Answer:
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
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