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59PM Simulate Soil Consolidation
创建日期：2024-04-11 15:23:55
所属分类：物理Physical
标签： 59PM
Simulate Soil Consolidation

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Simulate Soil Consolidation

You are tasked to write the parts needed for the simulation of soil consolidation.
Changes to description
This document is being updated:
Students should refer to the Ed website for any clari�cations, or changes to being made to the
assignment description.
A brief list of changes is documented here.
9 May 2021
CLAY_COMPRESSIBILITY_RATE is 0.001. It is a constant de�ning water units moved per 1kN per
hour (no longer 100kN)
A message to all students about Academic Integrity
This is an assignment and sta� are not permitted to give speci�c guidance on your code, or how to
solve the speci�c problem. That is the purpose of the assessment that you are required to perform to
achieve the grade.
You may ask clari�cation questions about the assignment description. This is often necessary to
implement functionality that is otherwise ambiguous.
The assignment description is not intended to be complete and you can con�rm your assumptions in
a form of a question. In asking the question you should be quoting the description you are asking
about.
If you have a question to ask on Ed please search before asking. However, remember that you
should not be posting any assignment code publicly, as this would constitute academic
dishonesty.
Also, do not wait too long before starting. This assignment needs time and sustained e�ort.
Background
Consolidation is the gradual changes in volume of a partly or fully saturated soil when subject to a
sustained load. The changes are mainly due to the removal of gases, �uids and organic matter from
the soil. We simplify our model to consider this matter as water.
A sample of soil shows the particles arranged with voids between them.
Soil can be composed of many minerals, primary silica, but clay, sand, shale, rock. Some of these
have more water content than others. Water is present within soil and we can assume water is an
incompressible �uid, where any pressure applied will cause the water to move to a lower pressure.
The water is e�ectively squeezed out of the soil very slowly.
Soil consolidation has a huge impact on the planning and construction of buildings throughout
history. The leaning tower of Pisa is a great example to showcase the importance of soil consolidation
The purpose of the simulation is to determine:
how long consolidation will take for a given load and placement over a soil con�guration
how much water is displaced during consolidation
what are the changes in height of the soil after consolidation
Simulation Input and Output
functionality for �le input/output provided for you
The simulation will require information about the nature of the soil, the load, and the parameters
used to simulate. These are read from a �le using the three command line arguments.
$ simulate.py h>
Your program will read in a �le for the simulation parameters from argument 1
Your program will read in a �le for the soil geometry and composition from argument 2
Your program will write to a �le for the results of the simulation from argument 3
For example:
$ simulate.py tests/params_example1.in tests/soil_example1.in sim_results_p1_s1.txt
Modelling of the problem
The modelling of soil consolidation in this assignment makes the following assumptions:
soil particles have no air
soil particles initially have a capacity to hold water and is considered full
for example, if clay can hold 40% water, then at the beginning of the simulation, the clay
particle holds 40% water.
water is an incompressible �uid
water moving out of a particle will cause the particle to compress
any amount of water removed from a particle cannot be reintroduced
soil particle categorisation is limited to Clay and Shale and only relevant to the initial conditions
bedrock is an incompressible particle and will always provide an equal and opposite reactive
force
The void particle is a simple characterisation of representing a lower pressure area and it is
assume to have no volume or capacity. A sand column could be represented as a lower
pressure region, but as a void it has no capacity.
We model a particle of soil.
soil consists of a mixture of solid matter (aggregate) and water.
soil particle has a capacity to hold water. This is dictated by the soil type and the water capacity
value [0,1]
soil particle has pressure acting on it.
Initially the soil particle is at a rest state, in equilibrium with its neighbours.
adding force to the soil will cause change in pressure and result in a movement of water
assume a particle is 1 unit wide and 1 unit deep (square)
We model the movement of water.
water is an incompressible �uid.
the entire soil mass is considered as a body of �uid where the pressure is even throughout.
water reaching a void in the soil will cause it to leak out (sink). Void are explicitly de�ned in
every simulation.
pressure acting on the soil will cause the equivalent water mass to leave the exit points of the
soil.
There should always be a leak point, place for water to be displaced.
If soil has no leak points, the pressure will continue to rise and no water movement is
possible until there is a break in those soil barriers such as bedrock. This building of
pressure and breaking is not considered in the model or the simulation.
water can move up into a void, against gravity. This is to reduce complexity of the assignment.
We model the rate at which water moves out of soil:
it is constant (for simplicity)
Water volume moved per 1kN per hour
CLAY_COMPRESSIBILITY_RATE = 0.001
SHALE_COMPRESSIBILITY_RATE = 0.0005
More water cannot be removed from a particle:
if there is 0.04 water in a particle of clay, and 100kN is applied over one hour.
Water moved is 100,000 x 0.001 = 0.1 . 0.1 > 0.04 , and we expect the particle
of clay will have 0.00 units of water.
File formats
Applicable to parse_sim_parameters and parse_soil_data
The �le format contains pairs of Labels and Values , as well as Comments .
# comments
label1
value(s) for label1
label2
value(s) for label2
label3
value(s) for label3
# comments
...
Comments
A comment is identi�ed when the # symbol �rst appears in the line after any whitespace.
# a valid comment
# also a valid comment
### valid comment
-- # INVALID comment
NOT # a # comment
Comments can appear after or before a label/value(s) pair.
# comment1
# more of comment1
label1
value(s) for label1
# comment2
# comment2
label2
value(s) for label2
# comment3
label3
value(s) for label3
# comment4
# more comments...
Labels and Values
A label is the text used to identify the parameter. e.g. Load weight . A value is information for that
label. e.g. 1000
Load weight
1000
Once a label has been identi�ed, the value(s) are always in next lines that follow. A blank line is used
to separate de�nitions of labels and their values.
Labels can appear at any point within the �le. There is no speci�c ordering.
Label matching and identi�cation
Each �le format has a well de�ne set of labels.
All label matching is case insensitive.
Duplicate labels
Labels appearing more than once are permitted, however, only the last value is used.
For example:
Load weight
1000
Load weight
800
During the reading of the �le, the program should print a message to the console stderr Warning
label Load weight defined twice. Using last value of 800
Simulation Parameters �le format
Applicable to parse_sim_parameters
Load location, width
,
Load weight

