Algebraic Solutions to the Questions

ALGREBRAIC SOLUTIONS TO THE QUESTIONS
PART A
QN 1. For this question, we use the theorems provided in the question hence:
(a) Given P= P11 0.2 P13
P21 0.7 P23
P31 0.1 P33
Let state matrix be Q such that if:
Understocked, Q1= 0.7
0.2
0.1
If stocked, Q2= 0.2
0.7
0.1
If overstocked, Q3= 0.4
0.4
0.2
Therefore the possible future states are determined as follows:
Let future states be denoted as F:
(i) So if understocked in the previous week, F1= P11 0.2 P13 0.7
P21 0.7 P23 0.2
P31 0.1 P33 0.1
0.7P11+0.04+0.1P13
= 0.7P21+0.14+0.1P23
0.7P31+0.01+0.1P33

(ii) And if stocked the previous week, F2= P11 0.2 P13 0.2
P21 0.7 P23 0.7
P31 0.1 P33 0.1
(0.2P11+0.14+0.1P13)
= (0.2P21+0.49+0.7P23)
(0.1P31+0.01+0.1P33)
(iii) And if overstocked the previous week, F3= P11 0.2 P13 0.4
P21 0.7 P23 0.4 =
P31 0.1 P33 0.2
= (0.4P11+0.08+0.2P13)
(0.4P21+0.28+0.2P23
(0.4P31+0.04+0.2P33)
This is a Markov chain matrix since all entries in the system state matrix are positive
(b) X(0) = 0
1
0
Given theorem 2: X(1)= P11 0.2 P13 0
P21 0.7 P23 1
P31 0.1 P33 0

= 0.2P11+0.14+0.1P13
0.2P21+0.49+0.1P23
0.2P31+0.07+0.1P33
X(2) = P11 0.2 P13 0.2P11+0.14+0.1P13
P21 0.7 P23 0.2P21+0.49+0.1P23
P31 0.1 P33 0.2P31+0.07+0.1P33
Hence this is reached at the 2nd week
2. (a) Given P= P11 0.2 P13
P21 0.7 P23
P31 0.1 P33
(I-P)q= 0
I= 1 0 0
0 1 0
0 0 1
Let q = X1
X2
X3

1 0 0 P11 0.2 P13
0 1 0 - P21 0.7 P23 = 0
0 0 1 P31 0.1 P33
1-P11 -0.2 -P13 X1
P21 0.3 -P23 X2 = 0
-P31 -0.1 1-P33 X3
(b) If the elements of the state transition matrix above were known then we could determine
the state vector easily
PART B:
QN1: (a) The given equation is: y=ax2+bx+c
Now in order to determine the equation above, the given points within the curve are used:
That is: (0,7), (1,9), (3,19), (6,49)
When x=0, y= a(0)2+b(0)+c = 7 hence c= 7
X=1, y=9: 9= a+b+c......(ii)
X=3, y= 19: 19= 9a+3b+c.....(iii)
X=6, y= 49
49=36a+6b+c......(iv)
Subtracting equation (iii) from (iv), results in: 30= 27a+3b.....(v)
From equation (ii): 9= a+b+7->a=2-b.....(vi)
Substituting equation (vi) into (v), we have : 27(2-b)+3b= 30
54-27b+3b=30
-24b=-24, b= 1 and a=2-1= 1
Therefore the equation is determined as: y= x2+x+7