1. Unit commitment involves determining the optimal mix of generators to meet expected demand while satisfying operational constraints like minimum up and down times. It aims to minimize total costs which include start-up costs and variable running costs. 2. The example problem determines the lowest cost combination of 3 generators to produce 550MW of power. Various constraints like minimum generation levels and ramp rates must be considered. 3. Key constraints in unit commitment include minimum and maximum generation limits, minimum up and down times, and ramp rates for changing output. System constraints require matching generation to load while maintaining sufficient operating reserves. Environmental and network limits also factor into the optimization.

1. Unit commitment
Ahmed Mohamed abdel-hakeem Elkholy
Economic operation of power system
March 19, 2016

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1. Introduction
1.1 Definition
determining the mix of generators and their estimated output
level to meet the expected demand of electricity over a given
time horizon (a day or a week), while satisfying constraints such
as ramp rate limits, up-time and down-time constraints, reserve
and energy requirements. While the load profile is given and also
the unit available to work is given.
Example 1.1
To fully illustrate the uc study the following sample example is
shown. There are three unit with following data.
Unit 1:
PMin = 250 MW, PMax = 600 MW
C1 = 510.0 + 7.9 P1 + 0.00172 P12 $/h
Unit 2:
PMin = 200 MW, PMax = 400 MW
C2 = 310.0 + 7.85 P2 + 0.00194 P22 $/h
Unit 3:

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PMin = 150 MW, PMax = 500 MW
C3 = 78.0 + 9.56 P3 + 0.00694 P32 $/h
What combination of units 1, 2 and 3 will produce 550 MW at
minimum cost?
How much should each unit in that combination generate?
Solution:
Taking consideration of above given using excel sheet to solve all
possible scenarios of operations for illustration we will solve one
case manual
Case 6
Scenario is
Unit 1 is off
Unit 1 is on
Unit 1 is on
𝜆 =
𝑝 𝑑 + ∑
𝛽𝑖
2 ∗ 𝛾𝑖
𝑛
𝑖=1
∑
1
2 ∗ 𝛾𝑖
𝑛
𝑖=1
=
550 +
7.85
2 ∗ .00194
+
9.56
2 ∗ .00694
1
2 ∗ .00194
+
1
2 ∗ .00694
= 9.891369369
And the power will be
𝑝 =
𝜆 − 𝛽
2 ∗ 𝛾
Power generated from unit 2 =

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=
9.891369369 − 7.85
2 ∗ .00194
= 526.1261261 𝑀𝑊
To meet constrains the power of unit 2 will be 400 mw
P2=400 mw
The power generated from unit3
=
9.891369369 − 9.56
2 ∗ .00694
= 23.87387387 𝑀𝑊
Also to meet constrains of unit 3 will be 150 mw
P2=150 mw
Also we also cheek constrain of power demand that’s
Pload =p1+p2=400+150=550 mw
The total cost by substituted in cost equation (given)
Total cost will be = 5428.55
This is for one case.
All case in table below

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in above example for one loading case imagine that we can
consider above example for one hour in a day so we have to
calculate for 24 hour imagine for a week or month
so we find it possible to solve uc problem in that way not just for
long calculation but also considering constrains above example
not taking all constrain as we will illustrate in this report .
But we can learn from example 1.1 that
pmin pmax
U1 U2 U2 250 600 U1 U2 U2
α for units 510 310 78 200 400 510 310 78
β for units 7.9 7.85 9.56 150 500 7.9 7.85 9.56
γ for units 0.0017 0.0019 0.0069 0.002 0.002 0.007
case1 0 0 0 #DIV/0! 0 0 0 #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0! #DIV/0!
case2 0 0 1 17.194 150 500 0 0 0 0 0 0 0 0
case3 0 1 0 9.984 200 400 0 0 0 0 0 0 0 0
case4 1 0 0 9.792 250 600 1 550 0 0 5375.3 0 0 5375.3
case5 0 1 1 9.89136937 350 900 1 0 400 150 0 3760.4 1668.15 5428.55
case6 1 1 0 8.87936612 450 1000 1 284.7 265.3 0 2898.538 2529.155 0 5427.693
case7 1 0 1 9.74592148 400 1100 1 400 0 150 3945.2 0 1668.15 5613.35
case8 1 1 1 8.95839745 600 1500
0 0 0 0 0 0 0 0
cost
cost
Units data
λ opration p1 p2 p3

