NANYANG TECHNOLOGICAL UNIVERSITY
SEMESTER I EXAMINATION 2011-2012
MAS446 / MTH437 – Probabilistic Methods in OR
November 2012
TIME ALLOWED: 2 HOURS
INSTRUCTIONS TO CANDIDATES
1. This examination paper contains FIVE (5) questions and comprises
SIX (6) printed pages.
2. Answer all questions. The marks for each question are indicated at the
beginning of each question.
3. Answer each question beginning on a FRESH page of the answer book.
4. This IS NOT an OPEN BOOK exam.
5. Candidates may use calculators. However, they should write down
systematically the steps in the workings.
MAS446 / MTH437
QUESTION 1.
(15 marks)
A dental surgery has two operation rooms. The service times are assumed to
be independent and exponentially distributed with mean 15 minutes. Andrew
arrives when both operation rooms are empty. Bob arrives 10 minutes later
while Andrew is still under medical treatment. Another 20 minutes later
Caroline arrives and both Andrew and Bob are still under treatment. No
other patient arrives during this 30-minute interval.
(i) What is the probability that Andrew will be ready before Bob?
(ii) What is the probability that Caroline will be ready before Andrew?
(iii) What is the variance of the waiting time in the system for Caroline?
Solution:
(i)
1
2
(ii)
1
4
(iii) 152 + 7.52 = 281.25 minutes2
♦
2
MAS446 / MTH437
QUESTION 2.
(20 marks)
A two-server queueing system is in a steady-state condition and the steady1
1
, P1 = 41 , P2 = 38 , P3 = 14 , P4 = 16
, and Pn = 0
state probabilities are P0 = 16
if n > 4. The mean arrival rates are λ0 = 8, λ1 = 4, λ2 = 2, λ3 = 1, and
λn = 0 if n > 3. Find
(i) the expected number of customers in the queue,
(ii) the expected number of customers being served,
(iii) the expected waiting time in the queue per customer, and
(iv) the expected service time per customer.
Solution:
(i) Lq = (3 − 2) × 41 + (4 − 2) ×
(ii) 1 × 14 + 2 × ( 38 + 14 +
1
)
16
=
1
16
=
13
8
1
+ 4 × 41 + 2 × 38 + 1 ×
(iii) λ̄ = 8 × 16
3
Wq = Lq /λ̄ = 20
(iv) L = 2, W = 45 ,
1
µ
3
8
= W − Wq =
4
5
♦
3
1
4
−
= 52 ,
3
20
=
13
20
MAS446 / MTH437
QUESTION 3.
(20 marks)
Smith is in jail and has 3 dollars; he can get out on bail if he has 8 dollars.
A guard agrees to make a series of bets with him. If Smith bets x dollars, he
wins x dollars with probability 0.4 and loses x dollars with probability 0.6.
Smith may choose either of the following two betting strategies.
(a) He bets 1 dollar each time (timid strategy).
(b) He bets, each time, as much as possible but not more than necessary
to bring his fortune up to 8 dollars (bold strategy).
Determine which strategy gives Smith the better chance of getting out of
jail? (Hint: Find the probability that Smith wins 8 dollars before losing all
of his money for each betting strategy.)
Solution:
We formulate the problem as a Markov chain of nine states. Each state
represents the amount of dollars that Smith has.
State
State
State
State
State
State
State
State
State
0
1
2
3
4
5
6
7
8
$0
$1
$2
$3
$4
$5
$6
$7
$8
If the timid strategy is adopted, then the one-step transition matrix P is
0
1
2
3
4
5
6
7
8
0
1
2
3
4
5
6
7
8
1
0
0
0
0
0
0
0
0
0.6 0 0.4 0
0
0
0
0
0
0 0.6 0 0.4 0
0
0
0
0
0
0 0.6 0 0.4 0
0
0
0
0
0
0 0.6 0 0.4 0
0
0
0
0
0
0 0.6 0 0.4 0
0
0
0
0
0
0 0.6 0 0.4 0
0
0
0
0
0
0 0.6 0 0.4
0
0
0
0
0
0
0
0
1
4
MAS446 / MTH437
We are going to find the absorbing probability f38 . To do so, we solve the
P
following linear equations, which come from fik = M
j=0 pij fjk .
