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Job sequencing on 3 machines. Two conditions for a easy solution: –1. The smallest duration on machine #1 is at least as great as the largest duration on machine #2. –2. The smallest duration on machine #3 is at least as great as the largest duration on machine #2.

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Conditions for an easy solution: T 1,smallest >= T 2,largest AND/OR T 3,smallest >= T 2,largest

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The easy solution: how-to... Add: T 1 + T 2 & T 2 + T 3 Use these sums to do Johnsons rule.

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Example: 3 machines Duration (hrs.) JobMach 1Mach 2Mach 3 B537 A1359 C645 D726

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Example: 3 machines Duration (hrs.) JobMach 1Mach 2Mach 3 A1359 B537 C645 D726 One of the conditions is met.

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Example: 3 machines Duration (hrs.) JobMach 1+2Mach 2+3 A B C D7+22+6

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Example: 3 machines Duration (hrs.) JobMach 1+2Mach 2+3 A1814 B810 C109 D98

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Example: 3 machines Duration (hrs.) JobMach 1+2Mach 2+3 A1814 B810 C109 D98 B goes first.

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Example: 3 machines Duration (hrs.) JobMach 1+2Mach 2+3 B810 A1814 C109 D98 D goes last.

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Example: 3 machines Duration (hrs.) JobMach 1+2Mach 2+3 B810 A1814 C109 D98 C goes next-to-last.

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Example: 3 machines Duration (hrs.) JobMach 1+2Mach 2+3 B810 A1814 C109 D98 B goes 3rd-to-last.

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Ex: 3 machines: Gantt Chart Duration (hrs.) JobMach 1Mach 2Mach 3 B537 A1359 C645 D726 B B B Time (hrs.) =>

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Ex: 3 machines: Gantt Chart Duration (hrs.) JobMach 1Mach 2Mach 3 B537 A1359 C645 D726 B B B A A A Time (hrs.) =>

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Ex: 3 machines: Gantt Chart Duration (hrs.) JobMach 1Mach 2Mach 3 B537 A1359 C645 D726 B B B A A A C C C Time (hrs.) =>

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Ex: 3 machines: Gantt Chart Duration (hrs.) JobMach 1Mach 2Mach 3 B537 A1359 C645 D726 B B B A A A C C CD D D Time (hrs.) =>

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Ex: 3 machines: Gantt Chart Duration (hrs.) JobMach 1Mach 2Mach 3 B537 A1359 C645 D726 B B B A A A C C CD D D

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