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The below figure illustrates the BOM (Bill of Materials) for product A. The MPS (Material requirements Planning) start row in the master production schedule for product A calls for 50 units in week 2, 65 units in week 5, and 80 units in week 8. Item C is produced to make A and to meet the forecasted demand for replacement parts. Past replacement part demand has been 20 units per week (add 20 units to C's gross requirements). The lead times for items F and C are 1 week, and for the other items the lead time is 2 weeks. No safety stock is required for items B, C, D, E, and F. The L4L lot-sizing rule is used for items B and F; the POQ (Periodic Order Quantity) lot-sizing rule (P = 3) is used for C. Item E has an FOQ (Fixed Order Quantity) of 600 units, and D has an FOQ of 250 units. On-hand inventories are 50 units of B, 50 units of C, 120 units of D, 70 units of E, and 250 units of F. Item B has a scheduled receipt of 50 units in week 2. Develop a material requirements plan for the next 8 weeks for items B, C, D, E, and F.
Draw a B-tree of order 3 for the following sequence of keys: 2,4,9,8,7,6,3,1,5,10.and delete 8 and 10
Two linked lists are having information of the same type in ascending order. Write down a module to merge them to a single linked list that is sorted merge(struct node *p, stru
DEPTH FIRST SEARCH (DFS) The approach adopted into depth first search is to search deeper whenever possible. This algorithm frequently searches deeper through visiting unvisite
Q. Enumerate number of operations possible on ordered lists and arrays. Write procedures to insert and delete an element in to array.
/* the program accepts two polynomials as a input & prints the resultant polynomial because of the addition of input polynomials*/ #include void main() { int poly1[6][
A full binary tree with n leaves have:- 2n -1 nodes.
Readjusting for tree modification calls for rotations in the binary search tree. Single rotations are possible in the left or right direction for moving a node to the root position
N = number of rows of the graph D[i[j] = C[i][j] For k from 1 to n Do for i = 1 to n Do for j = 1 to n D[i[j]= minimum( d ij (k-1) ,d ik (k-1) +d kj (k-1)
Explain the concept of hidden lines The problem of hidden lines or surfaces was implicit even in 2-D graphics, but we did not mention it there, because what was intended to be
Q. Take an array A[20, 10] of your own. Suppose 4 words per memory cell and the base address of array A is 100. Find the address of A[11, 5] supposed row major storage.
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