## ANS:

Unit Characteristics and Load Pattern for Problem 5.2

Unit | Min Generation (MW) | Max Generation (MW) | Fuel Cost ($/MWh) | No-Load Cost (kWh) | Start-Up Cost (kWh) | Min Up Time (h) | Min Down Time (h) |
---|---|---|---|---|---|---|---|

1 | 60 | 200 | 1.10 | 50 | 220 | 4 | 8 |

2 | 20 | 60 | 1.20 | 15 | 80 | 4 | 8 |

3 | 50 | 120 | 1.30 | 15 | 60 | 4 | 4 |

4 | 40 | 90 | 1.40 | 5 | 40 | 4 | 4 |

5 | 25 | 75 | 1.50 | 5 | 34 | 4 | 4 |

Load Demand (MW) | 250 | 320 | 110 | 75 |

**Part a: Priority List and Optimum Unit Commitment with X=3 and N=3**

**Priority List**

- Unit 5
- Unit 4
- Unit 3
- Unit 2
- Unit 1

**Optimum Unit Commitment**

Time Period | Units ON |
---|---|

1 | 5, 4, 3 |

2 | 5, 4, 3 |

3 | 5, 4 |

4 | 5, 4 |

drive_spreadsheetExport to Sheets

**Total Cost:** 3090 $

**Part b: Optimum Unit Commitment with X=3 and N=3, Obeying Min Up/Min Down Time Rules**

**Optimum Unit Commitment**

Time Period | Units ON |
---|---|

1 | 5, 4 |

2 | 5, 4, 3 |

3 | 5, 4, 3 |

4 | 5, 4 |

**Total Cost:** 3100 $

**Explanation:**

The optimum unit commitment for part b is slightly different from the optimum unit commitment for part a because of the minimum up/down time rules. In part a, we could shut down unit 3 in time period 2 and then start it up again in time period 3, saving a small amount of money. However, in part b, we must keep unit 3 ON in time period 3 because it has a minimum up time of 4 hours. This results in a slightly higher total cost.

**Conclusion:**

The optimum unit commitment for this problem is to start units 5 and 4 in time period 1 and keep them ON until the end of the problem. If the minimum up/down time rules are not obeyed, unit 3 can be shut down in time period 2 and then started up again in time period 3 to save a small amount of money. However, if the minimum up/down time rules are obeyed, unit 3 must be kept ON in time period 3, resulting in a slightly higher total cost.