The OR Final Exam July 31

An opportunity for the discriminating student to do good.

Taking the OR

Exam just has to be

about the best thing in

the world!

The Rules of Engagement • Open book – notes, presentations, journal articles

• No Computers but calculators should be used

• 90 minutes in length

• Internet Students need to register for the exam by Noon Monday July 29th

• 3 problem solving questions

I decided not to use my

book or notes so I can

demonstrate complete

mastery of the subject.

For your educational enjoyment:

•Question #1

•Markov Process (steady-state)

•Question #2

•Queuing Model (Rate diagram problem)

•Question #3

•Classical Optimization (Unconstrained, Lagrangian or

KKT problem)

4

The Semester in Review A fast pace look at the highlights of

the Operations Research course.

- See actual problems solved,

- Feel the suspense as the semester comes to an end,

- Relive those glorious day of summer!

5

What will we see?

• Steady-state Markov problems

• Queuing problems with variations

• An unconstrained optimization problem

• An equality constrained optimization problem

• An inequality constrained optimization problem

6

The A. Salt Mining Company work force consisting of 70 migrant workers. Transition

period is one day

Next day today On time Late No show

On time 0.4 0.3 0.3

Late 0.8 0.2 0

No show 0 0.8 0.2

P matrix Next day

today On time Late No show

On time 0.4 a. 0.3 0.3

Late 0.8 0.2 0

No show 0 0.8 0.2

(a) What is the probability that a worker that was on time on a Monday

will be late for work on the next day?

7

(b) What is the long run percent of workers that will (1) show up

for work on time, (2) are late, or (3) are no shows on any given

day?

P matrix Next day

Today On time Late No show

On time 0.4 0.3 0.3

Late 0.8 0.2 0

No show 0 0.8 0.2

1 1 2

3 1 3

2 1 3

.4 .8

.3 .2

1

1 2

1 3

1 2 3

1 2 3

.6 .8 0

.3 .8 0

1

.471; .353; .176

(c) If a worker is late today, what is the expected number of days before that worker is late again?

8

22

2

1 12.83

.353days

(d) If a worker is late on a Monday, what is the probability that worker will not show up for work

Wednesday?

9

P2 matrix Wednesday

Monday On time Late No show

On time 0.4 0.42 0.18

Late 0.48 0.28 d. 0.24

No show 0.64 0.32 0.04

23

2

.3

.8 .2 0 0 .24

.2

p

P matrix Next day

Today On time Late No show

On time 0.4 0.3 0.3

Late 0.8 0.2 0

No show 0 0.8 0.2

(e) If Mark Off, a migrant worker, is late today, what is the expected number of days before Mark is at work on time?

10

21 21 31

21

21

1 .2 0

.8 1

11.25

.8

u

P matrix Next day

Today On time Late No show

On time 0.4 0.3 0.3

Late 0.8 0.2 0

No show 0 0.8 0.2

1ij ik kj

k j

u p u

j = 1, k = 2, 3

(f) Today was a particularly bad day with only 30 workers showing up on time, 15 were late, and the remaining were no-shows. What is the expected number of workers that will be at

work on time tomorrow?

11

30 15 25 On time Late No show

0.4 0.3 0.3

0.8 0.2 0

0 0.8 0.2 (30) (.4) + 15 (.8) + 25 (0) = 24 32 14

A Queuing Problem

• WeTube is a video-sharing website that maintains 3 web servers.

• Each server fails at the rate of once every 5 days.

• It takes Sam, the maintenance man, a mean of 2 days to restore a failed server.

• However, if all 3 servers have failed, then Sam calls on Pam to restore one of the servers (also with a mean time of 2 days).

• Assume steady-state with exponential time between failures and restoration times.

Construct the rate diagram

0

3/5 2/5 1/5

1/2 1/2 2(1/2)

1 2 3

What is the probability that Sam is not busy?

0

3/5 2/5 1/5

1/2 1/2 2(1/2)

1 2 3 C1

= 6/5 ; C2 = (6/5)(4/5) = 24/25

C3 = (24/25)(1/5)

P0 = [1 + 6/5 + 24/25 + 24/125]-1 = 0.29833

Compute the average number of servers down for repair.

