Answer:
The target inventory position is T= 713.6 gallons.
Explanation:
Given:
Average demand =per day = D = 152 Gallons
Standard deviation of demand = σ = 33 Gallons per day
Lead time for delivery = L = 4 days
Z value for 94.5% service level = 1.6
The target inventory position = (Average demand x Lead time) + Safety stock
= (D × L) + (Z× σ × )
= (152 × 4) + (1.6 × 33 × )
= (152 × 4) + (1.6 × 33 × 2)
= 608 + 105.6
= 713.6
Answer:
Step 1
Let us assume that x1 amount of Fliptop and x2 amount of Tiptop models are produced, then the objective function is to maximize profitability with the constraints on the production limited by the available plastic, ink and time. Hence the LP model is given by the objective function and the three constraints as shown below:
Objective function ($) (OF): maximize z = 1000x1 + 1000x2
Plastic material constraint ( Eqn. 1): 3x1 + 4x2 <= 36
Ink material constraint (Eqn. 2): 5x1 + 4x2 <= 40
Time constraint (Eqn. 3): 5x1 + 2x2 <= 30
Non negativity constraints: x1, x2 >= 0
Step 2
Since it is a 2 variable problem it can be solved graphically or using a model solver such as MS-Excel ®. The feasible region is defined by the corner points (boundary points) A, B, C, D, E and the boundary lines of the constraint equations 1,2,3 and the objective function OF as shown in the diagram.
The coordinates and the OF values at the corner points are given below:
A (0,0); OF = 0 (intersection of non-negativity constraints)
B (6,0); OF = 6000 (intersection of x2=0 and eqn 3)
C (4,5); OF = 9000 (intersection of eqn 2 and 3)
D (2, 7.5); OF = 9500 (intersection of eqn 1 & 2)
E (0,9); OF= 9000 (intersection of x1=0 and eqn 1)
Step 3
Hence the optimal solution is given by the point D where the OF equation touches the feasible region with the maximum value. There is an excess of 5 units of molding time available.
