Given f(x) = e^ax
∴ f(0) = e^0x = 1  (1)
Δf(x) = f(x + h) - f(x)  (2)
Δf(0) = f(1) - f(0) = e^a(1) - e⁰ = e^a-1 (2)
Δ²f(0) = Δ [Δf(0)]
= Δ[f(1) + f(0)] = Δf(1) - f(1) - Δ(0)
= f(2) - f(1) - (e^a - 1)
= e^2a - e^a - e^a + 1
Δ²f(0) = e^2a- 2e^a + 1
From (1), (2) & (3), the three terms are 1, e^a - 1, e^2a -2e^a+1
⇒ 1, e^a -1, (e^a - 1)² [∵ (a - b)² = a² - 2ab + b²]
Common ratio r = \(\frac { { e }^{ a }-1 }{ 1 } =\frac { ({ { e }^{ a }-a) }^{ 2 } }{ { e }^{ a }-1 } \) = e^a - 1
Since the common ratio is same throughout Δf(0) and Δ²f(0) forms a G.P.

The assignment problem is a special case of transportation problem where the number of sources and destinations are equal. Here, jobs represent sources and machines represent destinations.

The objective function is minimize Z = \(\overset { m }{ \underset { i=1 }{ \Sigma } } \overset { n }{ \underset { j=1 }{ \Sigma } } { { C }_{ ij } }{ x }_{ ij }\) subject to the constraints
\(\overset { n }{ \underset { j=1 }{ \Sigma } }{ x }_{ ij } = a_i, i=1,2,.....m\) (Supply constraints)
\(\overset { m }{ \underset { i=1 }{ \Sigma } }{ x }_{ ij } = b_j, j=1,2,.....n\) (demand constraints)
x_ij ≥, 0 for all i, j (non-negative restrictions)

The ratio between the total value of current period and total value of the base period is known as true value ratio.
(\( \frac{\sum P_{1} q_{1}}{\sum p_{0} q_{0}}\) is a true value tario)

The two normal equations are
\(\sum\) Y = n a + b \(\sum\) X
\(\sum\) XY  = a  \(\sum\) X + b  \(\sum\)X²  where n is the number of years given in the data.

It is a general tendency of time series to increase or decrease or stagnates during a long period of time. An upward tendency is usually observed in population of a country, production, sales, prices in industries, income of individuals etc., A downward tendency is observed in deaths,epidemics, prices of electronic gadgets, water sources, mortality rate etc. It is not necessarily that the increase or decrease should be in the same direction throughout the given period of time. This feature is known as secular trend.

Given h = 1
LHS = (E⁻¹Δ) x³
= Δ(E^-1(x³))
= Δ(x - h)³  [∵ E^-1f(x) = f(x - nh)]
= Δ(1 - h)³ [∵ h = 1]
= (x - 1+ 1)³ - (x - 1)³ [∵ Δf(x) =f(x + h) - f(x)]
= x³ - (x - 1)³
= x³ - (x³ - 3x² + 3x - 1)
[∵ (a - b)³ = a³ - 3a²b + 3ab² - b³]
= x³ - x³ + 3x² - 3x + 1
= 3x² - 3x + 1
= RHS
Hence proved

(i) Max (2000,3500,4000) = 4000
∴ Crop C is the best according to maximin criteria
(ii) Min (8000,5000,5000) = 5000
∴ Crop B and C are best according to minimax criteria

 First allocation:

Second allocation:

Third allocation:

 Fourth allocation:

 Final allocation:

The total transportation cost is
\( =(15 \times 9)+(10 \times 5)+(35 \times 4) +(15 \times 7)+(25 \times 6) \)
= 135 + 50 + 140 + 105 + 150 = Rs. 580

Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is

Here only 3 tasks can be assigned to 3 men.
Step 1: Its not necessary, since each row contains zero entry. Go to Step 2.
Step 2:

Step 3 (Assignment) :

Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is

The optimal assignment (minimum) cost = Rs. 35

Using family budget method,
  C.L.I =  \(\frac {\sum PV}{\sum V}\) = \(\frac {3284.72}{28}\) = 117.31

