This document summarizes a case study where the authors were tasked with creating an operating schedule for power generators to meet electrical load demands over the course of a day. They defined variables to represent the number of generators running by type and time period, the power output by type and time period, and costs. They formulated constraints and an objective function to minimize total costs. By solving the linear program, they found an optimal schedule that met demands at a minimum cost of $2,030,300.

1. CASE STUDY 2
POWER GENERATION
BRIANNE KLEINSCHMIDT, ROY PHELPS, AND CAMERON SHUGHRUE
1. Introduction
For this project, we were tasked with creating an operating schedule for generators to meet given
electrical load demands over the course of a day. The schedule also had to be able to accommodate
a surge potential of 15scheduled demand without the need to bring additional generators online.
Below are the data as given.
12 a.m. to 6 a.m. 15,000 megawatts
6 a.m. to 9 a.m. 30,000 megawatts
9 a.m. to 3 p.m. 25,000 megawatts
3 p.m. to 6 p.m. 40,000 megawatts
6 p.m. to 12 a.m. 27,000 megawatts
Table 1: Electricity load demands by time period
Minimum Maximum Cost/hour $/hour/MW Startup Number
level level at min above min Cost Available
Type 1 850 MW 2000 MW 2000 4 4000 12
Type 2 1250 MW 1750 MW 5200 2.6 2000 10
Type 3 1500 MW 4000 MW 6000 6 1000 5
Table 2: Generator Utilization by time period
2. Process
Our first task was to define our variables. This was done by examining what types of factors
incur cost. Those were noted to be (1) the number of generators started, (2) the number of hours
above minimum level a generator was run, and (3) the number of generators running. Next we
looked what needed to be known to find those costs. For the number of generators started we only
needed to know how many generators were running each time period. The costs varied dependant
on the type of generator. This became our first variable: R, defined as both a function of the type
of generator that was running i, and the time period it was running in, j. Obviously, a generator
Date: November 25, 2013.
1

2. 2 BRIANNE KLEINSCHMIDT, ROY PHELPS, AND CAMERON SHUGHRUE
is either running or not, thus R needs to be an integer variable. To find the number of megawatt-
hours above minimum level that generators were run, we needed to know how many generators
were running of each type, and how much power the generators of one type were putting out all
together. This led to our second variable for power produced per hour: P, also dependent on i,j.
Now we were able to define all of our costs in terms of variables:
generators started = Si,j in terms of Ri,j
num hours above min = Ai,j in terms of Pi,j and Ri,j
num of generators running = Ri,j
This gives us all of our variables with their definitions:
Ri,j = # of generating units of type i running in timeframe j: i ∈ [1, 3],
j ∈ [1, 5], integer
Si,j = # of generating units of type i started in timeframe j: i ∈ [1, 3],
j ∈ [1, 5], integer
Ai,j = MW/hr above minimum generator level for generating units of type
i during timeframe j: i ∈ [1, 3], j ∈ [1, 5]
Pi,j = hourly output from generating units of type i during timeframe j:
i ∈ [1, 3], j ∈ [1, 5]
Our objection function can now be written:
(1)
5
j=1
2000 · hours · R1,j + 5200 · hours · R2,j + 6000 · hours · R3,j
+ 4000 · S1,j + 2000 · S2,j + 1000 · S3,j
+ 4 · hours · A1,j + 2.6 · hours · A2,j + 6 · hours · A3,j
The constraints of the linear problem are broken up by time period, and are nearly identical for
each. The power generated must meet demand,
(2)
3
i=1
Pi,j ≥ demandi,
and possible power using running generators must meet 115% of demand:
(3) 2000R1,j + 1750R2,j + 4000R3,j ≥ 1.15demandi.
Hourly output must be within the generator’s minimum and maximum range:
(4) Pi,j ≥ (minimum output)i · Ri,j,
(5) Pi,j ≤ (maximum output)i · Ri,j
and the number generators of each type are limited:
R1,j ≤ 12,(6)
R2,j ≤ 10,(7)
R3,j ≤ 5.(8)

