The document describes an optimization problem to minimize the total generation cost of meeting a 500 MW demand across 3 generators, each with different cost functions, using MATLAB. The optimal dispatch found by the MATLAB algorithm was generator 1 at 135 MW, generator 2 at 150 MW, and generator 3 at 215 MW, meeting the total demand while minimizing costs. The total optimal generation cost was calculated to be $5,257.04. Lambda, the incremental cost of power, was also calculated to be $8.245/MW based on the optimal dispatch.

1. Manjot Singh
Sacramento State University
Generation Optimization
Problem Statement:
Bus 1 Bus 2 Bus 3
Bulk Load
Generator 1 Generator 2 Generator 3
Bus 4
In a power system with 3 generation units, the cost functions associated with each generator are:
Generator 1: 𝐶1 = 370 + 7.20 P1 + 0.0040 (P1)2
(MW)
Generator 2: 𝐶2 = 500 + 7.50 P2 + 0.0025 (P2)2
(MW)
Generator 3: 𝐶3 = 550 + 6.74 P3 + 0.0035 (P3)2
(MW)
The total system has a demand of 500 MW.
Design an algorithm in MATLAB to find the optimal dispatch, 𝜆 (lambda) the incremental cost
of power and total cost if generator limitations are as follows:
135 ≤ P1 ≤ 400 (MW)
150 ≤ P2 ≤ 600 (MW)
50 ≤ P3 ≤ 445 (MW) (MW = Megawatts)
Matrices Data Used in MATLAB:
𝐴𝑙𝑝ℎ𝑎(𝛼) = [370, 500, 550]
𝐵𝑒𝑡𝑎(𝛽) = [7.20, 7.50, 6.74]
𝐺𝑎𝑚𝑚𝑎(𝛾) = [0.0040, 0.0025, 0.0035]
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑀𝑖𝑛 = [135, 150, 50]
𝐺𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 𝑀𝑎𝑥 = [400, 600, 445]
𝑆𝑦𝑠𝑡𝑒𝑚 𝐷𝑒𝑚𝑎𝑛𝑑 = [500]

2. MATLAB Algorithm
MATLAB Results

3. Results Summary
Optimal Dispatch:
𝑃1 = 135 𝑀𝑊 𝑃2 = 150 𝑀𝑊 𝑃3 = 215 𝑀𝑊
𝑃1 + 𝑃2 + 𝑃3 = 𝑃𝐷𝑒𝑚𝑎𝑛𝑑
135 + 150 + 215 (𝑀𝑊) = 500 𝑀𝑊
As we can see from the above results obtained from MATLAB the optimal dispatch values are
within the generation limits of each generator. Also if we sum all of the generation values it is
equal to the total system demand, which insures that the solution obtained can meet system
demands while minimizing the costs of generation as much as possible.
Total Generation Cost:
𝐶1 = 370 + 7.20 P1 + 0.0040 (P1)2
= 370 + 7.20 (135) + 0.0040 (135)2
= $ 1,414.9
𝐶2 = 500 + 7.50 P2 + 0.0025 (P2)2
= 500 + 7.50 (150) + 0.0025 (150)2
= $ 1,681.25
𝐶3 = 550 + 6.74 P3 + 0.0035 (P3)2
= 550 + 6.74 (215) + 0.0035 (215)2
= $ 2,160.89
𝑇𝑜𝑡𝑎𝑙 𝐶𝑜𝑠𝑡 = 𝐶1 + 𝐶2 + 𝐶3 = 1,414.9 + 1,681.25 + 2,160.89 ($) = $ 5,257.04
As we can see from the above results and calculations that the total cost calculated matches the
result obtained from MATLAB. This further confirms our results that we obtained for the
Optimal dispatch.
Lambda (𝜆):
𝜆𝑖 = 𝑃(𝑖) ∗ 2𝛾(𝑖) + 𝛽(𝑖)
𝜆1 = 𝑃(1) ∗ 2𝛾(1) + 𝛽(1) = 135 ∗ 2(0.0040) + 7.20 = 8.28
𝜆2 = 𝑃(2) ∗ 2𝛾(2) + 𝛽(2) = 150 ∗ 2(0.0025) + 7.50 = 8.25
𝜆3 = 𝑃(3) ∗ 2𝛾(3) + 𝛽(3) = 215 ∗ 2(0.0035) + 6.74 = 8.245
𝑆𝑦𝑠𝑡𝑒𝑚 𝜆 = 𝑀𝑖𝑛𝑖𝑚𝑢𝑚(𝜆𝑖) = $ 8.245
𝑀𝑊�
Lambda the incremental cost of power is 8.245 dollars per Megawatt. The calculations above
confirm the results obtained from MATLAB and assure that we have found the ideal solution.
This means that as we increase or decrease the generation of each generator around the optimal
dispatch for this operating point, we can see a direct relationship in the cost. For example, if we
increase the generation at generator 1 from 135 MW to 136 MW the total cost will go up by
about $8.245 We must also keep in mind that varying the operating point by a drastic amount
will have a noticeable effect on lambda but for small changes around a local operating point it
can be assumed to be a constant.