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IGNOU MSTE-002 Solved Assignment 2024 | PGDAST

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MSTE-002 Assignment Question Paper 2024

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1.State whether the following statements are True or False and also give the reason in support of your answer.
a) The Set S = { ( x , y ) : 0 y 5 S = { ( x , y ) : 0 y 5 S={(x,y):0 <= y <= 5S=\{(x, y): 0 \leq y \leq 5S={(x,y):0y5 when 0 x 2 0 x 2 0 <= x <= 20 \leq x \leq 20x2 and 3 y 5 3 y 5 3 <= y <= 53 \leq y \leq 53y5 when 2 x 7 } 2 x 7 } 2 <= x <= 7}2 \leq x \leq 7\}2x7} is not a convex set.
b) If 10 is added to each of the entries of the cost matrix of a 3 × 3 3 × 3 3xx33 \times 33×3 assignment problem, then the total cost of an optimal assignment for the changed cost matrix will increase by 10 .
c) The solution to a transportation problem with m-rows (supplies) and n-columns (destinations) is feasible if number of positive allocations is m + n m + n m+n\mathrm{m}+\mathrm{n}m+n.
d) The Value d i 3 d i 3 d_(i) >= 3\mathrm{d}_{\mathrm{i}} \geq 3di3 indicates an outlying observation in regression analysis.
e) Variations which occur due to natural forces and operate in a regular and periodic manner over a span of less than or equal to one year are termed as cyclic variations.
2(a) Rewrite the following linear programming problem in Standard form:
Minimise Z = 2 x 1 + x 2 + 4 x 3  Minimise  Z = 2 x 1 + x 2 + 4 x 3 ” Minimise “Z=2x_(1)+x_(2)+4x_(3)\text { Minimise } Z=2 x_1+x_2+4 x_3 Minimise Z=2x1+x2+4x3
Subject to the Constraints:
2 x 1 + 4 x 2 4 x 1 + 2 x 2 + x 3 5 2 x 1 + 3 x 3 2 x 1 0 , x 2 0 , x 3 0 2 x 1 + 4 x 2 4 x 1 + 2 x 2 + x 3 5 2 x 1 + 3 x 3 2 x 1 0 , x 2 0 , x 3 0 {:[-2x_(1)+4x_(2) <= 4],[x_(1)+2x_(2)+x_(3) >= 5],[2x_(1)+3x_(3) <= 2],[x_(1) >= 0″,”x_(2) >= 0″,”x_(3) >= 0]:}\begin{aligned} & -2 x_1+4 x_2 \leq 4 \\ & x_1+2 x_2+x_3 \geq 5 \\ & 2 x_1+3 x_3 \leq 2 \\ & x_1 \geq 0, x_2 \geq 0, x_3 \geq 0 \end{aligned}2x1+4x24x1+2x2+x352x1+3x32x10,x20,x30
(b) Solve the following LPP using graphical method:
Maximize Z = 3 x 1 + 2 x 2  Maximize  Z = 3 x 1 + 2 x 2 ” Maximize “Z=3x_(1)+2x_(2)\text { Maximize } Z=3 x_1+2 x_2 Maximize Z=3x1+2x2
Subject to the Constraints:
2 x 1 + x 2 1 x 1 2 x 1 + x 2 3 x 1 , x 2 0 2 x 1 + x 2 1 x 1 2 x 1 + x 2 3 x 1 , x 2 0 {:[-2x_(1)+x_(2) <= 1],[x_(1) <= 2],[x_(1)+x_(2) <= 3],[x_(1)”,”x_(2) >= 0]:}\begin{gathered} -2 x_1+x_2 \leq 1 \\ x_1 \leq 2 \\ x_1+x_2 \leq 3 \\ x_1, x_2 \geq 0 \end{gathered}2x1+x21x12x1+x23x1,x20
3 Solve the following LPP using Simplex method:
Maximize Z = x 1 + 2 x 2 Subject to the Constraints: x 1 + 2 x 2 8 x 1 + 2 x 2 12 x 1 x 2 3 x 1 , x 2 0  Maximize  Z = x 1 + 2 x 2       Subject to the Constraints:       x 1 + 2 x 2 8      x 1 + 2 x 2 12      x 1 x 2 3      x 1 , x 2 0 {:[” Maximize “quad Z=x_(1)+2x_(2),],[” Subject to the Constraints: “,-x_(1)+2x_(2) <= 8],[,x_(1)+2x_(2) <= 12],[,x_(1)-x_(2) <= 3],[,x_(1)”,”x_(2) >= 0]:}\begin{array}{ll} \text { Maximize } \quad Z=x_1+2 x_2 & \\ \text { Subject to the Constraints: } & -x_1+2 x_2 \leq 8 \\ & x_1+2 x_2 \leq 12 \\ & x_1-x_2 \leq 3 \\ & x_1, x_2 \geq 0 \end{array} Maximize Z=x1+2x2 Subject to the Constraints: x1+2x28x1+2x212x1x23x1,x20
4 A department head has four subordinates, and four tasks to be performed. The subordinates differ in efficiency, and the tasks differ in their intrinsic difficulty. His estimate, of the time each man would take to perform each task, is given in the table below:
Tasks Subordinates
E F G H
A 18 26 17 11
B 13 28 14 26
C 38 19 18 15
D 19 26 24 10
Tasks Subordinates E F G H A 18 26 17 11 B 13 28 14 26 C 38 19 18 15 D 19 26 24 10| Tasks | Subordinates | | | | | :— | :—: | :—: | :—: | :—: | | | E | F | G | H | | A | 18 | 26 | 17 | 11 | | B | 13 | 28 | 14 | 26 | | C | 38 | 19 | 18 | 15 | | D | 19 | 26 | 24 | 10 |
How should the tasks be allocated, one to a subordinate, so as to minimise the total man hour?
