Operations Research : Applications and Algorithms
4th Edition
ISBN: 9780534380588
Author: Wayne L. Winston
Publisher: Brooks Cole
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Expert Solution & Answer
Chapter 4, Problem 16RP
a.
Explanation of Solution
- Consider the following variables:
- x1 = items of type 1
- x2 = items of type 2
- Weight of item is a1 lb and item 2 weighs a2 lb.
- Therefore, the constraints is a1x1+a2x2 ≤ b
- Each type 1 item carries a benefit of c1 units, each type 2 item carries a benefit of c2 units...
- x1 = items of type 1
- x2 = items of type 2
b.
Explanation of Solution
- from the result of part (a), it can be observed that the constraint in linear
programming problem is less than or equal to type.
- So, add the slack variable s1 to the constraint.
- Thus, the standard form of the linear programming problem is,
- Max z= c1x1+c2x2
- Subject to constraints,
- a1x1+a2x2+s1 = b
- The initial simplex table is shown below:
z x1 x2 s1 rhs Basic variable 1 -c1 -c2 0 0 z = 0 0 a1 a2 1 b s1 = b
- Now, there are two negative entries in the row0, suppose the most negative entry is in column 2.
- Therefore, choose entering variable to be x2.
- The leaving variable is calculated with the use of the following formula,
- Leaving variable = {(rhs/x1), x1 ≥ 0}
- Consider the below table:
z x1 x2 s1 rhs Basic variable 1 -c1 -c2 0 0 - 0 a1 a2 1 b (b/a2)→
- From the above table, the pivot table is intersection of both entering and leaving variables...
- Max z= c1x1+c2x2
- Subject to constraints,
- a1x1+a2x2+s1 = b
| z | x1 | x2 | s1 | rhs | Basic variable |
| 1 | -c1 | -c2 | 0 | 0 | z = 0 |
| 0 | a1 | a2 | 1 | b | s1 = b |
- Leaving variable = {(rhs/x1), x1 ≥ 0}
| z | x1 | x2 | s1 | rhs | Basic variable |
| 1 | -c1 | -c2 | 0 | 0 | - |
| 0 | a1 | a2 | 1 | b | (b/a2)→ |
c.
Explanation of Solution
- The following are the assumptions that are violated by this formula:
- Deterministic Nature
- Additivity
- Direct Proportionality
- Fractionally
- The follo...
- Deterministic Nature
- Additivity
- Direct Proportionality
- Fractionally
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Chapter 4 Solutions
Operations Research : Applications and Algorithms
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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