91-412-01 Operations Research II
Winter 2003
Solution of Assignment 1
Problem #2 p. 629
Let Hi=1 if i'th card drawn is a heart and Hi=0 otherwise.
If T= total number of hearts, then T= H1+ H2+ H3+ H4+ H5.
E(Hi) = 1/4 (1) +3/4 (0)=1/4,
V(Hi) = ¼ (1)2 + ¾ (0)2-(1/4)2=3/16
E(T) = 5(1/4)=5/4
Total payoff in $= 4T
E (total payoff in $) = E (4T)= 4*E(T)= $5
V(T) = 5 V (Hi) = 15/16
V(total payoff in $) = V (4T)= 42* V(T)= 16*15/16=$15
Problem #3 p. 629
x
1
3 a. F ( x )   e s ds   e s
0

3b. E( x )   xe x dx  
0
x
0
 1  e x , x  0

xe x
0

  e  x dx  0 
0

 e x
0
0
1
We have used integration by parts with u=x, dv=e-x , du= dx and v= -e-x

2 x
E( x )   x e
2
0
dx  

x 2e x
0

 2  xe x dx  2
0
Here we have used integration by parts with u=x2, dv=e-x , du= 2x and v=-e-x
Var X = E(x2) - E(x)2=1
3c. P( 1  X  2 )  F ( 2 )  F ( 1 )  e 1  e 2  0.23
Problem #3 p. 967
Let the states be the number of working machines at the beginning of a day; then we have
the states as: 0, 1, and 2 working machines,
0
0 working
1
2
0
1 
 0
 0 1 / 3 2 / 3


1 / 9 4 / 9 4 / 9
1 working
2 working
For example, if 0 machines are working at the beginning of the current day, then two
machines must have broken down during the previous day, and at the beginning of the
next day two machines will be working. If two machines are working at the beginning of
a day, let Wi = event that machine I doesn’t break down during the current day. Then next
day begins with 0 machines working with probability
P( w1  w2 )  ( 1 / 3 ) 2  1 / 9
Next day begins with 1 machine working with probability
P( w1  w2 )  P( w2  w1 )  2( 1 / 3 )( 2 / 3 )  4 / 9
Next day begins with 2 machine working with probability
P( w1  w2 )  ( 2 / 3 ) 2  4 / 9
Problem #1 p. 972
1. U=Urban S=Suburban R=Rural. Then
U
S
R
.80 .15 .05
S .06 .90 .04


R .04 .06 .90
U
P=
a. Puu ( 2 )  .80 .15 .05
Pus ( 2 )  .80 .15 .05
.80
.06  .65
 
.04
.15
.90  .258
 
.06
.05
Pur ( 2 )  .80 .15 .05 .04  .091
.90
Note that these probabilities sum to one; after all, after two years an urban deweller must
be somewhere.
b. q=[.40 .35 .25]. We seek
 .651
2
q( column1 of p )  .40 .35 .25 .104  .315
.072
c. The moving tendencies of Americans change overtime. After all, people used to move
into urban areas, and now people are moving out of urban areas. Thus we have, in all
probability, non-stationary Markov Chain.
Problem #3 p. 975
3a.
3b.
3c.
3d.
State 4 is transient.
States 1,2,3, 5 and 6 are recurrent.
{1,3,5} and {2,6} are closed sets.
Since states 4 and 1 do not communicate the chain is not ergodic.
Problem #9 p. 983
Let AA= had an accident during each of the last two years
NA= No accident two years ago but an accident last year
AN= Accident two years ago but no accident last year
NN= No accident during each of the last two years
We given that
AA
AA
NA
AN
NN
NA
AN NN
.10 0 .90 0 
.10 0 .90 0 


 0 .03 0 .97 


 0 .03 0 .97 
The steady state probabilities are  AA  1/ 310 ,  NA  9 / 310 ,  AN  9 / 310 , and
 NN  291 / 310 . Thus the expected premium per year is
(1/310)400+ (18/310)(300)+(291/310)(100)=$112.58.
Extra problem
Let:
state 1= a patient is in a critical condition
state 2= a patient is in a serious condition
state 3= a patient is in a stable condition
a) We need to calculate P13(3)
P13(3)=[P3]13= 0.19
0.526 0.284 0.19 
Where
P  0.472 0.308 0.22 
0.382 0.344 0.274
b) We need to calculate m23
3
From
mij  1 
 pik mkj , we have
k j
m23  1  p 21 m13  p 22 m23  1  0.4m13  0.4m23
m13  1  p11 m13  p12 m23  1  0.7m13  0.2m23
By solving the two equations, then we have
m23  7 and m13  8
c) We have to solve :
  P
and
 i  1
, i = 1, 2, 3
Then we have   0.4783 0.3043 0.2174