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## PhD计算机Lingo编程代写

1) Executive summary: a short description of the problem and summary of yourResults

A company manufactures three diﬀerent products at four plants. Our task is to find the optimal production and shipping plan for the company to minimize the cost of meeting customer demands with Linear Programming and LINGO. The cost of this company consists of production cost, transportation cost and inventory cost. By finding the optimal production and shipping plan, we can minimize the cost of this company. With Linear Programming and LINGO, we found global optimal solution of the cost of this company: \$836931.1. Specifically, production cost, transportation cost and inventory cost is \$750754.6, \$85035.13 and \$1141.382,respectively.

2) Problem description: rewrite the problem according to your understanding

A company manufactures three diﬀerent products at four plants. The cost of this company consists of production cost, transportation cost and inventory cost. The production cost is the time spent on machines multiplied by Machine cost per hour. In every plant, the time spent on every machine cannot exceed the available machine time. The transportation cost is the number of product transported from plant to customer multiplied by unit transportation cost. The number of product transported from all plant to a customer cannot exceed the demand of this customer. The inventory cost is the number of inventory product multiplied by unit inventory cost. In first month, the number of inventory product = the number of Initial inventory + the number of product produced in this month – the number of product transported to all customer. In other month, the number of inventory product = the number of inventory product on last month + the number of product produced in this month – the number of product transported to all customer. The number of inventory product must be greater than 0.

3) Formulation of the problem and LINGO implementation

sets:

! four plant, the unitInventoryCost is unit inventory cost of every plant;

plant/1..4/:unitInventoryCost;

! every plant have three machines;

machine/1..3/;

! every plant produce three type of products.

product/1..3/;

! find the optimal production and shipping plan in next four month;

month/1..4/;

! this company have six customer;

customer/1..6/;

! Machine time for product at plant (minutes/unit)

! MTplant is Total machine time at plant for the next four months (Hours);

! MachineCost is Machine cost at plant for the next four months (\$/Hour)

! InitialInventory is Initial inventory at plant (units)

! demand is Demand for product by customer in each month (units);

! transportsum is the number of product transported to customer;

! TransporationCostis Transporation cost between plant and customer (\$/unit)

!transport is the number of product transported from a plant to a customer in some month;

! prod is the number of product produced in a plant in some month;

! time is the time spent to produce product in a plant in some month;

! realtrans is the number of product transported in a plant in some month;

! inventory is the number of inventory product in a plant in some month;

EssayPhD--计算机lingo编程代写 endsets

! calculate the time spent to produce product in a plant in some month;

@FOR( month(M): @FOR( plant(I): @FOR( machine(K): time(M,I,K)=@SUM( product(P):MTproduct(I,P,K)*prod(M,I,P)))));

! the time spent to produce product in a plant in some month must be less than the available machine time in that plant

@FOR( month(M): @FOR( plant(I): @FOR( machine(K): time(M,I,K)<=MTplant(I,K,M)*60)));

! calculate the cost of production

productCost = @SUM( month(M): @SUM( plant(I): @SUM( machine(K): (time(M,I,K)/60)*MachineCost(I,K,M))));

! calculate the number of product transported to customer;

@FOR( month(M): @FOR( product(P): @FOR(customer(C): transportsum(P,C,M)=@SUM( plant(I):

transport(M,I,P,C)))));

! the number of product transported to customer must be equal to demand for product by customer;

@FOR( month(M): @FOR( product(P): @FOR(customer(C): transportsum(P,C,M)=demand(P,C,M))));

!calculate the number of product transported in a plant in some month;

@FOR( month(M): @FOR( plant(I): @FOR(product(P): realtrans(M,I,P)=@SUM( customer(C):transport(M,I,P,C)))));

! calculatetransporation cost between plant and customer

transportCost =@SUM(month(M):@SUM(product(P):@SUM(customer(C):@SUM(plant(I):transport(M,I,P,C)*TransporationCost(I,C)))));

! calculate the number of inventory product;

@FOR( month(M): @FOR( plant(I): @FOR( product(P): inventory(M,I,P) = @if(M #eq# 1,prod(M,I,P)+InitialInventory(P,I)-realtrans(M,I,P),prod(M,I,P)+inventory(M-1,I,P)-realtrans(M,I,P)))));

! The number of inventory product must be greater than 0.

@FOR( month(M): @FOR( plant(I): @FOR( product(P): inventory(M,I,P)>=0)));

! calculate the cost of inventory

inventoryCost = @SUM( month(M): @SUM( plant(I): @SUM( product(P): inventory(M,I,P)*unitInventoryCost(I))));

! Object function, minimize the cost of this company

min=productCost +transportCost +inventoryCost ;

4) interpretation of the solution report

With Linear Programming and LINGO, we found global optimal solution of the cost of this company: \$836931.1. Specifically, production cost, transportation cost and inventory cost is \$750754.6, \$85035.13 and \$1141.382, respectively. The prod is production plan for the company. The transport is shipping plan for the company.

5) Summary: short description of your solution approach and main results The cost of this company consists of production cost, transportation cost and inventory cost. Our task is to find the optimal production and shipping plan for the company to minimize the cost of meeting customer demands with Linear Programming and LINGO. We found global optimal solution of the cost of this company: \$836931.1. Specifically, production cost, transportation cost and inventory cost is \$750754.6, \$85035.13 and \$1141.382, respectively.

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