Let the missing entries be y₂ and y₄
Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero
⇒ (E - 1)⁴y_k = 0
⇒ (E₄- 4E₃ + 6E₂ - 4E + 1) y_k = 0  (1)
Put k = 0 in (1) we get,
y₄ - 4y₃ + 6y₂ - 4y₁ + y₀ = 0
y₄ - 4(18) + 6y₂ - 4(11) + 7 = 0
⇒ y₄ - 72 + 6y₂ - 44 + 7 = 0
⇒ y₄ + 6y₂ = 109    (2)
Put k = 1 in (1) we get,
(E₄- 4E₃ + 6E₂ - 4E + 1) y₁ = 0
⇒ y₅ - 4y₄ + 6y₃ - 4y₂ +y₁ = 0
⇒ 32 - 4 (y₄) + 6 (18) - 4y₂ + 11 = 0
32 - 4y₄ + 108 - 4y₂ + 11 = 0
⇒ -4y₄ - 4y₂ + 151 = 0

Adding,           
⇒ y₂ = 14.25
Substituting y₂ = 14.25 in (2) we get,
y₄ + 6 (14.25) = 109
⇒ y₄ + 85.5 = 109
⇒ y₄ = 109 - 58.5
⇒ y₄ = 23.5

Given

 Find f(2.8)
Since the required value 2.8 is at the end of the table, apply Newton's backward interpolation formula.
x_n + nh = 2.8 ⇒ 3 + n (1) = 2.8
⇒  n = 2.8 - 3 = -0.2
The difference table is

Newton's backward interpolation formula is
y(x = x_n + nh) = \(\frac { n }{ 1! } { \triangledown y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }\)
⇒ y(2.8) = 34 + (-0.2) (23) + \(\frac { (-0.2)(-0.2+1) }{ 2 } (14)+\frac { (0.2)(-0.2+1)(-0.2+2) }{ 6 } \)(16)
⇒ y(2.8) = 34 - 4.6 + (-0.2) (0.8) (7) + (-0.2) (0.8) (1.8)
⇒ y(2.8) = 34 - 4.6 - 1.12- 0.288
⇒ y(2.8) = 27.992

The forward interpolation formula is
\({ y }_{ ({ x=x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
Here x₀ + nh = x ⇒ x₀ = 0, h = 1
∴ 0 + n = x ⇒ n = x.
The difference table is

\({ y }_{ (n=x) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
\(y=1+\frac { n }{ 1! } (1)+\frac { n(n-1) }{ 2! } (-2)+\frac { n(n-1)(n-2) }{ 6 } (12)\)
⇒ y = 1 + x + (x² - x)(-1) + x(x² - 3x + 2) (2)
⇒ y = 1 + x - x² + x + 2x³ - 6x² + 4x
⇒ y = 1+ 6x - 7x² + 2x³
Hence, the cubic polynomial is 2x³ - 7x²+ 6x + 1.

Here total supply = 25 + 35 + 40 = 100
total requirement = 30 + 25 + 45 100
total supply = total requirement
∴ The given problem is a balanced transportation problem.
Hence, there exists a feasible solution for the given transportation problem.
North West Corner Rule:
I - allocation:

[∵ min (25, 30) = 25]
II - allocation:

[∵ min (5, 35) = 5]
III - allocation:

[∵ min (25, 30) = 25]
IV - allocation:

[∵ min (5, 45) = 5]
V - allocation:

[∵ min (40, 40) = 40]
Thus, the allocations are

∴ The transportation schedule is
S₁ → D₁, S₂ → D₁, S₂ → D₂, S₂ → D₃, S₃ → D₃
Hence, the total transportation cost
= 25(9) + 5(6) + 25(8) + 5(4) + 40(9)
= 225 + 30 + 200 + 20 + 360
= Rs. 835

Here, the number of rows and columns are equal.
∴ The given assignment problem is balanced.
Step 1 : Select a smallest element in each row and subtract this from all the elements in its row.
∴ The cost matrix: of the given assignment problem is

Column 1 contains no zero. Go to step 2.
Step 2 : Select the smallest element (1) in column 1 and subtract this from all the elements in its column.

