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INFR08010 INFORMATICS 2D: REASONING AND AGENTS
创建日期：2024-04-08 15:03:35
所属分类：数学mathematics
标签： INFR08010
INFORMATICS 2D: REASONING AND AGENTS

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SCHOOL OF INFORMATICS
INFR08010 INFORMATICS 2D: REASONING AND AGENTS

09:30 to 11:30
INSTRUCTIONS TO CANDIDATES
1. Answer Parts A, B and C.
2. The multiple choice questions in Part A are worth 50% in total and
are each worth the same amount. Mark one answer only for each
question - multiple answers will score 0. Marks will not be deducted
for incorrect multiple choice exam answers.
3. Parts B and C are each worth 25%. Answer ONE question from Part
B and ONE question from Part C.
4. Use the special mark sheet for Part A. Answer Parts B and C each in
a separate script book.
CALCULATORS ARE PERMITTED.
Convener: I. Simpson
External Examiner: I. Gent
THIS EXAMINATION WILL BE MARKED ANONYMOUSLY
Part A
ANSWER ALL QUESTIONS IN PART A. Use the special mark sheet.
1. Which one of the following cannot be deemed an agent?
(a) Laptop computer
(b) Air conditioner
(c) Aeroplane
(d) Roulette wheel
(e) Toy robot
2. Which of the following types of agents does not fit into the standard classification
of agents that was used in the course?
(a) Model-Based Reflex Agent
(b) Schema-Based Agent
(c) Simple Reflex Agent
(d) Utility-Based Agent
(e) Goal-Based Agent
3. Which of the following sets of first-order literals cannot be unified?
(a) {P (C, x1), P (x2, B(x2))}
(b) {P (C1, x1), P (x2, C2)}
(c) {P (C,B(x1)), P (x2, x2)}
(d) {P (C1, x), P (C1, C2)}
(e) {P (C, x1), P (x2, B(C))}
4. Which of the following is not part of the DPLL algorithm, as given in Russell &
Norvig and the lectures?
(a) A sentence is false if any of its clauses is false.
(b) A pure symbol can be set to false.
(c) A clause is true if any of its symbols is true.
(d) A pure symbol can be set to true.
(e) A symbol in a unit clause can be set to true.
5. Consider depth-first and breath-first search for finite maximum branching factor.
Which of the following statements is true?
(a) Both are optimal.
(b) Both are complete.
(c) Both have exponential worst-case time complexity.
(d) Neither is optimal.
(e) Iterative deepening search is the only practical way to use them (in combina-
tion).
Page 1 of 9
6. If the domain of a variable becomes empty, what does the AC-3 algorithm do?
(a) The CSP is indicated as satisfiable.
(b) The domain is refilled by default values.
(c) Nothing, because this cannot happen.
(d) The algorithm returns a truth value.
(e) The last arc is removed and the algorithm proceeds with other arcs.
7. Which statement on adversarial search using the α-β algorithm is correct?
(a) The α-β algorithm fails, if the opponent does not play optimally.
(b) In the α-β algorithm, pruning may cause the final result to be suboptimal.
(c) With perfect ordering, time complexity is quadratic.
(d) Backing up the minimax values through the tree can be omitted in most
practical applications.
(e) Space complexity is better than iterative deepening search.
8. Which of the following statements about Hierarchical Task Network (HTN) Plan-
ning is correct?
(a) All refinements of high-level actions (HLAs) are a sequence of only primitive
actions.
(b) A refinement of an HLA always has the same preconditions as the HLA.
(c) If no states in the pessimistic description of the reachable states of a high-level
plan (HLP) satisfy the goal, then that HLP is guaranteed to fail.
(d) If A is in the Effect of one refinement of HLA h, and ¬A is in another
refinement of h, then +˜A is in Reach+(s, h).
(e) An HLA that achieves a goal has the downward refinement property only if
all its implementations also achieve that goal.
9. Which of the following is not equivalent to the formula A⇔ B ∨ C?
(a) (A ∨ ¬B ∨ ¬C) ∧ (B ∨ ¬A) ∧ (C ∨ ¬A)
(b) (¬A ∨B ∨ C) ∧ (¬B ∨ A) ∧ (¬C ∨ A)
(c) (¬A ∨B ∨ C) ∧ ((¬B ∧ ¬C) ∨ A))
(d) (A⇒ (B ∨ C)) ∧ ((B ∨ C)⇒ A))
(e) (¬A ∨B ∨ C) ∧ (¬(B ∨ C) ∨ A)
10. Which of the following is true of the WALKSAT algorithm?
(a) It deletes clauses that contain pure literals.
