cubano.

Question 10. (5 points) Let e, l and c be the number of espressos, lattes and caf´es cubanos manufactured, and let g, m and s be the amounts of grounds, milk and sugar left over when they are done. Let p be the amount of money they take in. Record the linear equations relating e, l, c, g, m, s and p.

Question 11. (15 points) Start at the point where no drinks are made (so e = l = c = 0). Exchange one of these variables, in order to increase p. Repeat the process of exchanging a basis variable to increase p until there are no exchanges which will make p larger. How many of each drink should be made?

Answer 1

Solution:- the given data can be written as follows

                 7 kg of coffee ground                  1 gallon milk               10cups of sugar

espressos 8 gms of ground                       0.0625 gallons milk        0.125 cups of sugar

Lattes       15 gms of ground                      no milk                         0.125cups of sugar

CafeCubanos 7.5gms of ground                 no milk                         0.125 cups of sugar

Given that, e represents the espressos, l represents the Lattes, and c represents cafe's cubanos.

Let g represents the amount of grounds, m represents the amount of milk, and s represents the amount of sugar.

price at espressos is $2, and $4 for lattes and $5 for Cafecubanos.

Let p be the amount of money they takein.

Let x , y ,z be the amounts of coffeeground, milk and sugar to be used for manufacture of espressos, lattes and cafcubanos.

So, the matrix form can be written as,

So, this can be written as, 8x+0(y)+0(z)=e

15x+0.0625y+0.125z=l

7.5x+0(y)+0.125z=c

Total left over coffee ground when e=l=c=1 is

(7kg) -(8+15+7.5)=7000-30.5=6969.5

Total left over sugar is, 10 -(0.125+0.125)=1-0.250=0.75

Total left over milk is 1-0.0625=0.9375

So, for one quantitiy that is e=l=c=1, the left over quatitiy will be

6969.5g+9.750s+0.935m

NOw, cost of each espressos is $2

Cost of each lattes is $4 and cost of each cafe cabons is $5

So, p=2e+4l+5c

for e=l=c=0, the price p is 0.

If e=l=c=1, then P=2+4+5=11

Similarly, by giving the values like this, we will get different value at different values of e,l,c

e=8x,

l=15x+0.0625y+0.125z

c=7.5x+0(y)+0.125z

Choose the values of x, y and in such a way that, x7000,y1,z10