The master production schedule to manufacture a product is given below:
If the company follows Wilson’s model to place an order. The EOQ is given as 300, the stock-on-hand is 100, lead time is 1 month. Find the minimum number of orders required to fulfill the demand arise up to month-8
Minimum number of orders required is 4
The product structure of product ‘P’ is shown in the figure below. The assembly of 2 units of ‘B’ and 4 units of ‘C’ to produce 1 unit of ‘A’ takes a week.
The ordering time of B, C and D are 2, 1 and 2 weeks. The master production schedule for product ‘P’ is as follows:-
The net requirement of ‘C’ if initial inventories of A, B, C and D are 10, 20, 15 and 40 is _____________
Total demand = 80 units
For producing 1 unit of P, 80 units of ‘A’ is required but in inventory 10 units of ‘A’ is there so a total of 70 units of ‘A’ is required.
It means 4 × 70 = 280 units of C is required. But in inventory 15 units of C is there.
Therefore 280 − 15 = 265 units of ‘C’ are required.
Item P is made from components Q and R. Item Q, in turn is made from S and T. The lead times for items P, Q, R, S and T are 3, 4, 11, 6 and 7 weeks respectively. The lead time(in weeks) needed to respond to a customer order for item P is
P−Q−S = 6+4+3 = 13
P−R−T = 7+4+3 = 14
P−R = 11+3 = 14
Longest lead time is 14
The MRP table is shown below, the POR is
Consider the following MRP sheet
The inventory at the end of 5th week
Inventory at the end of 5th week = 100 units
The material requirement planning for an item is shown in the table below: The ordering quantity is 80 units. The 2nd planned order release happens in the week of _______________
(Assume LT = 3 weeks)
A company contains two alternative methods for manufacturing a product. The estimated time and cost for both are given below
The number of units up to which milling is justified economically is
Milling operation is justified as long as the total cost is less than the lathe.
TC)Milling ≤ TC)Lathe
10 Q ≤ 187.5
Q ≤ 18.75
Q ≤ 19
The plant has a capacity of producing 50,000 units per year. The annual fixed cost is Rs. 80,000. The variable cost per unit is Rs.18. The price of the product is Rs. 20 per unit. Then the breakeven point in terms of the capacity of the plant is
Fixed cost = 80,000
Variable cost = Rs. 18/unit = V
price (P) = Rs. 20/unit
Capacity of the plant = 50,000 units/year
% of BEP = 40,000/50,000 = 0.8
Find the total cost using Lot for Lot technique LT = 0, CO = Rs. 100/order, CC = Rs. 1/unit/week
Ordering quantity = demand in a particular period
TC = Stock × CC + N × CO = 0 + 4 × 100 = 400
Find the total cost using minimum cost period technique.
LT = 0, CO = Rs. 100/order, CC = Rs. 1/unit/week
In this method cost period is used to decide the optimum combination