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1. When the few units committed then the units can’t meet
the demand.
2. When the units is not enough to be committed then some
units operate above optimum.
3. When the units is many to be committed then some units
below optimum.
4. When the units is too many to be committed then minimum
generation exceeds demand.
5. No-load cost affects choice of optimal combination.
1.2 Type of units or stations.
If we take considerable load profile like
Load
Time
1260 18 24
500
1000

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And solve for every hour then
LOAD UNIT 1 UNIT 2 UNIT 3
1100 On On On
1000 On On Off
900 On On Off
800 On On Off
700 On On Off
600 On Off Off
500 On Off Off
From solution we find that
1. There is unit will operate all time so we called it base unit.
2. There is unit will operate part time so we called it
intermediate unit
3. There is unit will operate few time so we called it peak unit
As coming figure describe

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2. Constrains
2.1Definition
Constrain is limitations in power system avoiding it cause serious
problem. This limitation can be technical for unit or technical
limitation for power system or can be environmental limitations.
We can classified into
 Unit constrains
 System constrains
 Environmental constraints
 Network Constraints
Load
Time
1260 18 24
Unit 1 or base unit
Unit 2 or intermediate unit
Unit 3 or peak unit

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2.2Unit constrains
2.2.1 Maximum generating capacity
That constrain stat that the power generated from the unit
mustn’t exceed specific value because of thermal stability
of the unit exceeding this constrain cause damage to the
unit. Represented in mathematical formula as below.
𝑥( 𝑖, 𝑡) < 𝑝𝑚𝑎𝑥
Where
𝑥( 𝑖, 𝑡) 𝑖𝑠 𝑜𝑢𝑡𝑝𝑢𝑡 𝑝𝑜𝑤𝑒𝑟 𝑜𝑓𝑡ℎ𝑒 𝑢𝑛𝑖𝑡 𝑖 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑖𝑚𝑒 𝑡
2.2.2 Minimum stable generation
As above constrain the power outage from the unit
mustn’t fall down specific value because of technical
limitation like flam stability in the gas and steam units.
Represented in mathematical formula as below.
𝑥( 𝑖, 𝑡) > 𝑝𝑚𝑖𝑛
2.2.3 Minimum up time
This constrain stat that once the unit is running mustn’t
shut down immediately due technical limitation and
mechanical characteristic of the unit. This constrain
represented mathematically as:
Where
If u(i,t) =1 and ti
up
< ti
up,min
then u(i,t +1) =1

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2.2.4 Minimum down time
This constrain stat that once the unit is running
mustn’t shut down immediately due technical
limitation and mechanical characteristic of the unit.
This constrain represented mathematically as:
2.2.5 Ramp rates
2.2.5.1 Definition
To avoid damaging the turbine, the
electrical output of a unit cannot change by
more than a certain amount over a period of
time.
2.2.5.2 Ramp Up rate
2.2.5.2.1 Start-up ramp rate
According to this constrain the unit
can’t start immediately but taking
time this time called start up time.
2.2.5.2.2 Running up ramp rate
According to this one in case of
running condition. the unit can’t
u(i,t): Status of unit i at period t
Unit i is on during period tu(i,t) =1:
Unit i is off during period tu(i,t)= 0:
If u(i,t) = 0 and ti
down
< ti
down,min
then u(i,t +1) = 0