(b) It uses the min-conflict heuristic to minimize the number of satisfied clauses.
(c) It is complete for finite problems.
(d) It initially assigns True to all symbols
(e) With a certain probability, it flips the sign of a symbol in order to maximize
the number of satisfied clauses.
Page 2 of 9
11. Forward chaining vs. backward chaining: Which of the following is usually wrong?
(a) If the branching factor in backward direction is relatively large, use forward
chaining.
(b) If a large amount of data is given, backward chaining is not an option.
(c) If some data are missing, backward chaining is a reasonable option.
(d) If goals are to be suggested by the system, use forward chaining
(e) If the goal is known, use backward chaining
12. Which one of the following best describes the closed-world assumption in plan-
ning?
(a) Nothing changes without a known reason.
(b) Every plan will ultimately lead to (one of) the goal(s).
(c) Every change is due to the agent.
(d) Known facts cannot change at the next time step.
(e) There is a one-to-one correspondence between states and situations.
13. Which is the negation of the sentence “All birds are black”?
(a) There exists a bird and it is black.
(b) There exists a bird and it is white.
(c) If a bird exists, then it is not black
(d) There exists a bird and it is not black.
(e) All birds are not black.
14. You are given four action descriptions with conditional effects:
Action (A,Precond : {P,X} , Effect : {(whenZ : ¬X)})
Action (B,Precond : {¬Y } , Effect : {(whenP : Z,X)})
Action (C,Precond : {Z} , Effect : {(whenP : ¬Y ), X})
Action (D,Precond : {¬X} , Effect : {(whenP : Z,X)})
What state would result from executing the action sequence [D,C,B,A] in the
state {P, Y }?
(a) The plan is not executable because the preconditions for action A are not
met.
(b) {P,Z}
(c) {P,X,Z}
(d) {P,¬X,¬Y, Z}
(e) {P,X, Y, Z}
15. Which of the following is not generally true if the relation KB α holds
(a) KB contains a sentence that implies α
(b) Every model of KB is also a model of α
(c) KB ⇒ α
(d) Resolution of some sentences in KB and ¬α produces the empty clause.
(e) ¬α entails ¬KB
Page 3 of 9
16. In a vocabulary with only 3 propositional symbols A, B, and C, how many models
are there for the sentence A⇒ B?
(a) 0
(b) 1
(c) 2
(d) 3
(e) 6
17. The following formula describes the “forward” part of the algorithm used for
prediction in temporal probabilistic models:
f1:t+1 = αP (et+1|Xt+1)
∑
xt
P (Xt+1|xt)P (xt|e1:t)
Which of the following statements is incorrect?
(a) α is the normalisation factor
(b) P (xt|e1:t) is the current state distribution
(c) P (xt+1) depends on both P (xt) and P (et)
(d) P (et+1|Xt+1) is obtained from the sensor model
(e) P (Xt+1|xt) is obtained from the transition model
18. Consider the following PDDL action description in a Blocks World application:
Action(Move(b, x, y),
Precond: On(b, x) ∧ Clear(b) ∧ Clear(y)
Effect: On(b, y) ∧ Clear(x) ∧ ¬On(b, x) ∧ ¬Clear(y))
and the state S1 defined as follows:
S1 : {On(A,B), On(B,C), Clear(A), Clear(D)}
where
• Move(b, x, y) is the action of moving block b from block x to block y.
• On(b, x) means block b is on block x.
• Clear(x) means block x has nothing on top of it.
Which of the following statements is incorrect?
(a) Applying Move(A,B,D) in S1 would result in state
{On(A,D),¬On(A,B),¬Clear(D), Clear(B)}
(b) Move(A,B,D) can be applied in S1
(c) You need a sequence of at least two actions to achieve the goal On(B,D).
(d) [Move(A,B,D),Move(B,C,A)] can be applied in S1
(e) Move(B,C,D) cannot be applied in S1
Page 4 of 9
19. Which of the following is assumed to be true for naive Bayes classifier?
(a) It applies to connected nodes in a Bayesian network.
(b) Effects are assumed to be independent.
(c) Different causes are assumed to be independent.
(d) It is exact for binary random variables.
(e) A common cause fully explains the dependencies among the effects.
