11th Standard Syllabus & Materials
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TN 11th Tamil இயற்கை வேளாண்மை,சுற்றுச்சூழல் -செய்யுள் - மனோன்மணீயம் Important Questions And Answers Study Material - QB365 Set A
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Published on: 13/05/2022
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1.
The following table use the activities in a building project.
| Activity | 1-2 | 1-3 | 2-3 | 2-4 | 3-4 | 4-5 |
|---|---|---|---|---|---|---|
| Duration (days) | 21 | 26 | 11 | 13 | 5 | 11 |
Draw the network for the project, calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity and find the critical path. Compute the project duration.
2.
Calculate the earliest start time, earliest finish time, latest start time and latest finish time of each activity of the project given below and determine the critical path of the project and duration to complete the project.
| Activity | 1-2 | 1-3 | 1-5 | 2-3 | 2-4 | 3-4 | 3-5 | 3-6 | 4-6 | 5-6 |
| Duration (in week) | 7 | 6 | 11 | 3 | 9 | 2 | 4 | 9 | 6 | 3 |
3.
Compute the earliest start time, earliest finish time, latest start time and latest finish time of each activity of the project given below:
| Activity | 1-2 | 1-3 | 2-4 | 2-5 | 3-4 | 4-5 |
|---|---|---|---|---|---|---|
| Duration (in days) | 2 | 1 | 10 | 5 | 3 | 6 |
4.
Minimize and maximize Z =x1+ 2x2 . Subject to the constraints x1 + 2x2 ≥ 100,2x1 − x2 ≤ 0,2x1 + x2 ≤ 200 and x1,x2 ≥ 10.
5.
Solve graphically: Minimize Z = 20x1 + 40x2.
Subject to the constraints 36x1 + 6x2 ≥ 108,3x1 + 12x2 ≥ 36,20x1 + 10x2 ≥ 100 and x1,x2≥0
1.
| E1= 0 | L5= 48 |
| E2= 0+21=21 | L4= 48 -11 = 37 |
| E3 =(21 + 11) or (0 + 26) Whichever is maximum =32 |
L3= 37 - 5 = 32 |
| E4= (32 + 5) or (21 + 13) =Whichever is maximum = 37 |
L2= (37 - 13) or (32 - 11) Whichever is minimum =21 |
| E5=37 + 11 = 48 | L1=(21 - 21) or (32 - 26) Whichever is minimum = 0 |
| Activity | Duration | EST | EFT = EST + tij | EFT = EST - tij | LFT |
|---|---|---|---|---|---|
| 1-2 | 21 | 0 | 21 | 21-21=0 | 21 |
| 1-3 | 26 | 0 | 26 | 32-26=6 | 32 |
| 2-3 | 11 | 21 | 32 | 32-11=21 | 32 |
| 2-4 | 13 | 21 | 34 | 37-13=24 | 37 |
| 3-4 | 5 | 32 | 37 | 37-5=32 | 37 |
| 4-5 | 11 | 37 | 48 | 48-11=37 | 48 |
EFT and LFT are same in the activities.
1 - 2, 2 - 3, 3 - 4 and 4 - 5
Hence, the critical path is 1 - 2 - 3 - 4 - 5 and the duration of project completion is 48 days .
2.
