Question

Quiz 4

A manufacturer makes and sales four types of products:  Product X, Product Y, Product Z, and Product W.  

The resources needed to produce one unit of each product and the sales prices are given in the following Table.

Resource	Product X	Product Y	Product Z	Product W
Steel (lbs)	2	3	4	7
Hours of Machine Time (hours)	3	4	5	6
Sales Price ($)	4	6	7	8

• Currently, 4,600 pounds of steel and 5,000 machine hours are available.

• To meet customer demands, exactly 950 total products must be produced.

• Customers also demand that at least 400 units of Product W be produced.

Formulate an LP that can be used to maximize sales revenue for the manufacturer.

LP Formula

Let Pi be the number of product type i produced by the manufacturer, where i = X, Y, X, and W.

MAXIMIZE  4 PX + 6 PY + 7 PZ + 8 PW

Subject To

2 PX + 3 PY + 4 PZ + 7 PW <= 4600   ! Available Steel

3 PX + 4 PY + 5 PZ + 6 PW <= 5000   ! Available Machine Hours

PX + PY + PZ + PW    = 950               ! Total Demand

                            PW >= 400               ! Product W Demand

PX >=0

PY >=0

PZ >=0

PW >=0

Suppose manufacturer raises the price of Product Y by 50¢ per unit. What is the new optimal solution to the LP?

Objective Function Value:
PX:
PY:
PZ:
PW:

Answer 1

% by using the below matlab code the change in price of y is not affecting the solution of the linear program

x = optimvar('x','LowerBound',0);
y = optimvar('y','LowerBound',0);
z = optimvar('z','LowerBound',0);
w = optimvar('w','LowerBound',400);
prob = optimproblem('Objective',4*x + 6.5*y+ 7*z + 8*w,'ObjectiveSense','min');
prob.Constraints.c1 = x + y +z + w == 950;
prob.Constraints.c2 = 2*x + 3*y +4*z+7*w<= 4600 ;
prob.Constraints.c3 = 3*x+ 4*y+5*z+6*w <= 5000;
prob.Constraints.c4 = w >= 400;
problem = prob2struct(prob);
[sol,fval,exitflag,output] = linprog(problem);
sol
fval
prob.Variables

the output of matlab is

Objective Function Value:	5400
Px	550
Py	0
Pz	0
Pw	400