Since each row and column contains atleast one zero, assignments can be made.
Step 3 : Examine the rows with only one zero.
Row P, Q and S contains exactly one zero, mark them by 0 and mark the other zeros in the column byX.
Thus, all the four assignments have been made.
∴ The optimal assignment schedule and total cost is

\(\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|} \hline \text { Sample number } & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & \text { Total } \\ \hline \bar{X} & 49 & 45 & 48 & 53 & 39 & 47 & 46 & 39 & 51 & 45 & 462 \\ \hline \mathbf{R} & 7 & 5 & 7 & 9 & 5 & 8 & 8 & 6 & 7 & 6 & 68 \\ \hline \end{array}\)
\(\overline{\bar{X}}=\frac{\sum \bar{X}}{10}=\frac{462}{10}=46.2 \)
\(\bar{R}=\frac{\sum R}{10}=\frac{68}{10}=6.8 \)
\(\text { The control limits for } \overline{\mathrm{X}} \text { chart is }\)
\(\mathrm{UCL}=\overline{\overline{\mathrm{X}}}+\mathrm{A}_{2} \overline{\mathrm{R}}=46.2+(0.58)(6.8)=50.14\)
\(\mathrm{CL}=46.2\)
\(\mathrm{LCL}=\overline{\mathrm{X}}-\mathrm{A}_{2} \overline{\mathrm{R}}=46.2-(0.58)(6.8)=42.26 \) 

The control limits for range chart is
\(\mathrm{UCL}=\mathrm{D}_{4} \overline{\mathrm{R}}=(2.115)(6.8)=14.38 \)
\(\mathrm{CL}=\overline{\mathrm{R}}=6.8 \)
\(\mathrm{LCL}=\mathrm{D}_{3} \overline{\mathrm{R}}=0(6.8)=0\)

From the \(\overline X\) chart, we see that 4 points are outside the control limit lines. So we say that the process is out of control.
Conclusion: The above diagram shows all the three control lines with the data points plotted, since 2 points fall out of the control limits, we can say that the process is out of control.

Fisher's price index number 
\(P^{P}_{01}\) =  \(\sqrt \frac {\sum p_{1}q_{0}\times \sum p_{1}q_{1}}{{\sum p_{0}q_{0}\times \sum p_{0}q_{1}}}\) × 100
= \(\sqrt \frac{2140\times1880}{1560\times1344} \times 100\)
= \(\sqrt \frac{4023200}{2096640} \times 100\)
= \(\sqrt {1.9188} \times 100\)
\(P^{F}_{01}\) = 138.5
Time reversal test : 
P₀₁ × P₁₀ =  \(\sqrt \frac{\sum p_{1}q_{0}\times{p_{1}q_{1}}\times{p_{0}q_{1}}\times{p_{0}q_{0}}}{{\sum p_{0}q_{0}\times{p_{0}q_{1}}\times{p_{1}q_{1}}\times{p_{1}q_{0}}}}\)
= \(\sqrt \frac{2140\times1880 \times 1344\times1560}{1560\times1344 \times 1880\times2140 }\)
= \(\sqrt 1\) = 1
Time reversal test = 1
Factor reversal test: 
P₀₁ × Q ₀₁  =   \(\sqrt \frac{\sum p_{1}q_{0}\times{p_{1}q_{1}}\times{p_{0}q_{1}}\times{p_{0}q_{0}}}{{\sum p_{0}q_{0}\times{p_{0}q_{1}}\times{p_{1}q_{1}}\times{p_{1}q_{0}}}}\)
\(=\sqrt \frac{2140\times1880 \times 1344\times1560}{1560\times1344 \times 1880\times2140 }\)
= \(\sqrt{(\frac{1880}{1560})^2}\)
=  \(\frac {1880}{1560}\) 
P₀₁ × Q₀₁   =  \({\frac {\sum p_{1}q_{1}}{\sum p_{0}q_{0}}}\)

Smce \(\sum\)X = 0, a =\(\frac {\sum Y}{n}\) = \(\frac {244}{5}\) = 48.8
b = \(\frac {\sum XY}{\sum X^2}\) = \(\frac {20}{10}\) = 2
∴ The required equation of the straight line trend is given by
Y = a + bX  \(\Rightarrow \) Y = 48.8 + 2X
\(\Rightarrow \) Y = 48.8 +2 (X - 1994) .... (1)
∴ Number of salesmen in 1997 is put X = 1997 in (1)
∴ Y = 48.8 + 2 (1997 - 1994) 
= 48.8 + 2 (3) 
= 48.8 + 6 = 54.8
∴  Number of salesmen in 1997 is 54.8