IndE311 Winter 2007
Practice Final
Solution
Part 1: Multiple-choice
1.
Which of the following two lotteries would a risk-averse decision maker choose?
0.5
10
1
(A)
0.5
(a) A
2.
5
0
(b) B
(c) Indifferent
(d) Not enough information to know
Which of the following two lotteries would a risk-seeking decision maker choose?
0.5
8
1
(A)
0.5
(a) A
3.
(B)
5
(B)
0
(b) B
(c) Indifferent
(d) Not enough information to know
Which of the following information is not required for making a decision according to the
minimax regret criterion?
(a)
States of nature
(b)
Prior probabilities
(c)
Set of alternative actions
(d)
Payoff/loss function
For questions 4-6, refer to the Markov chain below.
1
B
A
0.2
0.5
0.2
0.5
C
D
1
0.3
1
4.
F
1
(b) 2
(c) 3
(d) 4
Which of the following is a periodic state?
(a) A
6.
0.3
How many communicating classes does the Markov chain have?
(a) 1
5.
E
(b) C
(c) F
(d) None
Which of the following is a recurrent state?
(a) A
(b) C
(c) F
(d) None
Page 1 of 9
IndE311 Winter 2007
Practice Final
Solution
Part 2: True/False
1.
An M/M/1/K system with =2 and =2 will never reach steady state.
T
F
T
F
T
F
Why: Because n = 0 if n ≥ K, this system will eventually reach the steady state.
2.
In a Markov chain, if state A is accessible from state B, then states A and B
are in the same communicating class.
Why: In order to have states A and B in the same communicating class, state B
needs to be accessible from state A.
3.
The Pn values can be derived from the flow balance equations for any
queueing system.
Why: The queueing system has to reach steady state before the Pn values can be
derived from the flow balance equations.
4.
In a Markov chain, stationary probabilities imply the transition probabilities
are independent of time.
T
F
5.
All Markov chains can reach steady state.
T
F
Why: Only irreducible and ergodic Markov chains can reach steady state.
6.
For a risk-neutral decision maker, the utility function for money is linear.
T
F
7.
A finite-capacity queueing system has a finite number of potential
customers.
T
F
Why: It is possible to have an infinite number of potential customers and still have a
finite-capacity queueing system, as in the M/M/s/K.
8.
An irreducible Markov chain with five states cannot have any absorbing
states.
T
F
9.
In a single-server queueing system, the utilization rate is 1 – P0.
T
F
Why: it doesn’t hold for M/M/1/K or when the system doesn’t have steady state.
Page 2 of 9
IndE311 Winter 2007
Practice Final
Solution
Part 3: Short Answer
1. Briefly explain (in words) what the Markovian property means for a Markov chain.
Markovian property implies that the conditional probability of any future event given any past
events and the present state is equal to the conditional probability of any future event given the
present state. Another explanation is that the future is independent of the past events and it does
depend only on the present state.
2. Put the appropriate relationship sign (, <, =, >, , or “?” for unknown) for the below pairs of
measures:
Expected value of experimentation (EVE) _______ Exp. value of perfect information (EVPI)
Expected value of experimentation (EVE) ___?____ Cost of experiment ()
3. Explain the difference between L and Lq in a queueing system.
L is the expected number of customers in the queueing system (includes customers being
served).
Lq is the expected queue length (excludes customers being served).
4. Consider an M/M/5 queueing system with five identical servers. Explain why the service rate
going from state 4 to state 3 is 4, where  is the service rate for an individual server.
This is because 4 servers are serving the 4 customers in the system.
5. Define what a state is in a queueing model.
The number of customers in the queueing system.
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IndE311 Winter 2007
Practice Final
Solution
6. Suppose you have a one-step transition probability matrix P. Consider the 100-step transition
probability matrix, P(100). What could it mean if the rows of P(100) are not identical? Give at least
two possible explanations.
1) The system doesn’t have a steady state.
2) The system hasn’t reached the steady state.
7. Briefly describe why the exponential distribution is not appropriate for service times for some
queueing systems.
The actual service time distribution may deviate greatly from the exponential form, particularly
when the service requirements of the customers are quite similar.
8. Explain what is unusual in the “decision-making” problems described by the following decision
trees.
(a)
No decision is needed to make in (a).
(b)
We are not facing any uncertainty in (b).
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IndE311 Winter 2007
Practice Final
Solution
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IndE311 Winter 2007
Practice Final
Solution
2. Parking on campus is very limited. My favorite lot is limited to only 3 spaces. Cars wanting
to use this area arrive according to a Poisson input process at the rate of 2 cars per hour. If a
car cannot find an empty space immediately, it may pull into a temporary space to wait until a
parked car leaves. The temporary space can hold only 2 cars. All other cars must go
elsewhere. The time cars tend to park has an exponential distribution with a mean of 1.5
hours.
(a) Draw a rate diagram for this situation, and clearly label the arrival and service rates.
(b) Write the balance equations.
Rate in = Rate out
(c) Suppose P0 is 1/22. Use this to determine the rest of the steady state probabilities.
P1  0.13636, P2  P3  P4  P5  0.20455
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IndE311 Winter 2007
Practice Final
Solution
(d) A car waiting in the temporary space can be considered waiting in a queue. Find Lq.
Lq  0 * P3  1* P4  2 * P5  0 * 0.20455  1* 0.20455  2 * 0.20455  0.61364
(e) Find the average number of cars in the entire parking lot, L.
L  0 * P0  1* P1  2 * P2  3 * P3  4 * P4  5 * P5
 0 * 0.04545  1* 0.13636  2 * 0.20455  3 * 0.20455  4 * 0.20455  5 * 0.20455
3
(f) Find the average time a car spends in the parking lot, W.
W 
L