| E1 =0 | L6 =22 |
| E2 = 0+7=7 | L5 =(22 - 3) = 19 |
| E3 =(7 + 3) or (0 + 6) Whichever is maximum = 0 |
L4 =22 - 6 = 16 . |
| E4 =(7 + 9) or (10 + 2) Whichever is maximum=16 |
L3 =(22 - 9) or (16 - 8) Whichever is minimum=13 |
| E5 =(0+ 11)or(10+4) Whichever is maximum=14 |
L2 =16 - 9 = 7 |
| E6 =(16 + 6) or (10 + 9) or (14 + 3) Whichever is maximum = 22 |
L1 =7-7=0 |
| Activity | Duration | EST | EFT=EST+tij | LST | LFT |
|---|---|---|---|---|---|
| 1-2 | 7 | 0 | 7 | 7-7=0 | 7 |
| 1-3 | 6 | 0 | 6 | 13-6=7 | 13 |
| 1-5 | 11 | 0 | 11 | 19-11=8 | 19 |
| 2-3 | 3 | 7 | 10 | 13-3=10 | 13 |
| 2-4 | 9 | 7 | 16 | 16-9=7 | 16 |
| 3-4 | 2 | 13 | 15 | 19-4=15 | 16 |
| 3-5 | 4 | 13 | 17 | 19-4=15 | 19 |
| 3-6 | 9 | 10 | 19 | 22-9=13 | 22 |
| 4-6 | 6 | 16 | 22 | 22-6=16 | 22 |
| 5-6 | 3 | 14 | 17 | 22-3=19 | 22 |
EFT and LFT are same in the activity, 1 - 2, 2 - 4 and 4 - 6.
Hence the critical path is 1 - 2 - 4 - 6 and the project completion time is 22 Weeks.
3.
| E1 =0 | L5 =18 |
| E2 =0+2=0 | L4 =18-6=12 |
| E3 =0+1=1 | L3 =(12 - 3) = 9 |
| E4 =(1+3)or(2+10) Whichever is maximum =12 |
L2 =(5 - 2) or (12 - 10) Whichever is minimum=2 |
| E5 =(2 + 5) or (12 + 6) Whichever is maximum =18 |
L1 =(2 - 2) or (9 - 1) Whichever is minimum = 0 |
| Activity | Duration | EST | EFT=EST+tij | EFT=EST-tij | LFT |
|---|---|---|---|---|---|
| 1-2 | 2 | 0 | 2 | 2-2=0 | 2 |
| 1-3 | 1 | 0 | 1 | 9-1=8 | 9 |
| 2-4 | 10 | 2 | 12 | 12-10=2 | 12 |
| 2-5 | 5 | 2 | 7 | 18-5=13 | 18 |
| 3-4 | 3 | 1 | 4 | 12-3=9 | 12 |
| 4-5 | 6 | 12 | 18 | 18-6 | 18 |
EFT and LFT are same in the activities.
1 - 2, 2 - 4 and 4 - 5.
Hence the critical path is 1 - 2 - 4 - 5 and the project completion time is 18 days.
4.
Consider the equations
\({ x }_{ 1 }+2{ x }_{ 2 }=100\)
| \({ x }_{ 1 }\) | 1 | 100 |
| \({ x }_{ 2 }\) | 50 | 0 |
\({ 2x }_{ 1 }-{ x }_{ 2 }=0\)
| \({ x }_{ 1 }\) | 0 | 10 |
| \({ x }_{ 2 }\) | 0 | 20 |
\({ 2x }_{ 1 }+{ x }_{ 2 }=200\)
| \({ x }_{ 1 }\) | 0 | 100 |
| \({ x }_{ 2 }\) | 200 | 0 |
The feasible region is ABCD and its co-ordinates are A(0, 50), D(0, 200) B is the point of intersection of the lines
x1+ 2x2 = 100 ... (1)
and 2x1-x2 = 0 ... (2)
Verification of B:
\((1)\times 2 \Rightarrow 2{ x }_{ 1 }+{ 4x }_{ 2 }=200\)
\( (-)\quad (-)\quad \quad (-) \)
\((2)\Rightarrow 2{ x }_{ 1 }-{ x }_{ 2 }=0\)
\(--------------\)
\( 5{ x }_{ 2 }=200 \Rightarrow { x }_{ 1 }=40 \)
\(From(2),2{ x }_{ 1 }-40=0 \Rightarrow { x }_{ 1 }=20\)
∴ B is (20,40)
Verification of C:
Also, C is the point of intersection of the lines
\(2{ x }_{ 1 }-{ x }_{ 2 }=0 ...(3)\)
\(and \quad 2{ x }_{ 1 }+{ x }_{ 2 }=200 ...(4)\)
\( ------------\)
\((3)+(4)\Rightarrow 4{ x }_{ 1 }=200 \Rightarrow { x }_{ 1 }=50\)
\(From(3),\quad 100-{ x }_{ 2 }=0 \Rightarrow { x }_{ 2 }=100\)
∴ C is (50,100)
| Corner Points | Z=x1+ 2x2 |
|---|---|
| A(0,50) | 100 |
| B(20,40) | 100 |
| C(50,100) | 250 |
| D(0,200) | 400 |
Minimum of Z occurs at A(0, 50) and B(20, 40), Zmin= 100
Maximum of Z occurs at D(0, 200) and Zmax= 400
5.