20. Suppose that we model a rational agent as a Markov Decision Process (MDP).
Then which of the following statements is incorrect?
(a) Identifying the action that will be performed next may be non-deterministic.
(b) The agent’s expected utilities change over time only because his beliefs do.
(c) Agents that are modelled as MDPs will value long-term rewards over imme-
diate rewards.
(d) MDPs require that all possible actions and all their possible outcomes be
known right from the start of the dynamic process.
(e) The value iteration algorithm calculates the (current) expected utility of a
state for a rational agent.
Page 5 of 9
Part B
ANSWER ONE QUESTION FROM PART B
1. Resolution and Modus Ponens
(a) State the binary resolution rule and explain how it can be applied to first
order clauses. [5 marks ]
(b) Given the following axioms
A ∨B ∨ C,
A ∨B ∨ ¬C,
A ∨ ¬B ∨ C,
A ∨ ¬B ∨ ¬C,
¬A ∨B,
¬A ∨ ¬B ∨ C,
show by resolution that A ∧B ∧ C. [6 marks ]
(c) Assuming that Modus Ponens is sound, prove that Generalized Modus Po-
nens (GMP) is also a sound rule of inference for first order logic. [4 marks ]
(d) Give an algorithm for forward-chaining using Generalised Modus Ponens. [4 marks ]
(e) Describe two possible ways in which the efficiency of the forward-chaining
algorithm from the previous question can be improved. In each case, you
need to explain briefly what the issue is and how the improvement addresses
it. [6 marks ]
2. Planning
In a system for robotic office delivery, two robots A and B have to transport
packages P and Q between different rooms X, Y and Z given the following
initial (left) and goal (right) states:
Each robot can carry any number of packages, the robots can move simultane-
ously, and they can both be in any room or any corridor at the same time.
QUESTION CONTINUES ON NEXT PAGE
Page 6 of 9
QUESTION CONTINUED FROM PREVIOUS PAGE
We use the following predicates to describe the domain:
• In(o, r) – robot or package o is in room r
• Holds(r, p) – robot r is carrying package p
• OnFloor(p) – package p is on the floor (i.e. no robot is carrying it)
• Robot(r) denotes that r is a robot
• Package(p) denotes that p is a package
• Room(rm) denotes that rm is a room
• Corridor(rm1; rm2) – there is a corridor between rm1 and rm2
(a) Using the above PDDL vocabulary, define the initial state and goal. [4 marks ]
(b) Specify action schemata for the following actions:
i. Go(r; rm1; rm2): robot r moves from room rm1 to room rm2 connected
by a corridor, moving anything it is potentially carrying to rm2 with it [2 marks ]
ii. Drop(r, p): robot r drops package p; resulting in the package being on
the floor [2 marks ]
iii. Pickup(r, p): the robot picks up a package from the floor; as a result,
the robot is holding the package [2 marks ]
(c) Define the actions, orderings, links and open preconditions of a partial-order
plan that solves the above planning problem. What is the main advantage
of using partial-order planning in this particular domain? [6 marks ]
(d) Now suppose that there is a light switch in each room, and the robots can
pick up a package if the light in the room is on. Furthermore, to move out
of a room (and into another), the room’s light must be on. The robots can
also flip a light switch if they are in the same room as the switch.
i. Define the predicates you need to express the light in the room being
on (or off, which you can assume is the same as not on). [4 marks ]
ii. Define the action schema Flip, of a robot flipping a light switch in the
room it’s in, which results in the light being on if it was off, and off if
it was on. [2 marks ]
iii. Assume that in the initial state, the lights are off in all rooms, and the
switches in rooms X, Y and Z are respectively SX , SY and SZ . Then
define a total ordered plan that solves the above planning problem. [3 marks ]
Page 7 of 9
Part C
ANSWER ONE QUESTION FROM PART C
1. Adversarial search
Consider a variation of the Tic-Tac-Toe game which is identical in terms of legal
moves, but is extended by the following rules:
• For a pair, i.e. two symbols in a row (horizontally or vertically), the player
obtains 2 points. For a triplet, i.e. three symbols in a row (horizontally,
vertically, or diagonally), the player obtains 5 points in addition to the two
times 2 points for the two pairs, i.e. a total of 9 points.
• The game terminates when a triplet has been completed or all squares have
been filled. The winner is the player who accumulates the highest number
of points.