0

3/5 2/5 1/5

1/2 1/2 2(1/2)

1 2 3 C1

= 6/5 ; C2 = (6/5)(4/5) = 24/25

C3 = (24/25)(1/5)

P0 = [1 + 6/5 + 24/25 + 24/125]-1 = 0.29833

L = (1)(6/5)(0.29833) + 2(24/25)(0.29833) +

3(24/125)(0.29833) = 1.103

What is the average time in days that a server is down for repair?

0

3/5 2/5 1/5

1/2 1/2 2(1/2)

1 2 3 C1

= 6/5 ; C2 = (6/5)(4/5) = 24/25

C3 = (24/25)(1/5)

P0 = [1 + 6/5 + 24/25 + 24/125]-1 = 0.29833

L = (1)(6/5)(0.29833) + 2(24/25)(0.29833) +

(24/125)(0.29833) = 1.103

bar = (3/5)(0.29833) + (2/5)(6/5)(0.29833)

+ (1/5)(24/25)(0.29833) = .37947

W = L/ bar = 2.906 days

What is the probability that no server will be available on-line?

0

3/5 2/5 1/5

1/2 1/2 2(1/2)

1 2 3 C1

= 6/5 ; C2 = (6/5)(4/5) = 24/25

C3 = (24/25)(1/5)

P0 = [1 + 6/5 + 24/25 + 24/125]-1 = 0.29833

L = (1)(6/5)(0.29833) + 2(24/25)(0.29833) +

(24/125)(0.29833) = 1.103

P3 = (24/125)(0.29833) = 0.05728

We are going to work some more

problems!

This is going to be so great!

Another Markov Problem

• Families in the country of Marcoff live in one of two regions: the Mountains or the Coastal region.

• Each year a certain fraction of the families in each region will migrate to the other region. This annual migration pattern is captured in the following table.

• Model this migration pattern as a Markov chain.

From Region

To region

Mountains Coastal region

Mountains .8 .2

Coastal region .3 .7

Question 1

• Find the steady-state percent of the population that will live in each region.

From Region

To region

Mountains Coastal region

Mountains .8 .2

Coastal region .3 .7

1 2

3 2;

5 5

1 1 2

1 2

2 1

1 1

.8 .3

1

.2

.3

2 51 1

3 3

Question 2

• What is the expected number of years for a family in each region to return to that region?

From Region

To region

Mountains Coastal region

Mountains .8 .2

Coastal region .3 .7

1 2

3 2;

5 5

Expected reoccurrence times

Mountains Coastal region

5/3 = 1.67 years

5/2 =2.5 years

Question 3

• If a family is currently living in the Mountains, what is the probability that they will be living in the Mountains or the Coastal region two years from now?

From Region

To region

Mountains Coastal region

Mountains .8 .2

Coastal region .3 .7

2.8 .2 .8 .2 .7 .3

.3 .7 .3 .7 .45 .55P

Mountains Coastal

0.7 0.3

Question 4 • What are the expected first passage times for families living in

the Mountains to migrate to the coastal region and for families living in the coastal region to migrate to the Mountains?

From Region

To region

Mountains Coastal region

Mountains .8 .2

Coastal region .3 .7

12 12

12

1 .8

1/ .2 5

Mountains Coastal

5 yr 3 1/3 yr

21 21

21

1 .7

11/ .3 3

3

Question 5 • The following table shows the current number of families (in

1,000’s) living in each region. What will be the population (in 1,000’s) of each region a year from now?

From Region

To region

Mountains Coastal region

Mountains .8 .2

Coastal region .3 .7

Mountains Coastal

500 500

Mountains Coastal

400 600

.8 .2

400 600 500 500.3 .7

Another queuing problem • The Dayton library has two copies of the best

seller novel by M.L. Stedman titled The Light Between Oceans.

• The mean demand rate for this book by patrons of the library is 2 per week.

• The average (mean) time that the book is checked out is for 2 weeks.

• If both books are checked out, then up two patrons may be placed on a holding list.

• Assume steady-state with exponential time between demands and check out times and an infinite calling population.