Consider the equations
\(36{ x }_{ 1 }+6{ x }_{ 2 }= 108\)
| \({ x }_{ 1 }\) | 0 | 3 |
| \({ x }_{ 2 }\) | 18 | 0 |
\(3{ x }_{ 1 }+12{ x }_{ 2 }= 36\)
| \({ x }_{ 1 }\) | 0 | 12 |
| \({ x }_{ 2 }\) | 3 | 0 |
\(20{ x }_{ 1 }+10{ x }_{ 2 }= 100\)
| \({ x }_{ 1 }\) | 0 | 5 |
| \({ x }_{ 2 }\) | 10 | 0 |
The feasible region is ABCD and its co-ordinates are A(12, 0), D(0, 18), B is the point of intersection of the lines
Verification of B:
\(3{ x }_{ 1 }+12{ x }_{ 2 }=36 \Rightarrow { x }_{ 1 }+4{ x }_{ 2 }=12 ...(1)\)
\(20{ x }_{ 1 }+10{ x }_{ 2 }=100 \Rightarrow { 2x }_{ 1 }+{ x }_{ 2 }=10...(2)\)
\( (1)\times 2\Rightarrow 2{ x }_{ 1 }+8{ x }_{ 2 }=24\)
\( (-)\quad (-)\quad \quad (-)\)
\((2)\Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }=10\)
\(--------------\)
\(7{ x }_{ 2 }=14 \Rightarrow { x }_{ 2 }=2\)
\(From(1), { x }_{ 1 }+8=12 \Rightarrow { x }_{ 1 }=4\)
∴ B is (4,2)
Also C is the point of intersection of the lines
Verification for C:
\(3{ 6x }_{ 1 }+6{ x }_{ 2 }=108\Rightarrow 6{ x }_{ 1 }+{ x }_{ 2 }=18 ...(3)\)
\( (-)\quad (-)\quad \quad (-)\)
\(20{ x }_{ 1 }+10{ x }_{ 2 }=100\Rightarrow 2{ x }_{ 1 }+{ x }_{ 2 }=10 ...(4)\)
\( --------------\)
\(4{ x }_{ 1 }=8 \Rightarrow { x }_{ 1 }=2\)
\( From(4), 4+{ x }_{ 2 }=10 \Rightarrow { x }_{ 2 }=6\)
∴ C is (2,6)
| Corner Points | z = 20x1+ 40x2 |
|---|---|
| A(12,0) | 240 |
| B(4,2) | 160 |
| C (2, 6) | 280 |
| D(0, 18) | 720 |
Minimum of B occurs at B(4, 2)
Hence, the solution is x1 = 4, x2 = 2 and Zmin = 160.
11th Standard Syllabus & Materials
11th Standard
TN 11th Tamil பீடு பெற நில் - செய்யுள் - காவடிச்சிந்து Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil பீடு பெற நில் - உரைநடை - மலை இடப்பெயர்கள் : ஓர் ஆய்வு Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil மாமழை போற்றுதும் - துணைப்பாடம் - யானை டாக்டர் Important Questions And Answers Study Material - QB365 Set A
NEW11th Standard
TN 11th Tamil மாமழை போற்றுதும் - செய்யுள் - ஐங்குறுநூறு Important Questions And Answers Study Material - QB365 Set A
Tamilnadu Stateboard 11th Standard Subjects
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Economics
Biology
Business Maths and Statistics
Accountancy
Computer Science
Physics
Chemistry
Maths
Biology
Economics
Physics
Chemistry
History
Business Maths and Statistics
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