In the root node of the following game tree, for example, the Max and Min players
would both score 2 as each player has completed one pair (we write 2/2 to denote
this intermediate score for Max/Min). In the rightmost leaf node, Max wins with
a score of 13 points (one triplet and four pairs), and Min loses with 2 points for
a single pair.
(a) Label all nodes in the tree given above with the respective intermediate/final
scores using the Max/Min format. [8 marks ]
(b) Apply the minimax algorithm to compute the optimal strategy for player
Max. You can specify the strategy by the first choice of Left-Middle-Right
moves that is taken by Max. For full marks, your answer should explain the
reasoning applied by the algorithm in detail. [7 marks ]
(c) Consider an incomplete strategy where the search only expands the tree
from Max’s point of view to depth 1 (i.e. only the first three nodes below
the root node are explored).
QUESTION CONTINUES ON NEXT PAGE
Page 8 of 9
QUESTION CONTINUED FROM PREVIOUS PAGE
i. If the heuristic was based on counting the number of pairs achieved so
far (which we call the Pairs heuristic), would this lead Max to make
the right choice in the root node? Justify your answer. [5 marks ]
ii. As an alternative to the Pairs heuristic, consider the heuristic Oppo-
nentGaps which discounts the value of the state by 1 for every empty
square Min could fill in her move that is adjacent to one or more O
symbols (e.g. the leftmost node under the root node would obtain a
value of -1). Note, that diagonal neighbours are not considered adja-
cent here. Would this work better or worse for Max than Pairs when
making a decision in the root node? Your answer should include a brief
explanation. [5 marks ]
2. Bayesian Inference
Consider a population of women, 50% of whom are over 40. If they are over 40
years old, then 10% of them have breast cancer. But only 5% of women under
40 have breast cancer. 70% of the women with breast cancer will have a positive
mammography (meaning the test indicates she has cancer). 20% of women who
do not actually have breast cancer also get a positive mammography (meaning
that they are incorrectly diagnosed with cancer).
(a) Model the above description as a Bayes Net, using Boolean variables and
conditional probability tables. [7 marks ]
(b) Using your Bayes Net, calculate the likelihood that a woman under 40 who
has had a negative mammography has breast cancer. Be sure to show each
step of your calculation. Note: Give your answer to 3 decimal places. [9 marks ]
(c) Do you need to know the prior probability that a woman is under 40 to
compute the posterior probability that a woman having breast cancer gets
a positive mammography? Explain your answer. [2 marks ]
(d) Suppose that a woman, who is under 40, gets a positive mammography test,
M1, but goes on to have a second mammography, M2, that turns out to be
negative. Use the Naive Bayes assumption to compute the likelihood that
she has breast cancer. [7 marks ]
Page 9 of 9
Solutions A
1. d Roulette wheel
2. b Schema-Based Agent
3. c {P (C,B(x1)), P (x2, x2)}
4. b A pure symbol can be set to false.
5. c Both have exponential worst-case time complexity.
6. d The algorithm returns a truth value.
7. e Space complexity is better than iterative deepening search.
8. d If A is in the Effect of one refinement of HLA h, and ¬A is in another
refinement of h, then +˜A is in Reach+(s, h).
9. a (A ∨ ¬B ∨ ¬C) ∧ (B ∨ ¬A) ∧ (C ∨ ¬A)
10. e It flips the sign of a symbol in order to maximize the number of satisfied
clauses.
11. b If a large amount of data is given, backward chaining is not an option.
12. a Nothing changes without a known reason.
13. d There exists a bird, but it is not black.
14. b {P,Z}
15. a KB contains a sentence that implies α
16. e 6
17. c P (xt+1) depends on both P (xt) and P (et)
18. a ApplyingMove(A,B,D) in S1 would result in state {On(A,D),¬On(A,B),¬Clear(D), Clear(B)}
19. e A common cause fully explains the dependencies among the effects.
20. c Agents that are modelled as MDPs will value long-term rewards over
immediate rewards.
Page 10 of 9
Part B
1. Resolution and Modus Ponens
(a) Stating the binary resolution rule [1], propositionalisation [1], unification [1],
CNF [1], resolution or factoring or standardising-apart variables [1]
(b) Negated goal: ¬A ∨ ¬B ∨ ¬C
A∨B∨C,A∨B∨¬C
A∨B [1],
A∨¬B∨C,A∨¬B∨¬C
A∨¬B [1],
¬A∨¬B∨C,¬A∨¬B∨¬C
¬A∨¬B [1]
A∨B,A∨¬B
A
[1],
¬A∨B,¬A∨¬B
¬A [1],
A,¬A
[1]
Other solutions are possible.
bookwork We need to show that p′1, . . . , p
′
n, (p1 ∧ . . .∧ pn ⇒ q) qθ provided p′iθ = piθ
for all i and some most general unifier θ.
