2 HOURS INST - Nanyang Technological University
I i
NANYANG
TECHNOLOGICAL
UNIVERSITY
SEMESTERI EXAMINATION 2O7O-2OLI MAS446- Probabilistic Methods in OR
December2010
TIME ALLOWED: 2 HOURS
INSTRUCTIONSTO CANDIDATES
1. This examination paper contains FOUR FIVE (5) printed pages.
(4) questions and comprises
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 answerbook. 4. This IS NOT an OPEN BOOK exam. 5. Candidates may use calculators. However, they should write down systematically the steps in the workings. 6. Pleasejustify your answers. Show your work.
I MAS446 QUESTTON 1.
(25 marks)
A student has to perform an assignmentthat has four parts: 1, 2, 3, and 4 (to be done in this order), Each part takes one day to complete. In each part, he has a probability of 0.1 of dropping out (i.e., not finishing the assignment), a probability 0.2 of having to repeat the part, and a probability 0.7 of moving on to the next part (in the fourth part, moving on means finishing the assignment). (i) Find how long a student can expect to be doing this assignment(starting from the beginning). (ii) Find the probability that a student will finish the assignment(starting from the beginning). Soluti,on: (1) We formulate the problem as a Markov chain of five states. State 0 State 1 State 2 State 3 State 4
Leave the assignment(dropping out or finishing) Be in the first part Be in the secondpart Be in the third part Be in the fourth part
The one-step transition matrix hence is
0 1 2 3 4
0123 1 0000 0.1 0.2 0.7 0 0 0.1 0 0.2 0.7 0 0.1 0 0 0.2 0.7 0.8 0 0 0 0 . 2
We are going to find the expectedfirst passagetime from state 1 to state 0, which is denoted by pro. It's done by solving the following linear equations, which come from p,;i : Ln+i pt*ltt i. * 0.7p'2g, I,Lto: I -t 0.21.t1s : I + 0.2p'26* 0.7 p3s, l-Lzo pso: I + 0.211's6 * 0.7 p'as,
I MAS446 Pao:L*0.21-tas Therefore, Fto : 4.1382. (2) We formulate the problem as a Markov chain of six states.
State0 State 1 State2 State 3 State 4 State 5
Drop out Be in the Be in the Be in the Be in the finish
first part secondpart third part fourth part
The one-step transition matrix hence is
0 I 1 0.1 2 0.1 3 0.1 4 0.1 5 0
0 0.2 0 0 0 0
0 0.7 0.2 0 0 0
000 000 0.7 0 0 0.2 0.7 0 0 0.2 0.7 001
We are going to find the absorption probability fts of the processentering state 5, given the starting state 1. It's done by solving the following linear equations,which come from fil,:DY-op;if in.
fu :0'2ftu + 0.7f2s, + 0.7hs, fru : 0.2fzu fss:0.2ftu+0.7f4s, fn:0-2fnu+0.7 Therefore, fm :0.5862.
MAS446 QUESTTON 2.
(25 marks)
In a self-servicerestaurant, each customer serves himself. Note that this corresponds to having an infinite number of servers available. Customers arrive according to a Poisson processwith parameter ), and service times have an exponential distribution with parameter p. (i) Find Ln and W* (ii) Construct the rate diagram for this queueing system. (iii) Use the balance equations to find the expressionfor Pn in terms of Po. (iv) Find Ps. (v) Find L and W. Solution: (u) L, : Ws: 0 since arriving customers never need to enter a queue becausethey immediateiy begin service (from themselves)when they arrive. (b) The rate diagram is shown below. (c) From the general results for the birth-and-death process,we have P n : # P o w h e r ep : \ l l r . (d) SinceD1,:oPn:l and Pn is expressedin terms of P6 in part (c), we have
: e-P. p"lnt\ ' \'7--o' ) (e) W :11t,,, sincethe only waiting is during the servicetime; L:\W:p. P.: fi -
l l
i l
tl l l
ll
I MAS446 QUESTTON 3.
(25 marks)
(a) Customers arrive at a repair shop according to a Poisson process at a mean rate of 28 per hour. The shop needs to hire one repair mechanic to serve the customers. There are two candidates in the pool: (1) Candidate A (fast but expensive) and (2) Candidate B (slow but inexpensive). Both candidates wouid have an exponential distribution for service times with A having a mean of 1.5 minutes and B having a mean of 2 minutes. The shop's revenue per month is given by $5,0001W where tr4lis the expected waiting time (in minutes) of a customer in the system. Determine the upper bound on the differencein their monthly compensationsthat would justify hiring A rather than B. Solution: M lM lI system. ) : 28. Assume a unit of time as one hour.
w:rlfu-^).
