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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:**

T1,smallest >= T2,largest AND/OR T3,smallest >= T2,largest

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**The easy solution: how-to...**

Add: T1 + T2 & T2 + T3 Use these sums to do Johnson’s rule.

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**Example: 3 machines Duration (hrs.) Job Mach 1 Mach 2 Mach 3 B 5 3 7**

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**Example: 3 machines Duration (hrs.) Job Mach 1 Mach 2 Mach 3 A 13 5 9**

One of the conditions is met. Duration (hrs.) Job Mach 1 Mach 2 Mach 3 A B C D

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**Example: 3 machines Duration (hrs.) Job Mach 1+2 Mach 2+3 A 13+5 5+9**

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**Example: 3 machines Duration (hrs.) Job Mach 1+2 Mach 2+3 A 18 14**

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**Example: 3 machines Duration (hrs.) Job Mach 1+2 Mach 2+3 A 18 14**

B goes first.

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**Example: 3 machines Duration (hrs.) Job Mach 1+2 Mach 2+3 B 8 10**

D goes last.

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**Example: 3 machines Duration (hrs.) Job Mach 1+2 Mach 2+3 B 8 10**

C goes next-to-last.

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**Example: 3 machines Duration (hrs.) Job Mach 1+2 Mach 2+3 B 8 10**

B goes 3rd-to-last.

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**Ex: 3 machines: Gantt Chart**

Duration (hrs.) Job Mach 1 Mach 2 Mach 3 B A C D Time (hrs.) => B B B

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**Ex: 3 machines: Gantt Chart**

Duration (hrs.) Job Mach 1 Mach 2 Mach 3 B A C D Time (hrs.) => B A B A B A

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**Ex: 3 machines: Gantt Chart**

Duration (hrs.) Job Mach 1 Mach 2 Mach 3 B A C D Time (hrs.) => B A C B A C B A C

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**Ex: 3 machines: Gantt Chart**

Duration (hrs.) Job Mach 1 Mach 2 Mach 3 B A C D Time (hrs.) => B A C D B A C D B A C D

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**Ex: 3 machines: Gantt Chart**

Duration (hrs.) Job Mach 1 Mach 2 Mach 3 B A C D 31 33 43 24 28 37 5 8 15 18 23 32 B A C D B A C D B A C D

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