5 a) Use graphical method to minimise the time added to process the following jobs on the machines shown:
Job 1: Sequence A B C D E
Time 3 4 2 6 2
Job 2: Sequence B C A D E
Time 5 4 3 2 6
Job 1: Sequence A B C D E Time 3 4 2 6 2 Job 2: Sequence B C A D E Time 5 4 3 2 6| Job 1: | Sequence | A | B | C | D | E | | :— | :— | :— | :— | :— | :— | :— | | | Time | 3 | 4 | 2 | 6 | 2 | | Job 2: | Sequence | B | C | A | D | E | | | Time | 5 | 4 | 3 | 2 | 6 |
Calculate the total time elapsed to complete both the jobs.
b) The following data comprising the number of customers (in hundred) and monthly sales (in thousand Rupees):
Number of
Customers (in
hundred)
Number of Customers (in hundred)| Number of | | :— | | Customers (in | | hundred) |
4 6 6 8 10 14 18 20 22 26 28 30
Monthly Sales
(in thousand Rs)
Monthly Sales (in thousand Rs)| Monthly Sales | | :— | | (in thousand Rs) |
1.8 3.5 5.8 7.8 8.7 9.8 10.7 11.5 12.9 13.6 14.2 15
“Number of Customers (in hundred)” 4 6 6 8 10 14 18 20 22 26 28 30 “Monthly Sales (in thousand Rs)” 1.8 3.5 5.8 7.8 8.7 9.8 10.7 11.5 12.9 13.6 14.2 15| Number of <br> Customers (in <br> hundred) | 4 | 6 | 6 | 8 | 10 | 14 | 18 | 20 | 22 | 26 | 28 | 30 | | :— | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | Monthly Sales <br> (in thousand Rs) | 1.8 | 3.5 | 5.8 | 7.8 | 8.7 | 9.8 | 10.7 | 11.5 | 12.9 | 13.6 | 14.2 | 15 |
Calculate the residuals and determine the standardised residuals for the model
Y = 2.6185 + 0.4369 X Y = 2.6185 + 0.4369 X Y=2.6185+0.4369X\mathrm{Y}=2.6185+0.4369 \mathrm{X}Y=2.6185+0.4369X
6 a) A Statistician collected the data of 78 values with two independent variable X 1 X 1 X_(1)X_1X1 and X 2 X 2 X_(2)X_2X2, and considered the four models:
(i) Y = B 0 + e Y = B 0 + e Y=B_(0)+e\mathrm{Y}=\mathrm{B}_0+\mathrm{e}Y=B0+e; (ii) Y = B 0 + B 1 X 1 + e Y = B 0 + B 1 X 1 + e Y=B_(0)+B_(1)X_(1)+e\mathrm{Y}=\mathrm{B}_0+\mathrm{B}_1 \mathrm{X}_1+\mathrm{e}Y=B0+B1X1+e; (iii) Y = B 0 + B 1 X 1 + e Y = B 0 + B 1 X 1 + e Y=B_(0)+B_(1)X_(1)+e\mathrm{Y}=\mathrm{B}_0+\mathrm{B}_1 \mathrm{X}_1+\mathrm{e}Y=B0+B1X1+e and
(iv) Y = B 0 + B 1 X 1 + B 2 X 2 + e Y = B 0 + B 1 X 1 + B 2 X 2 + e Y=B_(0)+B_(1)X_(1)+B_(2)X_(2)+e\mathrm{Y}=\mathrm{B}_0+\mathrm{B}_1 \mathrm{X}_1+\mathrm{B}_2 \mathrm{X}_2+\mathrm{e}Y=B0+B1X1+B2X2+e.