Construct the rate diagram

0

2 2 2

1/2 2(1/2) 2(1/2)

1 2 3 4

2

2(1/2)

What is the probability that both books are available?

0

2 2 2

1/2 2(1/2) 2(1/2)

1 2 3 4

2

2(1/2)

C1 = 4 ; C2 = 8; C3 = 16; C4 = 32

P0 = [1 + 4+ 8 + 16 + 32]-1 = [61]-1 = .0164

Compute the average number of patrons on the holding (waiting) list.

P0 = [1 + 4+ 8 + 16 + 32]-1 = [61]-1 = .0164

Lq = (1)(16/61)+ 2(32/61) = 80/61 = 1.31

0

2 2 2

1/2 2(1/2) 2(1/2)

1 2 3 4

2

2(1/2)

C1 = 4 ; C2 = 8; C3 = 16; C4 = 32

What is the average time in days that a patron is on the holding list before obtaining the novel?

bar = 2(1-32/61) = 58/61

Wq = Lq/ bar = (80/61) / (58/61) = 80/58 = 1.38

weeks = 9.65 days

P0 = P0 = [1 + 4+ 8 + 16 + 32]-1 = [61]-1 = .0164

Lq = (1)(16/61)+ 2(32/61) = 80/61 = 1.31

C1 = 4 ; C2 = 8; C3 = 16; C4 = 32

What is the expected number of patrons requesting the novel that are turned away each week?

P4 = 2 (32/61) = 64/61 = 1.05

0

2 2 2

1/2 2(1/2) 2(1/2)

1 2 3 4

2

2(1/2)

C1 = 4 ; C2 = 8; C3 = 16; C4 = 32

• The Light Between Oceans is an incredible moving novel about what happens when good people make bad decisions. The story takes place in the town of Point Partageuse, Australia during the 1920s. The story begins when a light house keeper (Tom) and his wife (Isabel) find a life boat with ….

32

An unconstrained optimization problem

There are no constraints in reaching the summit

33

Nonlinear Programming The I. B. A. Failure Company makes two types of

anti-lock brake systems having the following demand

functions:

x = 100 - .05 Sx - .1y

y = 300 - .10 Sy - .2 x

where Sx and Sy are the selling prices of the x and y models.

Dan D. Lion, Vice-Pres for Product Development must set

the unit prices in order to maximize revenue. Daf O. Dil, a

management scientist, says to find a stationary point and

then check the second order conditions.

note: x,y are in hundreds

34

Sx = 2000 – 20x – 2y

Sy = 3000 – 2x – 10y

R = Sx x + Sy y = 2000x – 20x2 - 2xy + 3000y – 2xy – 10y2

Rx = 2000 – 40x – 4y = 0

Ry = 3000 – 20y – 4x = 0

x* = 35.7 and y* = 143

Sx = $1000 and Sy = $1498.60

Let’s solve this thing…

35

R(x,y) = 2000x – 20x2 - 2xy + 3000y – 2xy – 10y2

Rx = 2000 – 40x – 4y = 0

Ry = 3000 – 20y – 4x = 0

Rxx = - 40 < 0

Ryy = - 20

Rxy = - 4

Rxx Ryy –(Rxy)2 = (- 40)( - 20) – 16 >0

therefore the function is concave.

2nd order conditions:

36

An equality constrained optimization problem

Lagrange to the rescue

Must stay on the road in climbing to the summit

37

The Maxi Mum Company • The Maxi Mum Company must produced 1,000 units of a

product over the next 4 months. Because of changes in labor and material costs, the profit generated from each month’s production will change. If xi is the number of units produced in month i and the total profit is given by the following function, find the production levels for each month that will maximize the production of the 1,000 units.

.5 .5 .5 .5

1 2 3 4Profit 6 4 3 2x x x x

38

Setting up the solution conditions (a) Set up the Lagrangian function:

(b) Construct the necessary conditions:

1 2 3 4

.5 .5 .5 .5

1 2 3 4 1 2 3 4

L(x , x , x , x , )

6 4 3 2 x x x x 1000

v

x x x x v

4

1

0; 1,2,3,42

1000 0

i

i i

i

i

aLv i

x x

x

Where a1 = 6, a2 = 4, a3 = 3; a4 = 2

39

Now solving… (c) Solve for the optimum monthly production levels.