Proof : For any sentence p, we have p pθ by the Universal Instantiation
rule. Using this (and a few basic rules), we thus have:
i. (p1 ∧ . . . ∧ pn ⇒ q) (p1 ∧ . . . ∧ pn ⇒ q)θ = (p1θ ∧ . . . ∧ pnθ ⇒ qθ)
ii. p′1, . . . , p
′
n p′1∧. . .∧p′n (p′1∧. . .∧p′n)θ = p′1θ∧. . .∧p′nθ = p1θ∧. . .∧pnθ
since by definition of GMP, we have p′iθ = piθ for all i.
iii. From 1(b)i and 1(b)ii, qθ follows by Modus Ponens.
Mark breakage: 1 mark for stating what is being proved, 1 mark for each of
the steps 1(b)i to 1(b)iii in the proof.
(c) (Bookwork) The following algorithm is an acceptable answer:
Note that the algorithm does not need to be given exactly as above. Al-
ternative answers not in pseudo-code but describing the main aspects are
acceptable.
(d) Any two of the following three improvements are acceptable [3 marks each]:
Page 11 of 9
i. Incremental forward-chaining: In this, at each iteration t, we check a
rule only if its premise includes a conjunct pi that unifies with a fact pi
newly inferred at iteration t−1. The rule-matching step then fixes pi to
match with p′i from the FC algorithm, but allows the other conjuncts of
the rule to match with facts from any previous iteration. This algorithm
generates exactly the same facts at each iteration as FC, but is much
more efficient.

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MATH 239 QF5203 INFR08010 IT114105 FIT2093 ECON3124 ECON1195 MATH 337 STAT 3006 COMP202 COMP3130 MATH3609 COMP 250 CS6140 IEEE ELE00041I FIN40090 CIV 5899 MATH3066 IGNMENT MATH 2009 STAT3405 HIST 101 MATH 3281 CEGE0042 ICT532 LAB 7 LAB 5 MCO455 CSIT214 Q4 IFB104 A1 ECO220Y5Y DA5020 CMPSC473 FIN9007 DSCI553 MSCI534 COMP3308 CSE-112 CMPT 310 EE104 F 2C P20 ALY2010 BUSM060 STA258 ECO00030M R BIOSTAT 274 COS6012-B FIT3155 CMPUT 340 EC421 INFS4205 ECOM135 NATS 1740 COMP809 MBE 6005 STA465 CSSE2310 CIVL2700 DATA4700 CMPSC 473 ISE 529 CS5044 MAST30020 COMP2207 ENGG1001 COMP0046 FIT3165 STAT GR5241 MECH0007 vATHK1001 ANALYTIC ATHK1001 MATH352001 HW3 S1 BMAN 24662 COMP 636 Stat-461 STAT 302 STAT 2132 STAR534 ENG4137 CMP3752M CS3103 BISM3222 STA 135 ME155A IIMT3601 ECON3371 ECON 3371 HW4 FINA 5840 DT211C CTEC3902 MAT005 MATH2148 S261F Econ 332 MATH3734 MIS41270 7CCMFM13 SYST 664 CIS434 CENG10014 MT4111 MT4527 ECMM424 CMT219 Math 136 MATH 3120 COMP10002 MAST20009 MATH0058 BEEM117 ISYE 6420 MGMTMFE 405 25579 1ECE 273 M122 COMP4434 1ELEC ELEC5620M STAT0010 BFF2401 MXB361 MATH3811 CS330 CMT307 FIT1033 CptS 583 UP 431 MATH6005 AF5343 CS2435 ECOM30002 ES50061 CS 2506 ECMT6002 CODE 00099F MT4539 ESSAY ICM317 MATH1231 MFIT5010 BMA547 MATH334 CIVL2201 MATH39032 T-04 CIVE97119 EMESTER CAB302 CAB301 59PM STAT802 ISYS3412 RM3 COSC 2123 AcF 311 MATH2022 1FINAL 4CCS1PHC ICM 142 ENVS3023 Math 4023 I1 