Exam 1
Name_
1)
M325K
EID_
Dr. Berg
Fall 09
Use a truth table to establish that ( p q) p p .
2)
Use a truth table to establish that a conditional statement is not logically equivalent to its
converse. (Point out what makes them not equivalent.)
3)
Explain
Chapter 2: The Logic of Compound Statements
Section 2.1
9. (n k ) (n k ) 15. p T T T T F F F F q T T F F T T F F r T F T F T F T F q F F T T F F T T qr T F T T T F T T p ( q r) T F T T F F F F
The truth table shows that p (q r) and (p q ) (p r) always hav
M362M - Introduction to Stochastic Processes: Homework 2
Problem 1.8, p. 63:
(b) The equation matrix = P is equivalent to the following
system of equations
(a)
101 + 83 + 44 = 0
102 + 23 + 64 = 0
4
91 + 42 104 = 0
2
1
3
1 + 62 103 = 0
We add the first two
M362M - Introduction to Stochastic Processes: Homework 3
Problem 1.56, p. 72:
and A are transient:
The states F and B are recurrent, and G
G
A
B
F
In the canonical form (with rows and columns ordered as F,B,G,A)
the matrices Q and R are given by
!
!
.8 .1
M362M - Introduction to Stochastic Processes: Homework 4
Problem 2.10, p. 93: Let TA , TB and TC be the times Alice,
Bob and Carol, respectively, spend in service; they are independent
and exponentially distributed with parameter = 1/4. Also, set
T1 = min
Laboratory Fluid Study
Volatile Oil
The major tests and measurements normally
performed in the laboratory on volatile oil
samples are:
l
Composition measurement
l
Flash vaporization test
l
Differential vaporization test (Optional)
l
Constant volume deplet
Solutions of Hm4
5-1
a) It is a dry gas because the reservoir temperature is larger than the cricondentherm and the separator
condition is outside the envelope.
b) It is a wet gas since the reservoir temperature is larger than the cricondentherm and the s
Solution of Hw7
Gu
6-6
Z=0.39 (from Ex 3-7)
Bg=0.02827=0.02827=0.00693
6-8
Z=0.951 (from Ex 3-10)
Bg=0.02827=0.02827=0.004615
Vg_scf=43560=1.53e10 scf
6-15
Cg=
Ming
)
b
/
f
c
(
e
m
u
V
r
a
l
o
M
Cg=0.00538 psi-1
For ideal gas, B=0, Cg=1/P=0.005 psi-1
Rela
IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
Solutions to Homework Assignment 6 due on Tuesday, October 16
Problem 3.1 (a) Yes since it is a contrapositive statement of (3.2.1).
(b) No. Note that we can have Xn+1 = 0, so that Sn = Sn+k = 0 fo
IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
Solutions to Homework Assignment 3 due on Tuesday, September 25.
Problem 2.1 The conditions (i) and (ii) of denition 2.1.2 are apparent from the denition 2.1.1.
Hence it is sucient to show that den
IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
Solutions to Homework Assignment 12 due on Thursday, November 29
Problem 5.3 (a) Let N (t) denote the number of transitions be t. It is easy to show in this case
that
(M t)j
P(N (t) n)
eM t
j!
j=n
IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
Solutions to Homework Assignment 4.
Problem 2.14 Let us use Dj instead of Oj for the number of people getting o at oor j. Let
Di,j denote the number of people that get on at oor i and get o at oor
IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
Solutions to Homework Assignment 5, not to be turned in.
Problem 2.25 Just as in Problem 2.20, the key thing here is to apply the conditional distribution
of arrival times given a Poisson number of
IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
SOLUTIONS to Homework Assignment 7
Problem 3.12 Take h(t) = 1(0,a] (t).
Problem 3.13 Since the state circulates the state space cfw_1, 2, , n in the same order. Hence
we can dene the alternating re
IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
SOLUTIONS to Homework Assignment 8.
Problem 3.19
P cfw_SN (t) s =
G(t)
=
= G(t) +
=
n=0 P cfw_Sn s, Sn+1 > t
+ P cfw_Sn s, Sn+1 >
n=1
t
n=1 0 P cfw_Sn
s, Sn+1 > t|Sn = yd(G Fn1 )(s)
s
G(t) + F (t
IEOR 6711: Stochastic Models I
Fall 2012, Professor Whitt
Solutions to Homework Assignment 9 due on Thursday, November 8
Problem 4.1 Let Dn be the random demand of time period n. Clearly Dn is i.i.d. and independent of all Xk for k < n. Then we can repres
Solution of Hw 8
Ming Gu
8-2
API=(141.5/r)-131.5=(141.5/.875)-131.5=30.21 API
8-7
B0=227 res cc/ 167.4 ST cc=1.356 res bbl/STB
Rs=(0.537+0.059) scf)/(167.44 ST cc * 6.2898e-6 bbl/cc)=565.9 scf/STB
8-10
1.6
1.4
Rs(scf/STB)
Bo (bbl/STB)
1.5
1.3
1.2
1.1
Pb
1
The University of Texas at Austin
Fall 2011 - PGE 310: Formulation and Solution in Geosystems Engineering
HW #9: System of Linear Equations by Iterative Methods - System of Non-Linear Equations by Newtons Method
Due Date: Thursday Nov 10th 2011
Do NOT for
Chapter 1: Speaking Mathematically
Section 1.1
6. a. s is negative b. negative; the cube root of s is negative (Or : c. is negative; 3 s is negative (Or : the cube root of s is negative) 3 s is negative)
9. a. have at most two real solutions b. has at mos
Chapter 3: The Logic of Quantied Statements
Section 3.1
6. a. When m = 25 and n = 10,the statement m is a factor of n2 is true because n2 = 100 and 100 = 4 25. But the statement m is a factor of n is false because 10 is not a product of 25 times any integ
16
Solutions for Exercises: Elementary Number Theory and Methods of Proof
Chapter 4: Elementary Number Theory and Methods of Proof
Section 4.1
3. a. Yes, because 4rs = 2 (2rs) and 2rs is an integer since r and s are integers and products of integers are i
Section 6.1
71
Chapter 6: Set Theory
The rst section of this chapter introduces additional terminology for sets and the concept of an element argument to prove that one set is a subset of another. The aim of this section is to provide a experience with a
PGE 312
Data Sheet for Exams
1.
R = 10.732 psia cu ft/lb mole R
2.
1 barrel = 5.615 cu ft
3.
PV = ZnRT
4.
Density of fresh water = 62.4 lb/cu ft
5.
Alkanes
Alkenes
Cycloalkanes
Alkynes
Aromatics
CnH2n+2, for n = 1, 2, 3, .
CnH2n, for n = 2, 3, 4, .
CnH2n,
PGE 312
Data Sheet for Test 2
1.
R = 10.732 psia cu ft/lb mole R
2.
1 barrel = 5.615 cu ft
3.
PV = ZnRT
4.
Density of fresh water = 62.4 lb/cu ft
5.
Alkanes
Alkenes
Cycloalkanes
Alkynes
Aromatics
CnH2n+2, for n = 1, 2, 3, .
CnH2n, for n = 2, 3, 4, .
CnH2n