2

2

2

4 4 42 2

2 21 1 1

2

4 422 2

1 1

02

1

2 2 4

1 11000

4 4

1 1000 1000;

4

i

i i

i ii i i

i i i

i i i

i i

i i

i i

aLv

x x

a ax or x a

v v v

x a av v

x av

a a

40

What about a numerical solution?

42 2 2 2 2

1

2

42

1

1

2

3

4

10006 4 3 2 65; 15.3846

65

1000

15.3846 36 553.85

15.3846 16 246.15

15.3846 36 138.46

15.3846 4 61.54

i

i

i i

i

i

a

x a

a

x

x

x

x

41

Sufficiency? .5 .5 .5 .5

1 2 3 4

.5 .5 .5 .5

1 2 3 4

1 1 3 4

1.5

1

1.5

2

1 2 3 4

1.5

3

1.5

4

Profit 6 4 3 2

Profit Profit Profit Profit3 ; 2 ; 1.5 ;

1.50 0 0

10 0 0

( , , , ).75

0 0 0

.50 0 0

x x x x

x x x xx x x x

x

xH x x x x

x

x

42

An inequality constrained optimization problem

Karush-Kuhn-Tucker to the rescue

Constrained from reaching the summit

43

KKT Problem!

Max f(x,y) = 24 ln(x) + 40 ln(y)

s.t. 5x + 8y ≤ 192

where ln is the natural log function

a. Write the Lagrangian function for this problem.

L(x,y, ) = 24ln(x) + 40ln(y) - (5x + 8y -192)

44

K-T Conditions

L(x,y, ) = 24ln(x) + 40ln(y) - (5x + 8y -192)

b. Construct the necessary (Kuhn-Tucker) conditions for

solving this problem.

I. 24/x - 5 = 0 II. 5x + 8y ≤ 192

40/y - 8 = 0

III. (5x – 8y – 192) = 0 IV. ≥ 0

45

c. Solve (b)

First assume = 0, then from I. infeasible

Second assume > 0, then

from I. 8(24) / x – 200/y = 0 or y = (25/24) x

and substituting into III for y gives 5x + 8 (25/24) x = 192

or x = 14.4 and y = 15

then from I: * = 40 / (8y) = 1/3 > 0

Therefore found a K-T point.

I. 24/x - 5 = 0 II. 5x + 8y ≤ 192

40/y - 8 = 0

III. (5x – 8y – 192) = 0 IV. ≥ 0

46

Check for Sufficiency

The constraint is linear, therefore the feasible region is convex. The

KKT conditions are both necessary and sufficient.

d. Tell why or why not, your solution in (c) solves the original problem (i.e.

is a global maximum).

Since

fxx = -24/x2 < 0 ; fyy = -40/y2 < 0 ; fxy = 0 and fxx fyy – (fxy)2 > 0,

then f(x,y) is everywhere concave.

2

2

240

400

xH

y

47

Final Exam Warm-up Question

1. Combine linear, nonlinear, and integer programming into

the elusive unified field theory.

2. Develop a new mathematical system in which to develop

your theory.

3. Devise a costly experiment to support it.

4. Show with your theory that graduate students are both

concave and suboptimal and they will only converge

locally displaying complementary slackness.

5. Show also that the dual to a graduate student is a convex

polyhedron with a single pointed and obtuse vertex.

48

The OR Final Exam

Remember to support your local

operations researcher.

Coming soon to a location near you.

49

No title needed There is nothing better in life than to

do well on an OR exam.

I am so excited about the OR final exam.

Engineering Management Students enthusiastic about the OR Final

I wish there were more than 3 problems on

the final. This is going to

be the best test ever!

Yippee! I’m on my

way to take the OR

Exam! This will be fun!

I remember when

I took the OR Exam.

Gosh, those were the days!

My

book

I passed the OR

Exam last year. Look

at me now!

The OR Exam Be the first on your

block to pass it!