3. POWER GENERATION 3
We found the amount of power over minimum of each type of generator by subtracting the
minimum output of the running generators from the actual output that time period:
(9) Ai,j ≤ Pi,j − (minimum output)i · Ri,j
The final constraint is the definition of Si,j. This constraint has four cases. The first is the
special case of Period 1, in which there is not a previous time period and all generators running
must be started:
Si,1 = Ri,1
The other cases are:
Ri,j − Ri,j−1 = 0 no generators started,
Ri,j − Ri,j−1 ≥ 0 some generators started,
Ri,j − Ri,j−1 ≤ 0 some generators shut down.
For the equation Si,j = Ri,j − Ri,j−1 the first two cases pose no problem, however, the final case is
an issue as we cannot allow Si,j to be negative. This is solved by rewriting the equation as:
(10) Si,j ≥ Ri,j − Ri,j−1.
Minimizing this LP will still force Si,j to its minimum value in the first two cases causing no
change, however in the third case, Si,j will now equal 0.
3. Results
Upon solving the linear program, we found the optimal value to be $2,030,300 using the schedule
in Table 3.
Number running
Time Type 1 Type 2 Type 3
12 a.m. to 6 a.m. 12 2 0
6 a.m. to 9 a.m. 12 8 0
9 a.m. to 3 p.m. 12 8 0
3 p.m. to 6 p.m. 12 9 2
6 p.m. to 12 a.m. 12 9 0
Table 3: Number of generators of each type running during each time period
Table 4 contains a breakdown of all variables and associated costs by generator type and time
period for the duration of the problem.

4. 4BRIANNEKLEINSCHMIDT,ROYPHELPS,ANDCAMERONSHUGHRUE
# Fixed Output Total Maximum # Starting Total
Running Cost Above Min Cost Output Output Started Cost Cost
Period 1 - 12 a.m. to 6 a.m. (6 hours)
Type 1 12 144,000 1,300 31,200 11,500 24,000 12 48,000 223,200
Type 2 2 62,400 1,000 15,600 3,500 3,500 2 4,000 82,000
Type 3 0 0 0 0 0 0 0 0 0
Totals 14 206,400 2,300 46,800 15,000 27,500 14 52,000 305,200
Output Level Required: 15,000
15% Over Required Output Level: 17,250
Period 2 - 6 a.m. to 9 a.m. (3 hours)
Type 1 12 72,000 5,800 69,600 16,000 24,000 0 0 141,600
Type 2 8 124,800 4,000 31,200 14,000 14,000 6 12,000 168,000
Type 3 0 0 0 0 0 0 0 0 0
Totals 20 196,800 2,300 100,800 30,000 38,000 6 12,000 309,600
Output Level Required: 30,000
15% Over Required Output Level: 34,500
Period 3 - 9 a.m. to 3 p.m. (6 hours)
Type 1 12 144,000 800 19,200 11,000 24,000 0 0 163,200
Type 2 8 249,600 4,000 62,400 14,000 14,000 0 0 312,000
Type 3 0 0 0 0 0 0 0 0 0
Totals 20 393,600 2,300 167,700 25,000 38,000 0 0 475,200
Output Level Required: 25,000
15% Over Required Output Level: 28,750
Period 4 - 3 p.m. to 6 p.m. (3 hours)
Type 1 12 72,000 11,050 132,600 21,250 24,000 0 0 204,600
Type 2 9 140,400 4,500 35,100 15,750 15,750 1 2,000 177,500
Type 3 2 36,000 0 0 3,000 8,000 2 2,000 38,000
Totals 23 248,400 15,550 100,800 40,000 47,750 3 5,000 420,100
Output Level Required: 40,000
15% Over Required Output Level: 46,000
Period 5 - 6 p.m. to 12 a.m. (6 hours)
Type 1 12 144,000 1,050 25,200 11,250 24,000 0 0 169,200
Type 2 9 280,800 4,500 70,200 15,750 15,750 0 0 351,000
Type 3 0 0 0 0 0 0 0 0 0
Totals 21 424,800 5,550 95,400 27,000 39,750 0 0 520,200
Output Level Required: 27,000
15% Over Required Output Level: 31,050
Total Cost 2,030,300
Table 4: Extended table of all variables and costs

5. POWER GENERATION 5
4. Additional Considerations
This problem focused on the very first day of operations of the power generation station. If
demand was found not to vary from day to day, the only modification necessary to the linear
program would be the modification of the constraint to compute Si,1. This would now be based on
the generators running during period 5: Si,1 = Ri,1 − Ri,5.
Department of Mathematics and Statistics, UMBC
E-mail address: bklein2@umbc.edu, p19@umbc.edu, camshu1@umbc.edu