A: p":601I.5 : 40. W:IlI2:5min. Revenue:5000/5:1000. Bt p - 6012:30. W : I/2: 30 min. Revenue: 5000/30: 166.66 : 833.33. DifferencebetweenA and B is 1000-166.66 justify This is the upper bound that would hiring A rather than B. (b) Airplanes arrive for take-off at the runway of an airport according to a Poisson process at a mean rate of 20 per hour. The time required for an airplane to take off has an exponential distribution with a mean of 2 minutes, and this process must be completed before the next airplane can begin to take off. Because a brief thunderstorm has just begun, all airplanes which have not commencedtake-off havejust been grounded temporarily. However, airplanes continue to arrive at the runway during the thunderstorm to await its end. Assuming steady-stateoperation beforethe thunderstorm, determine the expected number of airplanes that will be waiting to take off at the end of the thunderstorm if it lasts 30 minutes. Solution: This is an MlMll queuewith ): 20 and F:30. Efr'o. of airplanes at the end of the thunderstorm ]
MAS446 : E[no. of airplanes before thunderstorm]+E[ no. arrived planes during the thunderstor
: Lqt \t : \(Ws+ r) : 2o(*-#ro + 0.5): 11.33.
ir
MAS446 QUESTION 4.
(25 marks)
A mobile phone making company wants to increasesales of its phone. The phone salesfluctuate between two levels- low and high - depending upon two factors: (1) whether they advertiseand (2) the advertising and marketing of new products by competitors. The secondfactor is out of the company's control, but it is trying to determine what its own advertising policy should be. For example, one policy is to advertise when sales are low but not to advertise when sales are high. Advertising in any quarter of a year has primary impact on sales in the followi,ngquarter. At the beginning of each quarter, the needed information is available to forecast accurately whether saleswill be low or high that quarter and to decidewhether to advertisethat quarter. The cost of advertising is $1 million for each quarter of a year in which it is done. When advertising is done during a quarter, the probability of having high salesthe next quarter is 112 or 314 depending upon whether the current quarter's sales are low or high. These probabilities go down fo Ll4 or 112 when advertising is not done during the current quarter. The company's quarterly profits (excluding advertising costs) are $4 million when salesare high but only $2 million when salesare low. Management now wants to determine the advertising policy that will maximize the company's (longrun) expected averagenet profit (profit minus advertising costs) per quarter. (i) Formulate this problem as a Markov decision process by identifying the states and decisionsand then finding the values for each p4@) and C,;u. (ii) Identify all the stationary deterministic policies. For eachone, find the transition matrix and write an expressionfor the (long-run) expectedaverage net profit per quarter in terms of the unknown steady-state probabilities (tro,rrtr...,Tu). (iii) Use the policy improvement algorithm to find an optimal policy when starting with an initial policy of never advertising. Solution: State 0 - low sales;State 1 - high sales. Decision 1 -
do not advertises;Decision 2 -
Initial Policy:
d o ( n l ): 1 d 1 ( . R 1 ): 1
advertise.
MAS446 Policy Improvement Algorithm: TTERATION 1 Value Determination:
- u0(R1) * 0.25u1(ft1) s(Rr) - -2+ 0.75u0(fi1) - u1(fi1) + 0.50u1(R1) s(Rr)- -4+ 0.50o0(R1) Solution of Value Determination Equations:
s(Rr): -2'67 u0( R1):2' 667 o 1 ( B t ): 6 Policy Improvement: State0: -2 + 0.75(2.667) + (O)- (2.667): -2.67 for decision1 -1 + 0.5(2.667) + (0) - (2.667): -2.33 for decision2 State1: -4+0.50(2.667)+ (O)- (0) : -2.67 for decision1 -3 + 0.25(2.667) + (0)(0): -2.33 for decision2 The minimumfor both statesis achievedby usingdecision1 (don't advertise). Sincethis policy is identicalto the precedingpolicy (the initial policy), it must be an optimal policy. Optimal Policy: d o ( R 2 :)1
dr(R2): r s(RI): -2.67 u0( R1):2' 667 t ' 1 ( R r ): 3 END OF PAPER
NANYANG
TECHNOLOGICAL
UNIVERSITY
SEMESTERI EXAMINATION 2O7O-2OLI MAS446- Probabilistic Methods in OR
December2010
TIME ALLOWED: 2 HOURS
INSTRUCTIONSTO CANDIDATES
1. This examination paper contains FOUR FIVE (5) printed pages.
(4) questions and comprises
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 answerbook. 4. This IS NOT an OPEN BOOK exam. 5. Candidates may use calculators. However, they should write down systematically the steps in the workings. 6. Pleasejustify your answers. Show your work.