Load type

Load timing

Load custom data
, , , ...
Format details
Load location, width
2 integers, representing columns starting and how many columns wide
comma separated
must each be positive integers from 0 to 100
Load weight
1 integer representing kilonewtons of weight
must be a positive integer between 0 and 1,000,000
Load type
1 string - Constant or Linear or Custom
Optional, default is Constant .
Load timing
1 integer representing the number of hours the load will apply for Linear
not used for Constant or Custom Load types
must be a positive integer between 0 and 1,000,000
Load custom data
pairs of integer values representing the weight of the load at the given time
Comma separated in the order ,
Used when Custom Load type is de�ned. Otherwise ignored.
Must have an even number of values (a whole number of pairs)
de�nes zero and positive integers
time values de�ned incrementally
no duplicate time values
Example �le
Load location, width
0, 1
Load weight
100
Load type
Linear
Load timing
100
Load location, width
1, 3
Load weight
3860
Load type
Linear
Load timing
56
Load custom data
1,10, 25,50, 75,55, 4,100, 5,150
Soil data �le format
Soil width, depth
,
Soil keys
, , , , ...
Soil data
...
...
... ...
... ...
... ...
...
...
The keys describe a symbol to use for a particular soil category. The width of the symbol is 1
character.
The key description is a string of any size
Example
Soil width, depth
30, 20
Soil keys
C,Clay, H,Shale, B,Bedrock, V,Void
Soil data
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
CCCCCCCCCCCCCCCCCCCCCCCCCVCCCC
HHHHHHHHHCCCCCHHHHHHHHHHHVHHHH
HHHHHHHHHCCCCCHHHHHHHHHHHVHHHH
HHHHHHHHHCCCCCHHHHHHHHHHHHHHHH
HHHHHHHHHHHHBBBBBBBBHCCCCHHHHH
HHCCCCCCHHHHHHHHHHHHHCCCCHHHHH
HHHHHHHHHHHHHHCCCCHHHHHHHHHHHH
HHCCCCCCCCCCCCCCCCCCCHHHHHHHHH
HHHHHHVVHHHCCCCCCCCCCCCCCCCHHH
HHHHHHHHHHHHHHHVVVHHHHHHHHVHHH
HHCCCCCCCCCCCCCCCCCCCHHHHHVHHH
HHHHHHHHHHHCCCCCCCCCCCCCCCCHHH
HHHHHCCCCCCCHHHHHHHCCCCCCCCHHH
CCCCCCCCCHHHHHCCCCHHHHHCCCCCCH
HHHHHCCCCCCCHHHHHHHCCCCCCCCHHH
CCCCCCCCCHHHHHCCCCHHHHHCCCCCCC
HHCCCVVVCCCCCCCCCCCCCHHHHHHHHH
CCCCCCCCCHHHHHCCCCHHHHHCCCCCCC
BBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
A special particle is the void. Void is represented by a V . It represents a space which will allows water
to move to (lower pressure).
Format details
Applicable to parse_soil_data
Soil width, depth
2 integers representing the number of columns and the number of layers (rows) of soil data to
follow
comma separated
must each be positive integers from 0 to 10,000
Soil keys
pairs of character,string for representing a soil particle type by a single character
comma separated
must have an even number of values
must de�ne single character followed by a string
no duplicate values (key or description)
Soil data
string(s)
must have the appropriate number of columns and rows previously de�ned
must be characters de�ned in the Soil keys
all rows are equal length
all columns are equal length
Simulation parameters
Applicable to check_simulation_data, calculate_applicable_load, calculate_current_load,
simulation_start
Load location, width
Where and how wide the load is placed.
The location is the column integer, where 0 is the left most column.
The width is the side of the load
For example:
Load location = 1 Load width = 1
In this document, we will represent load with L to illustrate the load location and width. However,
the symbol L will not be present in the �le formats, or simulation.
L
BCCV
BCCB
BCCB
BBBB
Load location = 2 Load width = 1
L
BCCV
BCCB
