##### ICT704 Non Relational Database Systems Assignments

Delivery in day(s): 4

This is the solution of linear programming assignment help in which we discuss case study of linear programming and mathematical model

- After formulation of given constraints we get

Max Z = 3500x + 5500 y

6x + 8.5 y <= 50 000

4x + 2y <= 25000

Y <= 3800

x,y>= 0 - By graphical methods we are solving the constraints

6x + 8.5 y = 50 000

4x + 2y = 25000

Y <= 3800 - We get following solutions

Z = 2090000 at (0, 3800)

Z = 21875000 at (6250, 0)

Z= 31225000 at (2950, 3800) Maximum profit

Z = 30397700 at (5113.64, 2272.72)

Total door used = 19400 and total door unused = 5600

All the labors hours are used - We invested $ 400000 on advertising to increased demand of 25% of silhouette

Now (25/100)x 38000 = 950

So net demand = 4750

Profit Z = 3500x2950 + 3500x 4750 = 36450000

We spent 40000 so net profit = Z = 36410000

so yes the campaign should be undertaken - Labourtime = 50000+50000x (30/100) = 65000

Max Z = 3500x+ 5500y

6x+8.5y <= 6500

4x+2y <= 25000

Y<=3800

Now solving the above constraints by graphical method we get the solution

(x,y)= (4350, 3800) solving above constrained graphically we got only one feasible point

Max Z = 36125000 and all doors are used with 6600 used labour hours - Since in part (c) the total laborhours required to pay is

58400 – 50000 = 8400

So he should pay for 8400 labour hrs. - Now updating the constraints and solving it by graphical methods we get

Max Z = 3500x+ 5500y

6x+8.5y <= 6500 (1)

4x+2y <= 25000 (2)

Y<=4750 (3)

Now solving the above constraints by graphical method we get the solution

(x,y)= (3750, 5000) solving (1) and (2)

(x,y)= (4104.16, 4750) solving (1) and (3)

(x,y)= (3875, 4750) solving (3) and (2)

No we can conclude that second one is the optimal one because it is satisfying the constraints.

Hence all doors used and 6600 labourhr are unused. - According to (e)

Z = 414.16x3500+ 4750x 5500= 40489560

Now subtracting some new investment

40489560 – 180000000 = 38689560

SO net profit = 38289560 which is greater than previous one

Hence we go for the condition in (e). - After formulation

Y< = 2500

Max z = 3500x+ 4950y

X<= 4700

x-2y<= 0

Solving graphically

The feasible optimal solution = (x,y) = (4700, 2350) and the production ratio will be 2:1. - After formulation

Max z = 3500x + 5500y (1)

4x+6y <= 50000 (2)

4x+2y<= 25000 (3)

Y<= 3800

Solving (1) and (2) (875000, -5000)

Solving (3) and (2) (4550, 3800)

Solving (1) and (3) (6800, 3800) max profit

Hence max Z = 258900000 - Since, max profit is less in the case of (h) so, he must not use robot machines.

Considering the conditions in (g) and (i) we got

Number of eagle = 6800

Number of silhouetters = 3800

So the new profit = 22890000

According to the conditions in (j)

Number of doors = 6800x4 + 3800x2 = 34800

Now since the cost of each doors is 800$

So total cost for the doors = 800x 34800 = 2784000 $

We can say that Mark does not procure the order for the doors because he is getting loss in this case.