Introduction to mathematical programming
Introduction to mathematical programming
4th Edition
ISBN: 9780534359645
Author: Jeffrey B. Goldberg
Publisher: Cengage Learning
bartleby

Concept explainers

Expert Solution & Answer
Book Icon
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...

Blurred answer
Students have asked these similar questions
Consider that there are four items that can fit inside a 13-pound-capacity backpack. The respective weights of the items are 5.7, 4, and 3 pounds. Their respective utilities are 8, 11, 6, and 4 utils. Determine the optimal solution that maximizes the overall usefulness of the backpack.
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.
Two investments with varying cash flows (in thousands of dollars) are available, as shown in the table below. At time 0, $10,000 is available for investment, and at time 1, $7,000 is available. Assuming that r 0.10, set up an LP whose solution maximizes the NPV obtained from these investments. Graphically find the optimal solution to the LP.

Chapter 4 Solutions

Introduction to mathematical programming

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