Download PROBABILITY IN MATHEMATICS and more Assignments Mathematics in PDF only on Docsity!
Activity part 01
χS1_WH1 : Number of panels delivered from WH1 to S1. χS1_WH2 : Number of panels delivered from WH2 to S1. χS2_WH1 : Number of panels delivered from WH1 to S2. χS2_WH2 : Number of panels delivered from WH2 to S2.
- Constraints:
Demand at each site:
- χS1_WH1 + χS1_WH2 = 625 (for S1)
- χS2_WH1 + χS2_WH2 = 900 (for S2)
Supply from each warehouse:
- χS1_WH1 + χS2_WH1 = 475 (from WH1)
- χS1_WH2 + χS2_WH2 = 1050 (from WH2)
Maximum transport limit:
- χS1_WH2 ≤ 550
- χS2_WH2 ≤ 550
- Objective Function
Minimum the number of panels delivered from WH1 to S1 due to transportation difficulties. Thus, we aim to minimize χS1_WH1.
Given these constraints and the objective,we can set up and solve the linear programming
problem.
Subject to:
- χS1_WH1+χS1_WH2 = 625
- χS2_WH1+χS2_WH2 = 900
- χS1_WH1+χS2_WH1 = 475
- χS1_WH2+χS2_WH2 = 1050
- χS1_WH2 ≤ 550
- χS2_WH2 ≤ 550
- χS1_WH1, χS1_WH2, χS2_WH1, χS2_WH2 ≥ 0
Activity part 02
- Given, T 1 = ( 10 + 20 + T 2 + T 4 )
4
4T 1 – T 2 - T 4 = 30 1 T 2 = ( T 1 + T 3 + 20 + 40 )
4
4T 2 – T 1 – T 3 = 60 2 T 3 = ( T 4 + T 2 + 40 + 30 )
4 4T 3 – T 4 – T 2 = 70 3
T 4 = ( 10 + T 1 + T 3 + 30 )
4 4T 4 – T 1 – T 3 = 40 4
Which are the system of your equations.
2. 4 - 1 0 - 1 T1 30
- 1 4 - 1 0 T 2 = 60
0 - 1 4 - 1 T3 70
- 1 0 - 1 4 T4 40
A B
- 1 4 - 1 0 0 1 0 0 R 1 / 4 → R 1
- 1 4 - 1 0 0 1 0 0 R 1 + R 2 → R 2
- 1 - 1/4 0 - 1 /4 1/4
- 0 15/4 - 1 - 1/4 1/4 0 0 0 R 1 + R 4 → R
- 0 - 1 4 -
- 1 - 1/4 0 - 1/4 1/4
- 0 15/4 - 1 - 1/4 1/4 1 0 0 (4/15) R 2 → R
- 0 - 1 4 -
- 0 - 1/4 - 1 15/4 1/4
- 1 - 1/4 0 - 1/4 1/4
- 0 1 - 4/15 - 1/15 1/15 4/15 0 0 R 1 + (1/4) R 2 → R
- 0 - 1 4 -
- 0 - 1/4 - 1 15/4 1/4
- 1 0 - 1/15 - 4/15 4/15 1/15
- 0 1 - 4/15 - 1/15 1/15 4/15 0 0 R3 + R2 → R
- 0 - 1 4 - 1 0 0 1 0 R4 + (1/4) R2 → R
- 0 - 1/4 - 1 15/4 1/4
- 1 0 - 1/15 - 4/15 4/15 1/15
- 0 1 - 4/15 - 1/15 1/15 4/15 0 0 (15/56) R 3 → R
- 0 0 56/15 - 16/15 1/15 4/15
- 0 0 - 16/15 56/15 4/15 1/15
- 1 0 - 1/15 - 4/15 4/15 1/15
- 0 1 - 4/15 - 1/15 1/15 4/15 0 0 R 1 + 1/15 R 3 → R
- 0 0 1 - 2/7 1/56 1/14 15/56
- 0 0 - 16/15 56/15 4/15 1/15
- 1 0 0 - 2/7 15/56 1/14 1/56
- 0 1 - 4/15 - 1/15 1/15 4/15 0 0 R 2 + (4/15) R 3 → R
- 0 0 1 - 2/7 1/56 1/14 15/56 0 R 4 + (16/15) R 3 → R
- 0 0 - 16/15 56/15 4/15 1/15
- 1 0 0 - 2/7 15/56 1/14 1/56
- 0 1 0 - 1/7 1/14 2/7 1/14 0 (7/24) R 4 → R
- 0 0 1 - 2/7 1/56 1/14 15/56 0 R 1 + (2/7) R 4 → R
- 0 0 0 24/7 2/7 1/7 2/7
- 1 0 0 0 7/24 1/12 1/24 1/
- 0 1 0 - 1/7 1/14 2/7 1/14 0 R 2 + (1/7) R 4 → R
- 0 0 1 - 2/7 1/56 1/14 15/56 0 R 4 + (2/7) R 4 → R
- 0 0 0 1 1/12 1/24 1/12 7/
- 1 0 0 0 7/24 1/12 1/24 1/
- 0 1 0 0 1/12 7/24 1/12 1/
- 0 0 1 0 1/24 1/12 7/24 1/
- 0 0 0 1 1/12 1/24 1/12 7/
A-^1 = 7/24 1/12 1/24 1/
T 1 1/24 1/12 1/24 1/12 30 T1 = 20
T 2 = 1/12 7/24 1/12 1/24 60 T2 = 27.