f38 = 0.6f28 + 0.4f48 ,
f28 = 0.6f18 + 0.4f38 ,
f18 = 0.6f08 + 0.4f28 ,
f48 = 0.6f38 + 0.4f58 ,
f58 = 0.6f48 + 0.4f68 ,
f68 = 0.6f58 + 0.4f78 ,
f78 = 0.6f68 + 0.4f88 ,
f08 = 0,
f88 = 1.
Finally, we obtain f18 = 0.0203, f28 = 0.0508, f38 = 0.0964, f48 = 0.1649,
f58 = 0.2677, f68 = 0.4219, and f78 = 0.6531. Therefore, the probability that
Smith will win 8 dollars before losing all of his money is 0.0964.
If the bold strategy is adopted, then the one-step transition matrix P is
0
1
2
3
4
5
6
7
8
0
1
0.6
0.6
0.6
0.6
0
0
0
0
1 2 3 4 5 6 7 8
0 0 0 0 0 0 0 0
0 0.4 0 0 0 0 0 0
0 0 0 0.4 0 0 0 0
0 0 0 0 0 0.4 0 0
0 0 0 0 0 0 0 0.4
0 0.6 0 0 0 0 0 0.4
0 0 0 0.6 0 0 0 0.4
0 0 0 0 0 0.6 0 0.4
0 0 0 0 0 0 0 1
We are going to find the absorbing probability f38 . To do so, we solve the
P
following linear equations, which come from fik = M
j=0 pij fjk .
f38 = 0.6f08 + 0.4f68 ,
f68 = 0.6f48 + 0.4f88 ,
5
MAS446 / MTH437
f48 = 0.6f08 + 0.4f88 ,
f28 = − − −,
f18 = − − −,
f58 = − − −,
f78 = − − −,
f08 = 0,
f88 = 1.
Finally, we obtain f38 = 0.2560, f48 = 0.4, and f68 = 0.64. Therefore, the
probability that Smith will win 8 dollars before losing all of his money is
0.2560.
♦
6
MAS446 / MTH437
QUESTION 4.
(20 marks)
Each year you have to decide whether or not to declare a particular part
of your income to the Inland Revenue. If you do declare you pay $90 in tax.
If you do not declare and the Inland Revenue audits your submission you
pay $200. The Inland Revenue splits the taxpayers into three groups, that
is, those whom they did not audit in the previous year, those whom they
did audit in the previous year and were found to have declared correctly,
and those whom they did audit in the previous year and were found to have
declared incorrectly. The Inland Revenue adopts a policy of inspecting each
of these groups with probabilities 0.5, 0.3 and 0.7, respectively, and these
probabilities are known to you. You also know whether your submission was
audited in the previous year. Now you try to determine a declaration/no
declaration policy that minimizes your expected total discounted payments
to the Inland Revenue with a discount factor α = 0.9.
(i) Formulate this problem as a Markov decision process by identifying the
states and decision and then finding the values for each pij (k) and Cik .
(ii) Formulate a linear programming model for finding an optimal policy
(Don’t need to solve this model for solution).
Solution:
three decisions.
State 0
State 1
State 2
Decision 1
Decision 2
We formulate a Markov decision process of three states and
Not audited in the previous year
Audited in the previous year and declaration correct
Audited in the previous year and declaration incorrect
Do not declare
Declare
By problem description, we have the following table that presents the
values of pij (k) and Cik .