BCCB
BBBB
Most of the problems will focus on a single load point, where Load width =1
Distributed load
The weight of the load implies a force. The width of the load implies a distribution of force over area,
pressure.
For example: Load location = 1 Load width = 3
LLL
BCCCV
BCCCB
BCCCB
BBBBB
When simulating, the load will only a�ect the body of water which are under load
For example: Load location = 0 Load width = 3
LLL
BCCCV
BCCCB
BCCCB
BBBBB
Load on Column 0 has no e�ect in this model Load on Column 1 has e�ect in this model Load on
Column 2 has e�ect in this model
For simpli�cation, you may assume that where a column is not all bedrock, the load will a�ect it.
For example: Load location = 1 Load width = 3
LLL
BCCCV
BBCBB
BCCCB
BBBBB
All columns a�ected
Hint: You should focus on load width of 1 to simulate �rst before moving to distributed loads
Load type and timing
Constant load is �xed. Every hour will impose a force of Load weight at the load point(s).
Linear. At time zero, there is zero load. At time Load timing , there will be 100% of the Load weight
at the load point(s). For example, if Load weight=100 and Load timing = 10 . It would take 10
hours for the load to reach 100kN.
t=0, load=0
t=1, load=10
t=2, load=20
t=3, load=30
t=4, load=40
t=5, load=50
t=6, load=60
t=7, load=70
t=8, load=80
t=9, load=90
t=10, load=100
This a�ects the rate of consolidation.
Custom load are user speci�ed times and load weights to be used in the simulation. For example, the
user may specify:
t=0, load=100
t=1, load=120
t=2, load=180
t=3, load=200
t=4, load=150
t=5, load=110
The user may alternatively specify sparse time sequences:
t=0, load=0
t=1, load=500
t=6, load=1300
t=14, load=2000
t=15, load=1500
t=23, load=400
For the simulation, you may assume that the times not speci�ed will use the same value as the
previous time.
t=0, load=0
t=1, load=500
t=2-5, load=500 # this is implicit
t=6, load=1300
t=7-13, load=1300 # this is implicit
t=14, load=2000
t=15, load=1500
t=16-22, load=1300 # this is implicit
t=23, load=400
t=24-end of simulation, 400 # this is implicit
The only error checking you need for custom data is where there are missing pairs of time/load
values. Otherwise, you may assume that the custom data is correct.
A group of connected soil particles as a body of water
Particles of clay that are adjacent will share the same pressure and push of water throughout.
Therefore, the simulation requires knowing which particles constitute a group, or a body of water.
Example: 1 body of water. All clay particles C are adjacent to all other clay C particles. A pressure
acting on any clay particle will be transmissible throughout the body of water.
BCCCCCCCCCV
BCCCBCCBCBB
BCCCBCCBCBB
BBBBBBBBBBB
Example: 3 bodies of water. There are bedrock particles creating a barrier between the clay particles.
BCCCBCCCCCB
BCCCBCCBBBB
BCCCBCCBCBB
BBBBBBBBBBB
Important:
You do not have to calculate a body of water
You do not have to consider the case where a load is applied to 2 or more bodies of water.
This is calculated for you using the function
leak_points, fluid_body_particles = find_leak_points(start_particle)
Every hour of the simulation the body of water will lose an amount of water related to the load
applied.
The formula for calculating the water moved in 1 hour relates to how much load is applied at the
time, how much is applicable to the soil (exclude bedrock), and the rate of water movement out of
soil particles (constant).
water_moved = CLAY_COMPRESSIBILITY_RATE x load_applicable