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immediate changing the power up
without taking time called ramp rate
running up time. The change here
means increasing outage power.
2.2.5.3 Ramp down rate
2.2.5.3.1 Shut down ramp rate
Look like previse constrain the unit take time to
shut down.
2.2.5.3.2 Running down ramp rate
According to this one in case of running condition.
the unit can’t immediate changing the power
down without taking time called ramp rate running
down time. The change here means decreasing
outage power.
2.2.5.4 Mathematical representation
ramp up rate
x i,t +1( )- x i,t( )£ DPi
up,max
ramp down rate
x(i,t)- x(i,t +1) £ DPi
down,max
2.3 Systemconstrains
This constrain that effect more than one unit divide into:
2.3.1 Load / generation balance
Stat as the power generated fromall unit mustbe equal the
load and the losses. Mathematically represented as
∑ 𝑢( 𝑖, 𝑡) ∗ 𝑥( 𝑖, 𝑡) = 𝑙( 𝑡)
𝑛
𝑖=1
𝑤ℎ𝑒𝑟𝑒 𝑙( 𝑡) 𝑖𝑠 𝑡ℎ𝑒 𝑙𝑜𝑎𝑑 𝑝𝑜𝑤𝑒𝑟 𝑎𝑡 𝑡𝑖𝑚𝑒 𝑡
2.3.2 Reserveconstrain
2.3.2.1 Reason to keep reservepower
 Sudden unexpected increasein the load demand
 Forceoutage of some generating units
 Forceoutage of supplementary equipment’s due to stability
problem
 Underestimating the load due to error in load for casting
 Local shortagein the generated power

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2.3.2.2 Type of reserve
2.3.2.2.1 Cold reserve
Cold reserveis unit kept reserved for servicebut they are
not available in the immediate loading.
2.3.2.2.2 Hot (spinning) reserve
Hot reserverefers to the extra amount of capacity that
unit can provideimmediately when require.
2.3.2.2.3 Operating reserve
Operating reserveis referring to the capacity that already
in service in excess of peak demand.
2.3.2.3 Other sourceof reserve
There is other sourceof reservelike
 Pumped hydro plants
 Demand reduction (e.g. voluntary load shedding)
2.3.2.4 Condition of reserve
 Reservemust be higher than largest unit
 Should be spread around the network
 Must be able to deploy reserveeven if the network
is congested
 The unit mustoperate at 80-85% of its rated
2.4 Network constrain
Transmission network may havean effect on the
commitment of units because of
 Some units mustrun to providevoltage support
 The output of some units may be limited becausetheir
output would exceed the transmission capacity of the
network
2.5Environmental constrain
uc study is effected by environmentalconstrains becauseof
 constraints on pollutants such SO2, NOxvarious forms:
o Limit on each plant at each hour
o Limit on plant over a year
o Limit on a group of plants over a year
 Constraints on hydro generation

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o Protection of wildlife
o Navigation, recreation
2.6Cost constrains
Cost constrain taking two type of costin consideration
2.6.1 Start-up cost
Start-up cost depends on varicosefactor like
 warming up becausethe unit can’t bring on line immediately
 Start-up cost depends on time unit has been off
2.6.2 Running cost
A balance between start-up costs and running costs is
important becauseof
 How long should a unit run to “recover” its start-up
cost?
 Start-up one more large unit or a diesel generator
to cover the peak?
 Shutdown one moreunit at night or run several
units’ part-loaded?
Example:
 Diesel generator: low start-up cost, high running
cost

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 Coal plant: high start-up cost, low running cost
3 Summary
The summary of all above that
• Some constraints link periods together
• Minimizing the total cost (start-up + running) mustbe done over the whole
period of study
• Generation scheduling or unit commitment is a more general problem than
economic dispatch
• Economic dispatch is a sub-problemof generation scheduling
4 Type of units according flexibility
4.1Flexible unit
Power output can be adjusting within limits
Example of this unit
– Coal-fired
– Oil-fired
– Open cycle gas turbines
– Combined cycle gas turbines
– Hydro plants with storage
4.2Inflexible unit
Power output cannot be adjusted for technical or commercial reasons
Example of this unit
– Nuclear
– Run-of-the-river hydro
– Renewables (wind, solar,…….)
– Combined heat and power (CHP, cogeneration)
4.2.1.1
4.2.1.2
4.2.2
4.3
4.4
4.4.1.1
4.4.1.2
4.4.1.2.1

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4.4.1.2.2
4.4.1.3