State Decision 0
0
1
0.5
2
0.5
1
1
0.7
2
0.7
2
1
0.3
2
0.3
pij (k)
1
0
0.5
0
0.3
0
0.7
7
2
Cik
0.5 0.5 × $200
0
$90
0.3 0.3 × $200
0
$90
0.7 0.7 × $200
0
$90
MAS446 / MTH437
Let yik be the discounted expected time of being in state i and making
decision k, when the probability distribution of the initial state is βj . Note
P
that M
j=0 βj = 1 and βj > 0 for all j. Without loss of generality, we assume
that βj = 31 , for each j. The linear programming model is hence to choose
the yik so as to
Maximize Z = 100y01 + 90y02 + 60y11 + 90y12 + 140y21 + 90y22 ,
subject to the constraints
y01 + y02 =
1
+ 0.9 [0.5y01 + 0.5y02 + 0.7y11 + 0.7y12 + 0.3y21 + 0.3y22 ]
3
1
+ 0.9 [0y01 + 0.5y02 + 0y11 + 0.3y12 + 0y21 + 0.7y22 ]
3
1
= + 0.9 [0.5y01 + 0y02 + 0.3y11 + 0y12 + 0.7y21 + 0y22 ]
3
yik ≥ 0, ∀i, k.
y11 + y12 =
y21 + y22
After simplification, the linear programming model becomes
Maximize Z = 100y01 + 90y02 + 60y11 + 90y12 + 140y21 + 90y22 ,
subject to the constraints
0.45y01 + 0.45y02 − 0.63y11 − 0.63y12 − 0.27y21 − 0.27y22 =
1
3
1
=
3
0y01 − 0.45y02 + y11 + 0.63y12 − 0y21 − 0.63y22 =
−0.45y01 + 0y02 − 0.27y11 − 0y12 + 0.27y21 + y22
yik ≥ 0, ∀i, k.
♦
8
1
3
MAS446 / MTH437
QUESTION 5.
(25 marks)
Two operators attend four automatic machines. After each machine completes a batch run, an operator must reset it before a new batch is started.
The time to complete a batch run is exponential with mean 30 minutes. The
setup time is also exponential with mean 15 minutes, which requires the
complete attention of an operator.
(i) Determine the average number of machines that are waiting setup or
are being set up.
(ii) Compute the probability that all machines are working (i.e., no machine
is waiting setup or being set up).
(iii) Determine the average percentage of time a machine is down (i.e., the
percentage of time a particular machine is waiting setup or being set
up).
(iv) Determine the average percentage of time an operator is idle (i.e., the
percentage of time a particular operator is not working on setup).
Hint: For the finite calling population variation of the M/M/s (s > 1)
queueing model we have the following results:
n

λ
N!


 (N −n)!n! µ n
Cn = 


N!
(N −n)!s!sn−s
λ
µ
0
for n = 0, 1, 2, . . . , s
for n = s, s + 1, . . . , N
forn > N
Solution:
The finite calling population variation of the M/M/s (s >
1) queueing model is used to solve this problem, where N = 4 and s = 2.
Let one hour be the unit time period to be considered. Hence, λ = 60
=2
30
60
3
3
3
and µ = 15 = 4. Moreover, C0 = 1, C1 = 2, C2 = 2 , C3 = 4 , and C4 = 16
,
resulting in C0 + C1 + C2 + C3 + C4 = 87
.
16
P0 =
1
16
= ,
C0 + C1 + C2 + C3 + C4
87
32
,
87
24
P2 = C 2 · P0 = ,
87
P1 = C 1 · P0 =
9
MAS446 / MTH437
12
,
87
3
P4 = C 4 · P0 = .
87
P3 = C 3 · P0 =
(i) L = P1 + 2 · P2 + 3 · P3 + 4 · P4 =
(ii) P0 =
16
87
128
87
= 1.4713
= 0.1839
(iii) λ̄ = λ0 · P0 + λ1 · P1 + λ2 · P2 + λ3 · P3 + λ4 · P4 = 440
,
87
128
32
L
W = λ̄ = 440 = 110 = 0.2909.
32/110
Finally, what we need is 1/2+32/110
= 32
= 0.3678. Alternatively, we
87
1
1
3
can compute it in this way: 4 P1 + 2 P2 + 4 P3 + P4 = 32
= 0.3678.
87
(iv) P0 + 21 P1 =
32
87
= 0.3678.
♦
END OF PAPER
10