I MAS446 QUESTTON 1.
(25 marks)
A student has to perform an assignmentthat has four parts: 1, 2, 3, and 4 (to be done in this order), Each part takes one day to complete. In each part, he has a probability of 0.1 of dropping out (i.e., not finishing the assignment), a probability 0.2 of having to repeat the part, and a probability 0.7 of moving on to the next part (in the fourth part, moving on means finishing the assignment). (i) Find how long a student can expect to be doing this assignment(starting from the beginning). (ii) Find the probability that a student will finish the assignment(starting from the beginning). Soluti,on: (1) We formulate the problem as a Markov chain of five states. State 0 State 1 State 2 State 3 State 4
Leave the assignment(dropping out or finishing) Be in the first part Be in the secondpart Be in the third part Be in the fourth part
The one-step transition matrix hence is
0 1 2 3 4
0123 1 0000 0.1 0.2 0.7 0 0 0.1 0 0.2 0.7 0 0.1 0 0 0.2 0.7 0.8 0 0 0 0 . 2
We are going to find the expectedfirst passagetime from state 1 to state 0, which is denoted by pro. It's done by solving the following linear equations, which come from p,;i : Ln+i pt*ltt i. * 0.7p'2g, I,Lto: I -t 0.21.t1s : I + 0.2p'26* 0.7 p3s, l-Lzo pso: I + 0.211's6 * 0.7 p'as,
I MAS446 Pao:L*0.21-tas Therefore, Fto : 4.1382. (2) We formulate the problem as a Markov chain of six states.
State0 State 1 State2 State 3 State 4 State 5
Drop out Be in the Be in the Be in the Be in the finish
first part secondpart third part fourth part
The one-step transition matrix hence is
0 I 1 0.1 2 0.1 3 0.1 4 0.1 5 0
0 0.2 0 0 0 0
0 0.7 0.2 0 0 0
000 000 0.7 0 0 0.2 0.7 0 0 0.2 0.7 001
We are going to find the absorption probability fts of the processentering state 5, given the starting state 1. It's done by solving the following linear equations,which come from fil,:DY-op;if in.
fu :0'2ftu + 0.7f2s, + 0.7hs, fru : 0.2fzu fss:0.2ftu+0.7f4s, fn:0-2fnu+0.7 Therefore, fm :0.5862.
MAS446 QUESTTON 2.
(25 marks)
In a self-servicerestaurant, each customer serves himself. Note that this corresponds to having an infinite number of servers available. Customers arrive according to a Poisson processwith parameter ), and service times have an exponential distribution with parameter p. (i) Find Ln and W* (ii) Construct the rate diagram for this queueing system. (iii) Use the balance equations to find the expressionfor Pn in terms of Po. (iv) Find Ps. (v) Find L and W. Solution: (u) L, : Ws: 0 since arriving customers never need to enter a queue becausethey immediateiy begin service (from themselves)when they arrive. (b) The rate diagram is shown below. (c) From the general results for the birth-and-death process,we have P n : # P o w h e r ep : \ l l r . (d) SinceD1,:oPn:l and Pn is expressedin terms of P6 in part (c), we have
: e-P. p"lnt\ ' \'7--o' ) (e) W :11t,,, sincethe only waiting is during the servicetime; L:\W:p. P.: fi -
l l
i l
tl l l
ll
I MAS446 QUESTTON 3.
(25 marks)
(a) Customers arrive at a repair shop according to a Poisson process at a mean rate of 28 per hour. The shop needs to hire one repair mechanic to serve the customers. There are two candidates in the pool: (1) Candidate A (fast but expensive) and (2) Candidate B (slow but inexpensive). Both candidates wouid have an exponential distribution for service times with A having a mean of 1.5 minutes and B having a mean of 2 minutes. The shop's revenue per month is given by $5,0001W where tr4lis the expected waiting time (in minutes) of a customer in the system. Determine the upper bound on the differencein their monthly compensationsthat would justify hiring A rather than B. Solution: M lM lI system. ) : 28. Assume a unit of time as one hour.
w:rlfu-^).
