I just need help with part a I have a formulation but i don

I just need help with part a) . I have a formulation, but i don\'t know how to input the formulation into Lingo program / similar to matlab/ you have to create a maximization equation, followed by constraint equations, plug them in, and the program solves it.

This is what i have, but it\'s not correct, i checked the solution manual and the answer is different ( m is suppose to be a large constant, but i wasn\'t sure how to represent that on a problem solving software).

! Problem 12.4-6;

! Objective function;

max = 10*x1 + 15*x2 -50000*y1 - 80000*y2;

! constraints s.t

[st1] x1 <= m*y1;

[st2] x2 <= m*y2;

[st3] (x1/50) + (x2/40) <= 500 + m*y3;

[st4] (x1/40) + (x2/25) <= 700 + m*(1-y3);

[st5] x1 >= 0;

[st6] x2 >= 0;

@Bin(y1);

@Bin(y2);

@bin(y3);

@gin(x1);

@gin(x2);

12.3-4. The Toys-R-4-U Company has developed two new toys for possible inclusion in its product line for the upcoming Christmas season. Sting up the production facilities to begin production would cost $50,000 for toy1 and $80,000 for toy 2. Once these costs are covered, the toys would generate a unit profit of $10 for toy 1 and S15 for toy 2 The company has two factories that are capable of producing these toys. Ilowever, to avoid doubling the start-up costs, just onc factory would be used, where the choice would be based on max imizing profit. For administrative reasons, the same factory would be used for both new toys if both are produced. Toy 1 can be produced at the rate of 50 per hour in factory l and 40 per hour in factory 2. Toy 2 can be produced at the rate of 40 per hour in factory 1 and 25 per hour in factory 2. Factories 1 and 2, respectively, have 500 hours and 700 hours of production time available before Christmas that could be used to produce these toys. It is not known whether these two toys would be continued after Christmas. Therefore, the problem is to determine how many

Solution

a) MIP model will be

For factory 1

1) Maximize 50*10*x + 40*15*y - 50,000 - 80,000

subject to conditions x >= 1 && y >= 1 && x+y <= 500

(produce at least 1 of both toys)

Answer:- $169,000 at x,y = (1, 499)

2) Maximize 50*10*x - 50,000

subject to conditions x >= 1 && x <= 500  (only produce 1st toy)

Answer:- $200,000

3) Maximize 40*15*y - 80,000

subject to conditions y <= 500  (only produce 2nd toy)

Answer:- $220,000

For factory 2

1) Maximize 40*10*x + 25*15*y - 50,000 - 80,000

subject to conditions x >= 1 && y >= 1 && x+y <= 700

Answer:- $149,575 at x,y = (699, 1) (produce at least 1 of both toys)

2) Maximize 40*10*x - 50,000

subject to conditions x >= 1 && x <= 700 (only produce 1st toy)

Answer:- $230,000

3) Maximize 25*15*y - 80,000

subject to conditions y <= 700  (only produce 2nd toy)

Answer:- $130,00

We can see the best option is producing only toy 1 in factory 2.