Introduction to mathematical programming
4th Edition
ISBN: 9780534359645
Author: Jeffrey B. Goldberg
Publisher: Cengage Learning
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Expert Solution & Answer
Chapter 4, Problem 10RP
Explanation of Solution
Optimal solution:
- The LP for the Dakota problem is given as:
- Maximize z= 60x1+30x2+20x3
- Such that,
- 8x1+6x2+x3 ≤ 48 Lumber constraint
- 4x1+2x2+1.5x3 ≤ 20 Finishing constraint
- 2x1+1.5x2+0.5x3 ≤ 8 Carpentry constraint
- x1,x2,x3 ≥ 0 Non-negativity constraint
- Now it is known that the concept based on the theory of Linear
Programming suggests that for an LP with n decision variables to be degenerate, n+1 or moreof the LP’s constraints must be binding at extreme point...
Expert Solution & Answer
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1. Consider an instance of the Knapsack Problem without repetitions with 4 items, having
weights and values as follows. The weights (in pounds) are w1=2, w2 =7, w3 =10, w4 =12.
The dollar values of these items are respectively v1 = 12, v2 = 28, v3 = 30, v4 = 5. The
capacity of the knapsack is 12.
(a) Find the optimal solution for Fractional Knapsack.
(b) Find the optimal solution for 0-1 Knapsack.
Suppose that there are four items available which can be put into a knapsack that has a capacity of 13 pounds. The weights of the items are 5,7,4, and 3 pounds respectively. Their utilities are 8,11,6 and 4 respectively. Find the optimal solution that maximizes the total utility of the knapsack.
Consider the following 0-1 knapsack problem, the items, their weights and their profits appear in table below. If the knapsack has capacity 11 select the items that appear in the solution? i.e. select answers 1, 2,and 3 if those are the items in the optimal
knapsack.
Item
2
3
5
6
Weight
1
2
3
4
5
6
Profit
10
13
15
U item 1
U item 2
U item 3
U item 4
U item 5
U item 6
O O O O O O
Chapter 4 Solutions
Introduction to mathematical programming
Ch. 4.1 - Prob. 1PCh. 4.1 - Prob. 2PCh. 4.1 - Prob. 3PCh. 4.4 - Prob. 1PCh. 4.4 - Prob. 2PCh. 4.4 - Prob. 3PCh. 4.4 - Prob. 4PCh. 4.4 - Prob. 5PCh. 4.4 - Prob. 6PCh. 4.4 - Prob. 7P
Ch. 4.5 - Prob. 1PCh. 4.5 - Prob. 2PCh. 4.5 - Prob. 3PCh. 4.5 - Prob. 4PCh. 4.5 - Prob. 5PCh. 4.5 - Prob. 6PCh. 4.5 - Prob. 7PCh. 4.6 - Prob. 1PCh. 4.6 - Prob. 2PCh. 4.6 - Prob. 3PCh. 4.6 - Prob. 4PCh. 4.7 - Prob. 1PCh. 4.7 - Prob. 2PCh. 4.7 - Prob. 3PCh. 4.7 - Prob. 4PCh. 4.7 - Prob. 5PCh. 4.7 - Prob. 6PCh. 4.7 - Prob. 7PCh. 4.7 - Prob. 8PCh. 4.7 - Prob. 9PCh. 4.8 - Prob. 1PCh. 4.8 - Prob. 2PCh. 4.8 - Prob. 3PCh. 4.8 - Prob. 4PCh. 4.8 - Prob. 5PCh. 4.8 - Prob. 6PCh. 4.10 - Prob. 1PCh. 4.10 - Prob. 2PCh. 4.10 - Prob. 3PCh. 4.10 - Prob. 4PCh. 4.10 - Prob. 5PCh. 4.11 - Prob. 1PCh. 4.11 - Prob. 2PCh. 4.11 - Prob. 3PCh. 4.11 - Prob. 4PCh. 4.11 - Prob. 5PCh. 4.11 - Prob. 6PCh. 4.12 - Prob. 1PCh. 4.12 - Prob. 2PCh. 4.12 - Prob. 3PCh. 4.12 - Prob. 4PCh. 4.12 - Prob. 5PCh. 4.12 - Prob. 6PCh. 4.13 - Prob. 2PCh. 4.14 - Prob. 1PCh. 4.14 - Prob. 2PCh. 4.14 - Prob. 3PCh. 4.14 - Prob. 4PCh. 4.14 - Prob. 5PCh. 4.14 - Prob. 6PCh. 4.14 - Prob. 7PCh. 4.16 - Prob. 1PCh. 4.16 - Prob. 2PCh. 4.16 - Prob. 3PCh. 4.16 - Prob. 5PCh. 4.16 - Prob. 7PCh. 4.16 - Prob. 8PCh. 4.16 - Prob. 9PCh. 4.16 - Prob. 10PCh. 4.16 - Prob. 11PCh. 4.16 - Prob. 12PCh. 4.16 - Prob. 13PCh. 4.16 - Prob. 14PCh. 4.17 - Prob. 1PCh. 4.17 - Prob. 2PCh. 4.17 - Prob. 3PCh. 4.17 - Prob. 4PCh. 4.17 - Prob. 5PCh. 4.17 - Prob. 7PCh. 4.17 - Prob. 8PCh. 4 - Prob. 1RPCh. 4 - Prob. 2RPCh. 4 - Prob. 3RPCh. 4 - Prob. 4RPCh. 4 - Prob. 5RPCh. 4 - Prob. 6RPCh. 4 - Prob. 7RPCh. 4 - Prob. 8RPCh. 4 - Prob. 9RPCh. 4 - Prob. 10RPCh. 4 - Prob. 12RPCh. 4 - Prob. 13RPCh. 4 - Prob. 14RPCh. 4 - Prob. 16RPCh. 4 - Prob. 17RPCh. 4 - Prob. 18RPCh. 4 - Prob. 19RPCh. 4 - Prob. 20RPCh. 4 - Prob. 21RPCh. 4 - Prob. 22RPCh. 4 - Prob. 23RPCh. 4 - Prob. 24RPCh. 4 - Prob. 26RPCh. 4 - Prob. 27RPCh. 4 - Prob. 28RP
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