Question

I need the answers simple and in order please!!! go with the letter a,b,c etc.

Ahmadi, Inc. manufactures laptop and desktop computers. In the upcoming production period, Ahmadi needs to decide how many of each type of computers should be produced to maximize profit. Each computer goes through two production processes. Process I, involves assembling the circuit boards andprocess II is the installation of the circuit boards into the casing. Each laptop requires 24 minutes of process I time and 16 minutes of process II time. Each desktop requires 8 minutes of process I time and 32 minutes of process II time. In the upcoming production period, 240 minutes are available in process I and 320 minutes in process II. Each laptop costs $1,800 to produce and sells for $2,250. Each desktop costs $600 to produce and sells for $1,000.

      Let your decision variables be:

                  X1 = Number of laptops to produce

                  X2 = Number of desktops to produce

1. Formulate an LP problem to maximize profit. Write your problem formulation.
2. Graph the constraints and show the region of feasible solutions.
3. Determine the extreme points of the feasible region and find the profit at each extreme point.
4. Draw the isoprofit line and indicate the optimum point.
5. Are there any slacks at optimum?
6. If the selling price per desktop decreases to $700 per unit would there be any change in the optimum solution? If yes, what would be the new optimum solution, and would there be any slack?
7. Assume the company does not want to produce more than 9 laptops in this production period. Add this constraint to the original problem and show the region of feasible solutions.
8. Solve the problem as stated in part g and find the optimal solution. Let the profit per Laptop to be $450 and per Desktop to be $100.
1. Let the selling price of Desktops to be $750 and solve the problem as stated in part g.

Answer 1

ANSWER:

a)

decision variables be:

X1 = Number of laptops to produce

X2 = Number of desktops to produce

Objective Function

Maximize the profit from the production mix.

Profit from Laptop = Price – Cost = $2,250 - $1,800 = $450 per unit

Profit from desktop = Price – Cost = $1,000 - $600 = $400 per unit

Profit function is given as shown below:

Max. Z = $450X1 + $400X2

b)

Constraints:

Time available for process 1 is 240 minutes. Each unit of Laptops and desktop take 24 and 8 minutes respectively on process 1:

24X1 + 8X2 <= 240

Time available for process 2 is 320 minutes. Each unit of Laptops and desktop take 16 and 32 minutes respectively on process 2:

16X1 + 32X2 <= 320

Nonnegativity Constraints:

Let, X1, X2 >= 0

Graphical Solution:

Plotting of constraints on graph:

Convert the inequalities in the equality equation and find the X-axis and Y-axis intercepts to plot the equation lines.

Constraint	Equality Equation	X-axis coordinate (Y = 0)	Y-axis coordinate (X = 0)	Line and feasible region according to constraint type
1	24X1 + 8X2 = 240	X = 240/24 = 10 A (10, 0)	Y = 240/8 = 30 B (0, 30)	Green (towards origin)
2	16X1 + 32X2 = 320	X = 320/16 C (20, 0)	Y = 320/32 = 10 D (0, 10)	Blue (towards origin)

The feasible area is OADE, by extreme point method determine the profit at points O, A, D, and E

Coordinate of E:

At point E, Equation (1) - 24X1 + 8X2 = 240 and Equation (2) - 16X1 + 32X2 = 320 lines coincides, both the equation can be solved simultaneously as follows:

Eq (1) – (1/4) x Eq (2)

24X1 + 8X2 – ( (1/4) x (16X1 + 32X2) = 240 – (1/4)(320)

24X1 + 8X2 – 4X1 – 8X2 = 240 – 80

20X1 = 160

X1 = 8

Substitute, X1 = 8 in equation (1), we get

24(8) + 8X2 = 240

X2 = 48/8 = 6

Point E coordinates are (8, 6)

c)

Objective function value at the extreme points of feasible area:

Point	Objective function Value
O (0, 0)	Z = 0
A (10, 0)	Z = (450)(10) + (400)(0) = $4500
E (8, 6)	Z = (450)(8) + (400)(6) = $6000
D (0, 10)	Z = (450)(0) + (400)(10) = $4000

Thus, maximum profit is at point E (8, 6)

Thus, to maximize profit and achieve all constraint, 8 laptops and 6 desktops should be produced

Optimal production Plan: Laptops = 8 units and Desktop = 6 units

Optimal Profit = $6,000

d)

Iso-profit line:

Let the target profit is $8000, then determine coordinates of Isoprofit line of 8000.

Objective function for Isoprofit line of 8000is given as follows: 450X1 + 400X2 = 8000

If X2 = 0, the X-intercept is obtained as follows:

450X1 = 8000 or X1 = 17.78

X-intercept coordinate: (17.78, 0)

If X = 0, the Y-intercept is obtained as follows:

400X2= 8000 or Y = 20

Y-intercept coordinate: (0, 20)

Thus, the isoprofit line of $8000 passes through (17.78, 0) and (0, 20)

By moving the profit line towards the origin, it meets the feasible area at the extreme point E. At point E the solution is optimal solution.