Respuesta :
Answer:
Answer is explained in the explanation section below.
Explanation:
Solution:
The Production Planning of the Steel Case:
Demand (units):
Month 1: Â 2400
Month 2: Â 2200
Month 3: Â 2700
Month 4: Â 2500
Production Cost ($/unit):
Month 1: Â 74
Month 2: Â 75
Month 3: Â 76
Month 4: Â 76.5
Inventory cost ($/panel):
Month 1: Â 1.2
Month 2: Â 1.2
Month 3: Â 1.2
Month 4: Â 1.2
Starting Inventory = 1000
Ending Inventory at the end of 4th month = 1500
Starting Production Level = 1800
Cost of changing production level = $0.5/unit (increase)
                             = $0.3/unit (decrease)
Decision Variables:
Let Pn be the production in the month n.
[tex]I_{n}[/tex] be the inventory at the end of month n.
[tex]I_{o}[/tex] be the initial inventory at the start of month 1.
Objective Function:
The production cost is given below:
74[tex]P_{1}[/tex] + 75[tex]P_{2}[/tex] + 76[tex]P_{3}[/tex] + 76.5[tex]P_{4}[/tex]
And the holding cost is given below:
1.2( [tex]I_{1} + I_{2} + I_{3}[/tex] )
And,
Constraints:
In order to meet the demand, we have the following constraints:
[tex]P_{1}[/tex] + [tex]I_{o}[/tex] [tex]\geq[/tex] 2400
[tex]P_{2}[/tex] Â + [tex]I_{1}[/tex] [tex]\geq[/tex] 2200
[tex]P_{3}[/tex] Â + [tex]I_{2}[/tex] [tex]\geq[/tex] 2700
[tex]P_{4}[/tex] + [tex]I_{3}[/tex] [tex]\geq[/tex] Â 2500
Now, considering the inventory level at the end of each month:
[tex]I_{o}[/tex] = 1000
[tex]I_{1}[/tex] = [tex]P_{1}[/tex] - [tex]I_{o}[/tex] - 2400
[tex]I_{2}[/tex] = [tex]P_{2}[/tex] Â - [tex]I_{1}[/tex] - 2200
[tex]I_{3}[/tex] = [tex]P_{3}[/tex] Â - [tex]I_{2}[/tex] - 2700
[tex]I_{4}[/tex] = [tex]P_{4}[/tex] Â - [tex]I_{3}[/tex] - 2500
[tex]I_{4}[/tex] [tex]\geq[/tex] 1500
It is given that, at maximum 4000 units can be produced each month
So, we have the following constraints:
[tex]P_{n}[/tex] [tex]\leq[/tex] 4000 for n = 1, 2, 3 ,4
Also,
[tex]P_{n}[/tex] [tex]\geq[/tex] 0
[tex]I_{n}[/tex] [tex]\geq[/tex] 0