As the production planner for Xiangling Hu Products, Inc., you have been given a bill of material for a bracket that is made up
of a base, two springs, and four clamps. The base is assembled from one clamp and two housings. Each clamp has one handle and one casting. Each housing has two bearings and one shaft. There is no inventory on hand.
a. Design a product structure noting the quantities for each item and show the low-level coding.
b. Determine the gross quantities needed of each item if you are to assemble 50 brackets.
c. Compute the net quantities needed if there are 25 of the base and 100 of the clamp in stock.
a. Design a product structure noting the quantities for each item and show the low-level coding.
b. Determine the gross quantities needed of each item if you are to assemble 50 brackets.
c. Compute the net quantities needed if there are 25 of the base and 100 of the clamp in stock.
1 answer:
<h3>Explanation:</h3><h3>Part (a):</h3>
Solved part is attached as an image.
<h3>Part (b):</h3>Let us first determine the amount required of each item to produce 1 bracket.
From the attached diagram, we can see that to manufacture 1 bracket, quantity of each item needed is,
Base = 1
Spring = 2
Clamp = 1 + 4 = 5
Housing = 2
Handle = (1 * 1) + (4 * 1) = 5
Casting = (1 * 1) + (4 * 1) = 5
Bearing = 2 * 2 = 4
Shaft = 2 * 1 = 2
Hence, for 50 Brackets, quantity of each item required will be,
Base = 1 * 50 = 50
Spring = 2 * 50 = 100
Clamp = 5 * 50 = 250
Housing = 2 * 50 = 100
Handle = 5 * 50 = 250
Casting = 5 * 50 = 250
Bearing = 4 * 50 = 200
Shaft = 2 * 50 = 100
<em>NOTE: The above quantities give exclusive quantities required for each item. In actual practice, we won't have to purchase base, clamp & housing separately as the will be assembled from their components which are already procured.</em>
<h3>Part (c):</h3>As 25 bases are already in stock, parts for them will not be needed. I will refer the quantities subtracted due to this by indicating (B). Similarly, quantities subtracted due to clamps will be indicated as (C).
Base = 50 - 25 = 25
Spring = 50
Clamp = 250 - 100 - 25(B) = 125
Housing = 100 - (2*25)(B) = 50
Handle = 1 * 125 = 125
Casting = 1 * 125 = 125
Bearing = 2 * 50 = 100
Shaft = 1 * 50 = 50
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Answer: Graph of (A) (B) and {D) are attached accordingly.
Explanation:
A)
The critical region of the constraints can be seen in the following diagram -
(0,9) (0,5) (0,3) (0,0) (3,0) (5,0) (9,0) The feasible region is shown in white
The intersection points are found by using these equations -
Vertex Lines Through Vertex Value of Objective
(3,2) x+3y = 9; 2x+2y = 10 48
(9,0) x+3y = 9; y = 0 72
(2,3) 2x+2y = 10; 6x+2y = 18 52
(0,9) 6x+2y = 18; x = 0 108
So, we can see the minimum value of the objective function occurs at point (3,2) and the minimum value of the objective function is = 48.
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B)
When we change the coefficients of the variables in the objective function, the optimal solution may or may not change as the weights (coefficient) are different for each constraints for both the variabls. So, it all depends on the coefficient of the variables in the constraints.
In this case, the optimal solution does not change on changing the coefficient of X from 8 to 6 in the objective function.
The critical region would remain same (as shown below) as it is defined by the constraints and not the objective function.
(0,9) (0,5) (0,3) (0,0) (3,0) (5,0) (9,0) The feasible region is shown in white
However, the optimal value of the objective function would change as shown below-
Vertex Lines Through Vertex Value of Objective
(3,2) x+3y = 9; 2x+2y = 10 42
(9,0) x+3y = 9; y = 0 54
(2,3) 2x+2y = 10; 6x+2y = 18 48
(0,9) 6x+2y = 18; x = 0 108
So, we can see that the minimum value now has become 42 (which had to change obviously).
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C)
Now, when we change the coefficient of the variable Y from 12 to 6, again the critical region would remain same as earlier. But in this case, the optimal solution changes as shown below -
Vertex Lines Through Vertex Value of Objective
(3,2) x+3y = 9; 2x+2y = 10 36
(9,0) x+3y = 9; y = 0 72
(2,3) 2x+2y = 10; 6x+2y = 18 34
(0,9) 6x+2y = 18; x = 0 54
We can see that the minimum value now occurs at (2,3) which is 34, so both the optimal solution and optimal value have changed in this case.
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D)
When we limit the range of the variables as -
4 \leq X \leq 8 \:\: and\:\: 12\leq Y \leq 24,
the critical region now becomes -
So, the new critical points are (4,12), (4,24), (8,24) and (8,12).
So, the values of the objective function at these points can be calculated as -
Vertex Value of Objective
(4,12) 8*4+12*12 = 176
(4,24) 8*4+12*24 = 320
(8,24) 8*8+12*24 = 352
(8,12) 8*8+12*12 = 208
So, the new optimal solution is (4,12) and the optimal value is 176.
if we knew the range of the variables in the part B and C earlier, we could have just said that the optimal solution will not change as the value would have been no longer depended on the coefficients of variables in the constraints.