T 3 1/24 1/12 7/24 1/12 70 T3 = 30
T 4 1/12 1/24 1/12 7/24 40 T4 = 22.
Solution
From the physics of the problem, the maximum deflection would be between x = 0 and x = ʟ, where
ʟ = length of the boolshelf,
that is
0 ≤ x ≤ ʟ
0 ≤ x ≤ 29
Let us assume
X (^) ɭ = 0 2 χμ = 29
Check if the function changes sign between χɭ and χμ
ʄ ( χɭ ) = ʄ ( 0 )
= - 0.67665 × 10 -8^ (0)^4 – 0.26689 × 10 -5^ (0) + 0.12748 × 10 -3^ (0)^2 – 0.
= -0.
ʄ ( χμ ) = ʄ (29)
= -0.67665 × 10 -8^ (29)^4 – 0.26689 × 10 -5^ (29)^3 + 0.12748 × 10 -3^ (29)^2 – 0.
= 0.
Hence
ʄ (χɭ) ʄ (χμ) = ʄ (0) ʄ (29) = ( - 0.018507 ) ( 0.018826 ) < 0
So there is at least one root between χɭ and χμ that is between 0 and 29.
The absolute relative approximate error, | ∈ | at the end of Iteration 2 is
| 𝜖 | = χ new^ - χ old^ × 100
Χ new
= 21.75 – 14.5 × 100
None of the significant digits are at least correct in the estimated root
Χ ͫ = 21,
As the absolute relative approximate eror is greater than 5%.
Iteration 3
The estimate of the root is
Χ ͫ = χɭ + χμ
2
= 14.5 + 21. 2
= 18.
ʄ ( χ ͫ ) = ʄ ( 18.125 ) = - 0.67665 × 10-3^ (18.125)^4 - 0.26689 × 10-5^ (18.125)^3
- 0.12748 × 10-3^ (18.125)^2 – 0.
ʄ (χɭ) ʄ (χͫ) = ʄ (14.5) ʄ (18.125) = ( -1.4007 × 10-4^ ) ( 6.7502 × 10-3^ ) < 0
Hence, the root is bracketed between χɭ and χͫ that is, between 14.5 and 18.125.So, the lower and upper limits of the new bracket are
χɭ = 14.5 χμ = 18.
The absolute relative approximate error, | ∈ | at the end of Iteration 3 is
| ϵ | = χ new - χ old × 100
Χ new
= 18.125 – 21.74 × 100
Still none of the significant digits are at least correct in the estimated root of the equation as the absolute relative approximate error is greater than 5%.
Seven more iterations were conducted and these iterations are shows in Table 1.
Table 1 Root of ʄ (χ) = 0 as a function of the number of iterations for bisection method.
Iteration Χ (^) ɭ Χ μ Χ ͫ | ϵ | % ʄ (χ ͫ ) 1 0 29 14............ - 1.3992 × 10 -^4 2 14.5 29 21.75 33.333 0. 3 14.5 21.75 18.125 20 6.7502× 10-^3 4 14.5 18.125 16.313 11.111 3.3509× 10-^3 5 14.5 16.313 15.406 5.8824 1.6099× 10-^3 6 14.5 15.406 14.953 3.0303 7.3521× 10-^4 7 14.5 14.953 14.727 1.5383 2.9753× 10-^4 8 14.5 14.727 14.613 0.77519 7.8708× 10-^5 9 14.5 14.613 14.557 0.38911 - 3.0688× 10-^5 10 14.557 14.613 14.585 0.19417 2.4009× 10-^5
At the end of the 10th^ iteration.
Hence the number of significant digits at least correct is given by the largest value ofͫ for which
| ∈ | ≤ 0.5 × 10 2 - ͫ
0.19417 ≤ 0.5 × 10 2 - ͫ
0.38835 ≤ 10 2 - ͫ
log(0.38835) ≤ 2 - ͫ ͫ ≤ 2 – log(0.38835) = 2.
S
ͫ = 2
The number of significant digits at least correct in the estimated root 14.585 is 2.
From the charge curves, we can estimate the voltage and current at 2-minute intervals. For simplicity, let's use a few example points and you can fill in the rest similarly:
At t = 0 min: V = 3.2 V, I = 2000 mA
At t = 2 min: V ≈ 3.4 V, I ≈ 2000 mA
At t = 4 min: V ≈ 3.6 V, I ≈ 2000 mA
At t = 6 min: V ≈ 3.8 V, I ≈ 1800 mA
And so on, until t = 134 min.
Discussion
Using the final voltage of 4.2 V and the capacity of 2000 mAh (as given in Step 5):
Total Energy (Theoretical): E = 4.2 V × 2000 mAh = 8.4 Wh
Comparing this theoretical value with the calculated values from numerical integration methods:
The calculated values (1.495 Wh and 1.388 Wh) are significantly lower than the theoretical value of 8.4 Wh. This discrepancy might be due to several factors, including the estimation of data points from the graph, which might not cover all details accurately.