Consistency checking between parameters and soil
Having both parameters and soil data, we can test whether these conditions are satis�ed before
proceeding.
Load over soil
Load geometry must be positioned entirely over soil (Load geometry cannot be overhanging the
edge)
Example of not permitted:
L
VCC
BCC
BCC
Load not overhanging
Load must be placed over compressible soil particles (e.g. Clay)
Example of not permitted:
LLL
VCC
BCC
BCC
Simulation examples
Example 1
Consider a column of clay soil with a force applied vertically.
Let the load L be 100kN
The water volume and capacity for clay is 0.4 in this example.
The rate of water movement from clay is 0.001 per hour for every 1kN of force.
The sides of the column are empty, meaning water can leak there in�nitely
There are 4 leak points in this example Void (V) particles
Initial conditions can be represented as so:
L
VCV
VCV
BBB
Initial conditions, height of middle column = 3
0 1 0
0 1 0
1 1 1
Initial conditions, as numeric values of water capacity:
i 0.4 i
i 0.4 i
0 0 0
where i is in�nite
Initial conditions, as numeric values of water volume:
0 0.4 0
0 0.4 0
0 0 0
After 1 hour
There is 0.1 of water removed per hour on the entire soil mass. The soil pressure is even throughout,
and 0.1 water move to the 4 leak points.
Water leaked = 0.1 / 4 = 0.025
capacity (always the same)
i 0.4 i
i 0.4 i
0 0 0
water volume
0.025 0.35 0.025
0.025 0.35 0.025
0 0 0
0.1 water units removed from the soil in this hour 0.1 total water units removed since t=0
After 2 hours
water volume
0.050 0.3 0.050
0.050 0.3 0.050
0 0 0
0.1 water units removed from the soil in this hour 0.2 total water units removed since t=0
After 3 hours
water volume
0.075 0.25 0.075
0.075 0.25 0.075
0 0 0
0.1 water units removed from the soil in this hour 0.3 total water units removed since t=0
After 4 hours
water volume
0.100 0.20 0.100
0.100 0.20 0.100
0 0 0
0.1 water units removed from the soil in this hour 0.4 total water units removed since t=0
After 5 hours
water volume
0.125 0.15 0.125
0.125 0.15 0.125
0 0 0
0.1 water units removed from the soil in this hour 0.5 total water units removed since t=0
After 6 hours
water volume
0.150 0.10 0.150
0.150 0.10 0.150
0 0 0
0.1 water units removed from the soil in this hour 0.6 total water units removed since t=0
After 7 hours
water volume
0.175 0.05 0.175
0.175 0.05 0.175
0 0 0
0.1 water units removed from the soil in this hour 0.7 total water units removed since t=0
After 8 hours
water volume
0.200 0.00 0.200
0.200 0.00 0.200
0 0 0
0.1 water units removed from the soil in this hour 0.8 total water units removed since t=0
Final result
height of middle column = 2.2
0 0.6 0
0 0.6 0
1 1 1
0.8 units of water has been removed from the soil.
Example 2
So long as there is a lower force area, water can be pushed out.
Consider two columns of clay soil surrounded by bedrock with a vertical force applied on the left
column only.
Let the load L be 100kN
The water volume and capacity for clay is 0.4 in this example.
The rate of water movement from clay is 0.001 per hour for every 1kN of force.
There is one leak point in this example Void (V) particle. water �ows toward this point.
Initial conditions can be represented as so:
L
BCBCV
BCBCB
BCCCB
BBBBB
Initial conditions
5 columns:
height of column 1 = 4
height of column 2 = 4
height of column 3 = 4
height of column 4 = 4
height of column 5 = 3
Heights:
4 4 4 4 3
1 1 1 1 0
1 1 1 1 1
1 1 1 1 1
1 1 1 1 1
Initial conditions, as numeric values of water capacity:

