The decision variables represent the amounts of ingredients 1, 2, and 3 to put into a blend. The objective function represents profit. The first three constraints measure the usage and availability of resources A, B, and C. The fourth constraint is a minimum requirement for ingredient 3. Use the output to answer these questions. a. How much of ingredient 1 will be put into the blend? b. How much of ingredient 2 will be put into the blend? c. How much of ingredient 3 will be put into the blend? d. How much resource A is used? e. How much resource B will be left unused? f. What will the profit be? g. What will happen to the solution if the profit from ingredient 2 drops to 4? h. What will happen to the solution if the profit from ingredient 3 increases by 1? i. What will happen to the solution if the amount of resource C increases by 2? j. What will happen to the solution if the minimum requirement for ingredient 3 increases to 15? LINEAR PROGRAMMING PROBLEMMAX 4X1+6X2+7X3S.T. 1) 3X1+2X2+5X3<120 2) 1X1+3X2+3X3<80 3) 5X1+5X2+8X3<160 4) +1X3>10 OPTIMAL SOLUTIONObjective Function Value = 166.000 Variable Value Reduced Cost X1 0.000 2.000 X2 16.000 0.000 X3 10.000 0.000 Constraint Slack/Surplus Dual Price 1 38.000 0.000 2 2.000 0.000 3 0.000 1.200 4 0.000 -2.600 OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit X1 No Lower Limit 4.000 6.000 X2 4.375 6.000 No Upper Limit X3 No Lower Limit 7.000 9.600 RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit 1 82.000 120.000 No Upper Limit 2 78.000 80.000 No Upper Limit 3 80.000 160.000 163.333 4 8.889 10.000 20.000
The decision variables represent the amounts of ingredients 1, 2, and 3 to put into a blend. The objective function represents profit. The first three constraints measure the usage and availability of resources A, B, and C. The fourth constraint is a minimum requirement for ingredient 3. Use the output to answer these questions. a. How much of ingredient 1 will be put into the blend? b. How much of ingredient 2 will be put into the blend? c. How much of ingredient 3 will be put into the blend? d. How much resource A is used? e. How much resource B will be left unused? f. What will the profit be? g. What will happen to the solution if the profit from ingredient 2 drops to 4? h. What will happen to the solution if the profit from ingredient 3 increases by 1? i. What will happen to the solution if the amount of resource C increases by 2? j. What will happen to the solution if the minimum requirement for ingredient 3 increases to 15? LINEAR PROGRAMMING PROBLEMMAX 4X1+6X2+7X3S.T. 1) 3X1+2X2+5X3<120 2) 1X1+3X2+3X3<80 3) 5X1+5X2+8X3<160 4) +1X3>10 OPTIMAL SOLUTIONObjective Function Value = 166.000 Variable Value Reduced Cost X1 0.000 2.000 X2 16.000 0.000 X3 10.000 0.000 Constraint Slack/Surplus Dual Price 1 38.000 0.000 2 2.000 0.000 3 0.000 1.200 4 0.000 -2.600 OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit X1 No Lower Limit 4.000 6.000 X2 4.375 6.000 No Upper Limit X3 No Lower Limit 7.000 9.600 RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit 1 82.000 120.000 No Upper Limit 2 78.000 80.000 No Upper Limit 3 80.000 160.000 163.333 4 8.889 10.000 20.000
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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The decision variables represent the amounts of ingredients 1, 2, and 3 to put into a blend. The objective function represents profit. The first three constraints measure the usage and availability of resources A, B, and C. The fourth constraint is a minimum requirement for ingredient 3. Use the output to answer these questions.
a. | How much of ingredient 1 will be put into the blend? |
b. | How much of ingredient 2 will be put into the blend? |
c. | How much of ingredient 3 will be put into the blend? |
d. | How much resource A is used? |
e. | How much resource B will be left unused? |
f. | What will the profit be? |
g. | What will happen to the solution if the profit from ingredient 2 drops to 4? |
h. | What will happen to the solution if the profit from ingredient 3 increases by 1? |
i. | What will happen to the solution if the amount of resource C increases by 2? |
j. | What will happen to the solution if the minimum requirement for ingredient 3 increases to 15? |
LINEAR PROGRAMMING PROBLEM
MAX 4X1+6X2+7X3
S.T.
1) 3X1+2X2+5X3<120 | |
2) 1X1+3X2+3X3<80 | |
3) 5X1+5X2+8X3<160 | |
4) +1X3>10 | |
OPTIMAL SOLUTION
Objective Function Value = 166.000
Variable
|
Value
|
Reduced Cost
|
X1
|
0.000
|
2.000
|
X2
|
16.000
|
0.000
|
X3
|
10.000
|
0.000
|
Constraint
|
Slack/Surplus
|
Dual Price
|
1
|
38.000
|
0.000
|
2
|
2.000
|
0.000
|
3
|
0.000
|
1.200
|
4
|
0.000
|
-2.600
|
OBJECTIVE COEFFICIENT RANGES
Variable
|
Lower Limit
|
Current Value
|
Upper Limit
|
X1
|
No Lower Limit
|
4.000
|
6.000
|
X2
|
4.375
|
6.000
|
No Upper Limit
|
X3
|
No Lower Limit
|
7.000
|
9.600
|
RIGHT HAND SIDE RANGES
Constraint
|
Lower Limit
|
Current Value
|
Upper Limit
|
1
|
82.000
|
120.000
|
No Upper Limit
|
2
|
78.000
|
80.000
|
No Upper Limit
|
3
|
80.000
|
160.000
|
163.333
|
4
|
8.889
|
10.000
|
20.000
|
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