A: p":601I.5 : 40. W:IlI2:5min. Revenue:5000/5:1000. Bt p - 6012:30. W : I/2: 30 min. Revenue: 5000/30: 166.66 : 833.33. DifferencebetweenA and B is 1000-166.66 justify This is the upper bound that would hiring A rather than B. (b) Airplanes arrive for take-off at the runway of an airport according to a Poisson process at a mean rate of 20 per hour. The time required for an airplane to take off has an exponential distribution with a mean of 2 minutes, and this process must be completed before the next airplane can begin to take off. Because a brief thunderstorm has just begun, all airplanes which have not commencedtake-off havejust been grounded temporarily. However, airplanes continue to arrive at the runway during the thunderstorm to await its end. Assuming steady-stateoperation beforethe thunderstorm, determine the expected number of airplanes that will be waiting to take off at the end of the thunderstorm if it lasts 30 minutes. Solution: This is an MlMll queuewith ): 20 and F:30. Efr'o. of airplanes at the end of the thunderstorm ]
MAS446 : E[no. of airplanes before thunderstorm]+E[ no. arrived planes during the thunderstor
: Lqt \t : \(Ws+ r) : 2o(*-#ro + 0.5): 11.33.
ir
MAS446 QUESTION 4.
(25 marks)
A mobile phone making company wants to increasesales of its phone. The phone salesfluctuate between two levels- low and high - depending upon two factors: (1) whether they advertiseand (2) the advertising and marketing of new products by competitors. The secondfactor is out of the company's control, but it is trying to determine what its own advertising policy should be. For example, one policy is to advertise when sales are low but not to advertise when sales are high. Advertising in any quarter of a year has primary impact on sales in the followi,ngquarter. At the beginning of each quarter, the needed information is available to forecast accurately whether saleswill be low or high that quarter and to decidewhether to advertisethat quarter. The cost of advertising is $1 million for each quarter of a year in which it is done. When advertising is done during a quarter, the probability of having high salesthe next quarter is 112 or 314 depending upon whether the current quarter's sales are low or high. These probabilities go down fo Ll4 or 112 when advertising is not done during the current quarter. The company's quarterly profits (excluding advertising costs) are $4 million when salesare high but only $2 million when salesare low. Management now wants to determine the advertising policy that will maximize the company's (longrun) expected averagenet profit (profit minus advertising costs) per quarter. (i) Formulate this problem as a Markov decision process by identifying the states and decisionsand then finding the values for each p4@) and C,;u. (ii) Identify all the stationary deterministic policies. For eachone, find the transition matrix and write an expressionfor the (long-run) expectedaverage net profit per quarter in terms of the unknown steady-state probabilities (tro,rrtr...,Tu). (iii) Use the policy improvement algorithm to find an optimal policy when starting with an initial policy of never advertising. Solution: State 0 - low sales;State 1 - high sales. Decision 1 -
do not advertises;Decision 2 -
Initial Policy:
d o ( n l ): 1 d 1 ( . R 1 ): 1
advertise.
MAS446 Policy Improvement Algorithm: TTERATION 1 Value Determination:
- u0(R1) * 0.25u1(ft1) s(Rr) - -2+ 0.75u0(fi1) - u1(fi1) + 0.50u1(R1) s(Rr)- -4+ 0.50o0(R1) Solution of Value Determination Equations:
s(Rr): -2'67 u0( R1):2' 667 o 1 ( B t ): 6 Policy Improvement: State0: -2 + 0.75(2.667) + (O)- (2.667): -2.67 for decision1 -1 + 0.5(2.667) + (0) - (2.667): -2.33 for decision2 State1: -4+0.50(2.667)+ (O)- (0) : -2.67 for decision1 -3 + 0.25(2.667) + (0)(0): -2.33 for decision2 The minimumfor both statesis achievedby usingdecision1 (don't advertise). Sincethis policy is identicalto the precedingpolicy (the initial policy), it must be an optimal policy. Optimal Policy: d o ( R 2 :)1
dr(R2): r s(RI): -2.67 u0( R1):2' 667 t ' 1 ( R r ): 3 END OF PAPER
