OPERATIONS RESEARCH >INTERNATIONAL EDITI
4th Edition
ISBN: 9780534423629
Author: WINSTON
Publisher: CENGAGE L
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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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I need help to solve a simple problem using Grover’s algorithm, where the solution is not necessarily known beforehand. The problem is a 2×2 binary sudoku with two rules:
• No column may contain the same value twice.
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We want to design a quantum circuit that outputs a valid solution to this sudoku. While using Grover’s algorithm for this task is not necessarily practical, the goal is to demonstrate how classical decision problems can be converted into oracles for Grover’s algorithm.
Turning the Problem into a Circuit
To solve this, an oracle needs to be created that helps identify valid solutions. The first step is to construct a classical function within a quantum circuit that checks whether a given state satisfies the sudoku rules.
Since we need to check both columns and rows, there are four conditions to verify:
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using r language
I need help to solve a simple problem using Grover’s algorithm, where the solution is not necessarily known beforehand. The problem is a 2×2 binary sudoku with two rules:
• No column may contain the same value twice.
• No row may contain the same value twice.
Each square in the sudoku is assigned to a variable as follows:
We want to design a quantum circuit that outputs a valid solution to this sudoku. While using Grover’s algorithm for this task is not necessarily practical, the goal is to demonstrate how classical decision problems can be converted into oracles for Grover’s algorithm.
Turning the Problem into a Circuit
To solve this, an oracle needs to be created that helps identify valid solutions. The first step is to construct a classical function within a quantum circuit that checks whether a given state satisfies the sudoku rules.
Since we need to check both columns and rows, there are four conditions to verify:
v0 ≠ v1 # Check top row
v2 ≠ v3 # Check bottom row…
Chapter 4 Solutions
OPERATIONS RESEARCH >INTERNATIONAL EDITI
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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