0 0.4 0 0.4 i
0 0.4 0 0.4 0
0 0.4 0.4 0.4 0
0 0 0 0 0
Initial conditions, as numeric values of water volume:

0 0.4 0 0.4 0
0 0.4 0 0.4 0
0 0.4 0.4 0.4 0
0 0 0 0 0
After 1 hour
capacity (always the same)
i
0 0.4 0 0.4 0
0 0.4 0 0.4 0
0 0.4 0.4 0.4 0
0 0 0 0 0
Water volume decreases by 0.1 from all particles sharing this force. 7 particles: 0.1 / 7 = 0.014

0 0.386 0 0.386 0.1
0 0.386 0 0.386 0
0 0.386 0.386 0.386 0
0 0 0 0 0
0.1 water units removed from the soil in this hour 0.1 total water units removed since t=0
After 2 hours
water volume

0 0.371 0 0.371 0.2
0 0.371 0 0.371 0
0 0.371 0.371 0.371 0
0 0 0 0 0
0.1 water units removed from the soil in this hour 0.2 total water units removed since t=0
After 3 hours
water volume

0 0.357 0 0.357 0.3
0 0.357 0 0.357 0
0 0.357 0.357 0.357 0
0 0 0 0 0
0.1 water units removed from the soil in this hour 0.3 total water units removed since t=0
...
After 28 hours
water volume