ACSE8 FIN B385F CS-275 STA 4109-01 A31021 EE140 MT1 EE124 ECO-6004B Inf2-IADS CS126 IS2184 UL18 MATH5092 ECOM009 COMP 6211 ELEC202 MKT3019 IAB230 FIT5046 584-S21 COMP3210 CO250 A29401 MATH 136 MGT6174 ECMT6007 DATA2001 MAT 443 MATH10222 PHDE0071 ELE00123M COSC1295 MAT00045H MATH32032 1004GRC MAS223 COMP10001 DATA3404 CIVL2110 STAT2008 FNCE10002 EEE6431 COM6515 COMP9120 NBS8185 ECTE942 ECON 405 MATH314301 STAT 2011 COMPSCI 350 COMP20007 CMPSC-132 ECON5002 STA 35 MTH6108 FIT5147 MTH6150 100C FINA 5260 DESC9115 ECON4998 NBS8257 GMM ECON 6070 BSA2 ECE599 STATS 102B STA120 INFS601 BFC3241 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COMP2013 IB9X60 INVP001 EMATM0051 COMP220 BLAW1002 E20 TCP8001 ECON61001 EART40005 ECONM1022 MATH 365 MATH 367 ELEC6218 COMP 5812M EDUC 5061M ELEC ENG 1101 equation MATH6174 ST334 C4 EC9570 A33648 EBUS603 ECON47815 stata WFMS0001 PEM102 INF6060 ionically ELECTRICAL MATH 375 MMW 12 MKTG860 ECON60401 ALY6010 SOST20131 Introduction COM3524 MME2 GGR305 JNB 538 EECS101 COMM5030 IEOR E4601 2B CSC336 EN2315 LAW 432 CSC263 MATH2001 OLET1143 CSCB63 SIT772 COMM1170 STAT 441 MS930 INFT3050 GLBH0030 Stat 120B MAST10007 MGEC61 Machine GGR 252H1S ECON7030 Financial Marketing MGMT30019 ECON5120 SDSC 8009 CSC165H1S DSCI 551 3081W ACM41000 ECA 5304 MATH 1064 ECSE444 DATA 604 COMP7105 MIE1615 ECON4004 ENV200H1S EC481 MKT 306 EC1B1 MATH08051 ALY6140 CS260C CSE30 MSIN0226 CS4103 CSC3064 AM16 COMP11212 CSE4101 PSTAT 174 Math 380W CNT 5106C STAT 310 FIN2028 CS5014 STAT 425 COMP 4102A ELEC2140 MANF9544 QRM 9 QRM 8 ECON 457 COMPSCI 4091 MATH 523 DNSC-4211 PEAS 6000 MSIN0180 COMP2119 A2A HT 2022 IB93F0 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FN1024 MATH0094 MANG 6554 ECOS3003 ELEC6252 C2511C PHY328 EC1060 ECON32202 COMP4002-E1 MATH3063 EFIMM0124 QBUS2310 ECON4411 FUNDAMENTALS ACCT20001 18YX RPM COMP 2711 QBUS6600 BUSS5221 ST3189 FIT1047 MAT344 EGB375 MATH60013 MATH96011 COMP3557 MAST20029 ECM159 EAE1011 ENSCEN 105 SECTION A STA 141C SOSE 2022 PAM006 COMP0048 M40007 ISOM2500 CMPE 142 A06472 MCC150 ACST 8083 GEOG6087 ECON920 FINM7008 ENVENG 4110 COMPSCI 361 ENGRCEE 21 PHYS 1112 ACF2100 ECMM108 COMPS267F MAT00050M MAST30025 CSIT970 MSCI 224 INT202 PUBL0054 MSDM5004 BMAN 21020 BMAN21020 MGMT1002 MATH11150 LAWS7023 ACTL30002 CIVL3340 MATH1002 ECON 8069 COMPSCI 351 IBUS7302 STAT4528 CHEM1101 STAT3925 BUSN 7008 Comp 3120 ECOS2040 SWEN90010 ENGN6528 ECM158 ECON20022 COMP6452 Calculation CITS1401 S1889112 BUS 351 ACTL30008 FIT2091 ECON6002 IBUS6020 PHI 001 ECON30290 MATH7501 ECON10004 MFIN7017 ISYS90043 LAWS7012 ECON30025 COMP30023 COMM1190 ECOM20001 GSBS6009 STAT 337 EMET8002 ECOM90001 ECOM30001 FINC 2012 BUSINESS ENGCV512 ECO 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