Electronics produces memory units for specialized applications. Unit A provides 2 GB of high-bandwidth memory and 15 GB of low-bandwidth memory. Unit B provides 0.5 GB of high-bandwidth memory and 60 GB of low-bandwidth memory. Unit A costs $12.50 and Unit B costs $7.50. A particular application requires at least 35 GB of high-bandwidth memory, and 648 GB of low-bandwidth memory, and cannot use more than 28 individual units. How many of each memory unit should Electronics use to meet this requirement and minimize the cost of the product? Formulate a linear programming model for this situation. Solve this model using graphical analysis. Display your graph and the solution parameters. Create your graph using Microsoft Word or Excel.
Linear programming model: memory units for specialized applications.
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What are your thoughts?
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my work on questions regarding linear programming model
12.5x + 7.5y
Mini(12.5x + 7.5y)
Unit A: x<=28
Unit B: y<=28
High band: 2x + 0.5y > = 35
Low band: 15x + 60y > = 648
Unit 5: x is integer
Unit 6: y is integer
Unit A count is 16
Unit B count is 7
Total Cost: $252.50
High Band: 35.5 GB
Low Band: 660 GB
I also have a graph that I cannot seem to copy here. My professor said the graph was fine but then added this:
[FONT="]The graph should show three lines, intersecting at the extreme points (16, 6), (14, 14), and (24.57, 3.43). The feasible region is the area between these three lines. You need to find out The minimum value of the objective function occurs for 16 copies of Unit A and 6 copies of Unit B.
So I sent him this:
[/FONT]
[FONT="]I do not know what I am getting wrong....
[/FONT]
12.5x + 7.5y
Mini(12.5x + 7.5y)
Unit A: x<=28
Unit B: y<=28
High band: 2x + 0.5y > = 35
Low band: 15x + 60y > = 648
Unit 5: x is integer
Unit 6: y is integer
| Given | | | | Optimal Solution | | | |
| | High Band | Low Band | Cost | Units | Cost | High band | Low band |
| Unit A (x) | 2 | 15 | 12.5 | 16 | 200 | 32 | 240 |
| Unit B (y) | 0.5 | 60 | 7.5 | 7 | 52.5 | 3.5 | 420 |
| | | | | | 252.5 | 35.5 | 660 |
Unit A count is 16
Unit B count is 7
Total Cost: $252.50
High Band: 35.5 GB
Low Band: 660 GB
I also have a graph that I cannot seem to copy here. My professor said the graph was fine but then added this:
[FONT="]The graph should show three lines, intersecting at the extreme points (16, 6), (14, 14), and (24.57, 3.43). The feasible region is the area between these three lines. You need to find out The minimum value of the objective function occurs for 16 copies of Unit A and 6 copies of Unit B.
So I sent him this:
[/FONT]
| Unit A | Unit B | |||||||
| A | B | C | RHS | SLACK | ||||
| Cost | 12.5 | 7.5 | 248.7333 | |||||
| high-bandwidth memory | 2 | 0.5 | 35 | >= | 35 | 0 | ||
| low-bandwidth memory | 15 | 60 | 648 | >= | 648 | 0 | ||
| number of units | 1 | 1 | 22.64 | <= | 28 | 5.36 | ||
| solution values | 15.78667 | 6.853333 | ||||||
[/FONT]
| Unit A | Unit B | |||||||
| A | B | C | RHS | SLACK | ||||
| Cost | 12.5 | 7.5 | 248.7333 | |||||
| high-bandwidth memory | 2 | 0.5 | 35 | >= | 35 | 0 | ||
| low-bandwidth memory | 15 | 60 | 648 | >= | 648 | 0 | ||
| number of units | 1 | 1 | 22.64 | <= | 28 | 5.36 | ||
| solution values | 15.78667 | 6.853333 | ||||||
HallsofIvy
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So, letting "A" be the number of units A and "B" the number of units B, we want to minimize 12.5A+ 7.5B.
So we must have \(\displaystyle 2A+ 0.5B\ge 35\).
\(\displaystyle 15A+ 60B\ge 648\)
and \(\displaystyle A+ B\le 28\)
The simplest way to do this is, as the problem suggests, to graph the constraints,
The region in which \(\displaystyle 2A+ 0.5B\ge 35\) is bounded below by the line \(\displaystyle 2A+ 0.5B= 35\). Multiplying by 2, that is \(\displaystyle B= 4A+ 70\). When B= 0, A= 35/2= 17.5 and when A= 0, 0.5B= 35 so B= 70. The graph is the line through (17.5, 0) and (0, 70).
The region in which \(\displaystyle 15A+ 60B\ge 648\) is bounded below by the line \(\displaystyle 15A+ 60B= 648\). Dividing by 60, that is \(\displaystyle B= 0.25A+ 10.8\). When B= 0, A= 648/15= 43.2. When A= 0, B= 640/60= 10.8. The graph is the line through (0, 10.8) and (43.2, 0).
The region in which \(\displaystyle A+ B\le 28\) is bounded above by the line A+ B= 28 or B= 28- A. When B= 0, A= 28. When A= 0, B= 28.
Graphing those (I just did it by hand on a piece of scrap paper) the "feasible region" (the region in which all three constraits are satisfied) is the triangle bounded by those three lines. The basic idea behind 'linear programming' is that the max or min of a linear function on a polygonal region must be at one of the vertices of the region.
So you want to
1) Determine (A, B) such that B= 4A+ 70 and B= 0.25A+ 10.8.
2) Determine (A, B) such that B= 4A+ 70 and B= 28- A.
3) Determine (A, B) such that B= 0.24A+ 10.8 and B= 28.A
Then evaluate 12.5A+ 7.5B at each of those three points. Which makes it a minimum"