0 0 0 0 2.8
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0.1 water units removed from the soil in this hour 2.8 total water units removed since t=0
Final result
height of column 2 = 2.8
0.6
0.6
0.6
1
height of column 3 = 3.6 (cavity collapses)
1
1
0.6
1
height of column 4 = 2.8
0.6
0.6
0.6
1
2.8 units of water has been removed from the soil.
Example 3
This example will demonstrate the load linearly increasing to the maximum weight.
Consider a column of clay soil with a force applied vertically.
Let the load L be 2450kN
Let the load timing be 10 hours
The water volume and capacity for clay is 0.4 in this example.
The rate of water movement from clay is 0.001 per hour for every 1kN of force.
There are 3 leak points in this example Void (V) particles
Initial conditions can be represented as so:
L
BCCCCV
BCCCCV
BCCCCV
BBBBBB
Initial conditions t=0
5 columns:
height of column 1 = 4
height of column 2 = 4
height of column 3 = 4
height of column 4 = 4
height of column 5 = 4
height of column 6 = 1
Heights:
4 4 4 4 4 1
1 1 1 1 1 0
1 1 1 1 1 0
1 1 1 1 1 0
1 1 1 1 1 1
Initial conditions, as numeric values of water capacity:
0 0.4 0.4 0.4 0.4 i
0 0.4 0.4 0.4 0.4 i
0 0.4 0.4 0.4 0.4 i
0 0 0 0 0 0
Initial conditions, as numeric values of water volume:
0 0.4 0.4 0.4 0.4 0
0 0.4 0.4 0.4 0.4 0
0 0.4 0.4 0.4 0.4 0
0 0 0 0 0 0
After 1 hour t=1
capacity (always the same)
0 0.4 0.4 0.4 0.4 i
0 0.4 0.4 0.4 0.4 i
0 0.4 0.4 0.4 0.4 i
0 0 0 0 0 0
Force applied at this time is: (1/10) * 2450 = 245
Based on the formula: water_moved = CLAY_COMPRESSIBILITY_RATE x load_applicable
water_moved = ( 0.001 * 245 ) = 0.245
Water volume decreases by 0.245 from all particles sharing this force.
12 particles: 0.245 / 12 = 0.020416667 loss each
0 0.380 0.380 0.380 0.380 0.0817
0 0.380 0.380 0.380 0.380 0.0817
0 0.380 0.380 0.380 0.380 0.0817
0 0 0 0 0 0
0.245 water units removed from the soil in this hour 0.245 total water units removed since t=0
After 2 hours
Force applied at this time is: (2/10) * 2450 = 490
water_moved = ( 0.1 * 490 ) / 100 = 0.49
Water volume decreases by 0.490 from all particles sharing this force.
12 particles: 0.490 / 12 = 0.040833333 loss each
0 0.339 0.339 0.339 0.339 0.245
0 0.339 0.339 0.339 0.339 0.245
0 0.339 0.339 0.339 0.339 0.245
0 0 0 0 0 0
0.490 water units removed from the soil in this hour 0.735 total water units removed since t=0
After 3 hours
Force applied at this time is: (3/10) * 2450 = 735
water_moved = ( 0.001 * 735 ) = 0.735
Water volume decreases by 0.735 from all particles sharing this force.
12 particles: 0.735 / 12 = 0.06125 loss each
0 0.278 0.278 0.278 0.278 0.49
0 0.278 0.278 0.278 0.278 0.49
0 0.278 0.278 0.278 0.278 0.49
0 0 0 0 0 0
0.735 water units removed from the soil in this hour 1.47 total water units removed since t=0
After 4 hours
Force applied at this time is: (4/10) * 2450 = 980
water_moved = ( 0.001 * 980 ) = 0.98
Water volume decreases by 0.98 from all particles sharing this force.
12 particles: 0.98 / 12 = 0.081666667 loss each
0 0.196 0.196 0.196 0.196 0.816
0 0.196 0.196 0.196 0.196 0.816
0 0.196 0.196 0.196 0.196 0.816
0 0 0 0 0 0
0.98 water units removed from the soil in this hour 2.45 total water units removed since t=0
After 5 hours
Force applied at this time is: (5/10) * 2450 = 1225
water_moved = ( 0.001 * 1225 ) = 1.225
Water volume decreases by 1.225 from all particles sharing this force.
12 particles: 1.225 / 12 = 0.102083333 loss each
0 0.094 0.094 0.094 0.094 0.816
0 0.094 0.094 0.094 0.094 0.816
0 0.094 0.094 0.094 0.094 0.816
0 0 0 0 0 0
1.225 water units removed from the soil in this hour 3.675 total water units removed since t=0
After 6 hours
Force applied at this time is: (6/10) * 2450 = 1470
water_moved = ( 0.001 * 1470 ) = 1.47
Water volume decreases by 1.47 from all particles sharing this force.
12 particles: 1.47 / 12 = 0.1225 loss each
0 0.000 0.000 0.000 0.000 0.816
0 0.000 0.000 0.000 0.000 0.816
0 0.000 0.00 0.000 0.000 0.816
0 0 0 0 0 0
1.125 water units removed from the soil in this hour (remainder) 4.8 total water units removed since
t=0
Final result
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关键词一 关键词二 关键词三 关键词四 关键词五 和法规和的发更好的风格和 ECON 615 FINC 612 CSC148 COMP 639 ACCT 203 ACCT 211 PSYC 101 MTH2009 COMP52315 PSYCO 432 statistics MATH4063 11 xxx 1 MA1MSP MAST30028 MATH3888 CS341 EEE30002 ES2F5 MCD4160 CC0933 ECON 1101 MCHM3001 CHEM3118 COMMGMT 2511 COMP41280 STAT 631 24T1 IRE379 SENG365 ECON433 FIT3139 MATH4602 LING5621M FINS5516 ECON3208 MN2177 L2 SM EEET1415 Finance DS4023 COMP3004 7CCSMRTS K1S5B6 Stat230 CSCI4211 CSI2132 ECON30130 MAS6100 ENVS222 COM2107 INFO1112 STAT 4004 FIT2100 comp2022 COMP4691 Physics 307 Econ 659 probability CSCI316 fir29 CSCI203 COMP3160 COMP6714 INFO1113 Physics 7B Physics Photonics S203031 STAT2005 COMP90015 CSCI 2110 CSCC46 Information CSC477 COMP4650 Statistical COMP 346 COMP1017 ELEC340 MATH1007 Programming function STAT 244 CS 134 Java cse13s Math 108A FW20 ECON3389 CSE 130 Science 331 COMP9020 engineering COSC 445 CSE 5523 GAME2012 visualization MATH 235 Math 370 CSC343 COGS 108 EE524 COMP10200 STAT 416 3FK4 MSCI 311 ECON6113 ACC40700 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