L
 (1  P5 )

3
 1.8857
2(1  0.20455)
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IndE311 Winter 2007
Practice Final
3.
Solution
Machinists who work at a tool-and-die plant must check out tools from a tool center. An
average of ten machinists per hour arrive seeking tools. At present, the tool center is
staffed by a clerk who is paid $6 per hour and who takes an average of 5 minutes to handle
each request for tools. Since each machinist produces $10 worth of goods per hour, each
hour that a machinist spends at the tool center costs the company $10. The company is
deciding whether or not it is worthwhile to hire a helper for the clerk at $4 per hour. If the
helper is hired, the clerk will take an average of 4 minutes to process requests for tools.
Assume that service and interarrival time are exponential. Should the company hire a
helper? Suppose another option is to hire another clerk. Does it save more money to hire
another clerk as compared to hiring a helper?
λ=10 machinists/hr
Alternative 1 (status quo): M/M/1 queueing model
60
 12 machinists/hr
5

10
L

 5 machinists
   12  10
E (TC )  E ( SC )  E (WC )  Cs s  Cw L  $6(1)  $10(5)  $56

Alternative 2 (with helper): M/M/1 queueing model
60
 15 machinists/hr
4

10
L

 2 machinists
   15  10
E (TC )  E ( SC )  E (WC )  Cs s  Cw L  ($6  $4)(1)  $10(2)  $30
=
Yes, the company should hire a helper.
Alternative 3 (with another clerk): M/M/2 queueing model
60
 12 machinists/hr
5

10
5
=


s 2(12) 12
1
P0  s 1
n
( /  )
( /  ) s
1


n!
s ! 1   /( s  )
n 0
1
 1
 0.4118
n
(10 /12)
(10 /12) 2
1


n!
2! 1  10 /(2 *12)
n 0

Page 8 of 9
IndE311 Winter 2007
Practice Final
L  Lq 
Solution
 P0 ( /  ) s   0.4118(10 /12) 2 (5 /12)

 
 1.0084 machinists
 s !(1   )2 
2!(1  5/12) 2
E (TC )  E ( SC )  E (WC )  Cs s  Cw L  $6(2)  $10(1.0084)  $22.084
Yes, it saves more money to hire another clerk as compared to hiring a helper.
4. Optica, Ltd., makes prescription glasses according to orders received from customers. The
worker receives 30 orders per 8-hour day on average (according to a Poisson input process).
The time to complete a prescription is uniformly distributed between 14 and 16 minutes.
Determine the following:
This is a M/G/1 queueing model.
30 15

orders/hr
8
4
1 14  16
60

 15 minutes   
 4 orders/hr

2
15

2 
(16  14) 2 1
1
 minutes2 
hours2
12
3
3(60) 2
(a) The percentage of time the worker is idle.
P0  1    1 

15/ 4
 1
 1/16

4
(b) The expected number of orders for glasses waiting to be worked on (Lq)?
2
2
1
 15 
 15 
 


2 2
2
2
  
4 3(60)  16 
Lq 
 
 7.042 orders
2(1   )
 15 
2 1  
 16 
(c) The average time until a prescription is filled.
W  Wq 
1


Lq


1


7.042 1
  2.128 hours
15 / 4 4
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