Question

During the next two months, General Cars must meet (on time) the following demands for trucks and cars:

month 1—400 trucks, 800 cars;
month 2—300 trucks, 300 cars.
During each month, at most 1,000 vehicles can be produced.
Each truck uses 2 tons of steel, and each car uses 1 ton of steel.
During month 1, steel costs $400 per ton; during month 2, steel costs $600 per ton.
At most, 1,500 tons of steel may be purchased each month (steel may only be used during the month in which it is purchased).
At the beginning of month 1, 100 trucks and 200 cars are in inventory. At the end of each month, a holding cost of $150 per vehicle is assessed. Each car gets 20 mpg, and each truck gets 10 mpg. During each month, the vehicles produced by the company must average at least 16 mpg.
Formulate an LP to meet the demand and mileage requirements at minimum cost (include steel costs and holding costs).

Answer 1

the formulation of the given linear program is,

Minimize

z = 400S1 + 600S2 + 150 ( LT1 + LT2 + LC1 + LC2)

Subject to the constraints

T1 + C1 ≤ I000

T2 + C2 ≤ 1000

2T1 + C1 = S1

2T2 + C2 = S2

100 + T1 = 400 + LT1

LT1 + LT2 = 300 + LT2

200 + C1 = 800 + LC1

LC1 + C2 = 300 + LC2

S1,S2 ≤ 1 500

4C1 - 6T1 ≥ 0

4C2 - 6T2 ≥ 0

All variables ≥ 0

Step-by-step explanation:

Firstly;

Let T1 be number of trucks to be produced in month 1 and T2 be number of trucks to be produced in month 2.

Let C1 be number of cars to be produced in month 1 and C2 be number of cars to be produced In month 2.

Let S1 be tons of steel used in month 1 and S2 be tons of steel used in month 2.

Let LT1 be number of trucks in inventory at the end of month 1 and LT2 be number of trucks In inventory at the end of month 2.

Let LC1 be number of cars in inventory at the end of month 1 and LC2 be number of cars in inventory at the end of month 2.

Now the objective is to minimize the cost, so

z = [(cost of steel during month 1)(tons of steel used in month 1)] + [(cost of steel during month 2)( tons of steel used in month 2)] + [(holding cost at the end of each month )(trucks and cars in inventory at the end of each month)]

= 400S1 + 600S2 + 150 ( LT1 + LT2 + LC1 + LC2)

Thus, the objective function Is, Minimize

z = 400S1 + 600S2 + 150 ( LT1 + LT2 + LC1 + LC2)

Constraint1 Each month, the production capacity of the vehicle is 1000 vehicles.

Number of trucks produced in month 1 + number of cant produced in month 1 ≤ 1000

T1 + C1 ≤ 1000

Number of trucks produced in month 2 + number of cars produced in month 2 ≤ 1000

T2 + C2 ≤ 1000

Constraint2 Each month. each truck uses 2 tons of steel and each car uses 1 ton of steel.

[(Tons of steel used to produce truck in month 1) + (tons of steel used to produced cars in month 1)] = S1

2T1 + C1 = S1

[(Tons of steel used to produce truck in month 2) + (tons of steel used to produced cars in month 2)] = S2

2T2 + C2 = S2

Constraint3 At the beginning of month 1, 100 trucks are in inventory.

[100 trucks at the beginning are in inventory +

number of trucks produced in month 1] = [400 trucks are demanded in month 1 + number of trucks in inventory at the end of the month 1]

100 + T1 = 400 + LT1

[trucks at the beginning of month 2 in inventory + number of trucks produced in month 2 ] = [300 trucks are demanded in month 2 + number of trucks in inventory at the end of the month 2]

LT1 + T2 = 300 + LT1

Constraint 4 At the beginning of month 1, 200 cars are in Inventory

[200 cars at the beginning are in inventory + number of cars produced in month 1] = [800 cars are demanded in month 1 + number of cars in inventory at the end of the month 1]

200 + C1 = 800 + LC1

[cars at the beginning of month 2 in inventory + number of cars produced in month 2 ] = [300 cars are demanded in month 2 + number of cars in inventory at the end of the month 2]

LC1 + C2 = 300 + LC2

Constraint 5 At most, 1,500 tons of steel can be purchased each month.

S1 ≤ 1,500

S2 ≤ 1,500

Constraint 6 Each month, vehicle produced by company must average at least 16mpg.

[{{(mpg of trucks)(number of trucks produced in month 1)} + {(mpg of cars)(number of cars produced in month 1)}} / {(number of trucks produced in month 1 ) + ( number of Cars produced in month 1)}] ≥ 16

(10T1 + 20C1 / T1 + C1) ≥ 16

4C1 - 6T1 ≥ 0

[{{(mpg of trucks)(number of trucks produced in month 2)} + {(mpg of cars)(number of cars produced in month 2)}} / {(number of trucks produced in month 2 ) + ( number of Cars produced in month 2)}] ≥ 16

(10T2 + 20C2 / T2 + C2) ≥ 16

4C2 - 6T2 ≥ 0

Therefore, the formulation of the given linear program is,

Minimize

z = 400S1 + 600S2 + 150 ( LT1 + LT2 + LC1 + LC2)

Subject to the constraints

T1 + C1 ≤ I000

T2 + C2 ≤ 1000

2T1 + C1 = S1

2T2 + C2 = S2

100 + T1 = 400 + LT1

LT1 + LT2 = 300 + LT2

200 + C1 = 800 + LC1

LC1 + C2 = 300 + LC2

S1,S2 ≤ 1 500

4C1 - 6T1 ≥ 0

4C2 - 6T2 ≥ 0

All variables ≥ 0