The results obtained are: σ ^ 2 = 0.91 , SS ( B 0 ) = 652.42 , SS ( B 0 , B 1 ) = 679.34 σ ^ 2 = 0.91 , SS B 0 = 652.42 , SS B 0 , B 1 = 679.34 hat(sigma)^(2)=0.91,SS(B_(0))=652.42,SS(B_(0),B_(1))=679.34\hat{\sigma}^2=0.91, \operatorname{SS}\left(B_0\right)=652.42, \operatorname{SS}\left(B_0, B_1\right)=679.34σ^2=0.91,SS(B0)=652.42,SS(B0,B1)=679.34, SS ( B 0 , B 2 ) = 654.00 SS B 0 , B 2 = 654.00 SS(B_(0),B_(2))=654.00\operatorname{SS}\left(B_0, B_2\right)=654.00SS(B0,B2)=654.00, and SS ( B 0 , B 1 , B 2 ) = 687.79 SS B 0 , B 1 , B 2 = 687.79 SS(B_(0),B_(1),B_(2))=687.79\operatorname{SS}\left(B_0, B_1, B_2\right)=687.79SS(B0,B1,B2)=687.79. Find the additional contribution of (i) X 2 X 2 X_(2)X_2X2 over X 1 X 1 X_(1)X_1X1 and (ii) X 1 X 1 X_(1)X_1X1 over X 2 X 2 X_(2)X_2X2. Test whether their inclusion in the model is justified.
b) Fifteen successive observations on a stationary time series are as follows:
32 33 30 32 33 30 {:[32,33,30]:}\begin{array}{lll}32 & 33 & 30\end{array}323330
Calculate r 6 , r 7 , r 8 r 6 , r 7 , r 8 r_(6),r_(7),r_(8)\mathrm{r}_6, \mathrm{r}_7, \mathrm{r}_8r6,r7,r8 and r 9 r 9 r_(9)\mathrm{r}_9r9 and plot the correlogram.
7 Calculate seasonal indices by the ratio to moving average method from the following data:
Year
Quarter
Year Quarter| Year | | :— | | Quarter |
2001 2002 2003 2004
Q 1 Q 1 Q_(1)\mathrm{Q}_1Q1 750 860 900 1000
Q 2 Q 2 Q_(2)\mathrm{Q}_2Q2 600 650 720 780
Q 3 Q 3 Q_(3)\mathrm{Q}_3Q3 540 630 660 720
Q 4 Q 4 Q_(4)\mathrm{Q}_4Q4 590 800 850 930
“Year Quarter” 2001 2002 2003 2004 Q_(1) 750 860 900 1000 Q_(2) 600 650 720 780 Q_(3) 540 630 660 720 Q_(4) 590 800 850 930| Year <br> Quarter | 2001 | 2002 | 2003 | 2004 | | :— | :—: | :—: | :—: | :—: | | $\mathrm{Q}_1$ | 750 | 860 | 900 | 1000 | | $\mathrm{Q}_2$ | 600 | 650 | 720 | 780 | | $\mathrm{Q}_3$ | 540 | 630 | 660 | 720 | | $\mathrm{Q}_4$ | 590 | 800 | 850 | 930 |
8 Consider the following Transportation problem:
Factory Godowns
Stock
Available
Stock Available| Stock | | :—: | | Available |
1 2 3 4 5 6
A 7 5 7 7 5 3 60
B 9 11 6 11 5 20
C C C\mathrm{C}C 11 10 6 2 2 8 90
D D D\mathrm{D}D 9 10 9 6 9 12 50
Demand 60 20 40 20 40 40
Factory Godowns “Stock Available” 1 2 3 4 5 6 A 7 5 7 7 5 3 60 B 9 11 6 11 – 5 20 C 11 10 6 2 2 8 90 D 9 10 9 6 9 12 50 Demand 60 20 40 20 40 40 | Factory | Godowns | | | | | | Stock <br> Available | | :—: | :—: | :—: | :—: | :—: | :—: | :—: | :—: | | | 1 | 2 | 3 | 4 | 5 | 6 | | | A | 7 | 5 | 7 | 7 | 5 | 3 | 60 | | B | 9 | 11 | 6 | 11 | – | 5 | 20 | | $\mathrm{C}$ | 11 | 10 | 6 | 2 | 2 | 8 | 90 | | $\mathrm{D}$ | 9 | 10 | 9 | 6 | 9 | 12 | 50 | | Demand | 60 | 20 | 40 | 20 | 40 | 40 | |
It is not possible to transport any quantity from Factory B to Godown 5. Determine:
(a) Basic Feasible Solution by Vogel’s Approximation Method.
(b) Optimum solution using MODI method.
\(a^2=b^2+c^2-2bc\:Cos\left(A\right)\)

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