3.6. Algorithm of sequential gradient search
Step 1: Set specifications of the inductor
Step 2: Set the values of Bm , and no. of core steps. Step 3: 0.45≤ K ≤ 0.6 and 0.3 ≤ Rw ≤ 0.4
Step 4: i = 0 to 30 do: K 0.3 i / 100
Step 5: Calculate cost.
Step 6: If cost shows initially low and after that high, (concavity fails for K ) go to step 28 Step 7: i = 0 to 20 do: Rw 2 i /10
Step 8: Go to sub-routine, calculate the cost.
Step 9: If cost does not show low value and then high (concavity fails for Rw ) go to step 28
Step 10: i = 0 to 30 do: K 0.3 i / 100
Step 11: Go to sub-routine and calculate the cost.
Step 12: if present cost > previous cost go to step 14
Step 13: end for
Step 14: back to previous value of K
Step 15: For i = 0 to 20 do: Rw 2 i /10
Step 16: Go to sub-routine and calculate the cost.
Step 17: if present cost > previous cost go to step 19
Step 18: end for
Step 19: back to previous value of Rw
Step 20: Go to sub-routine and calculate cost, performance etc.
Step21: check for constraints violation (iron loss &copper loss), if it violets then go to step 25 Step 22: check for temperature rise, if it violets then go to step 25
Step 23: Print out results: go to step 26 Step 24: Stop
Step 25: End
Design optimization or optimal Design means effective and efficient design with minimum manufacturing cost within certain restriction imposed on it. Optimization is the process of searching highest and the least values of a given
First I will set up the apparatus as show above. I will add 1.5 grams
H. How would you prepare 10 mL of a 0.25M HCl solution if 1M HCl was available? How much
Rate= k [I-]1[H2O2]. (2.13675*10-5 ) = k [0.015] [0.015] then solve for k. For this trial, k=0.09497.
11. Tare the scale by pressing the Φ/T button so that the scale reads 0.0 g.
Total cost = (F * T/Q) + (H * Q / 2) = (80 * 200,000 / 10,000) + (1.00 * 10,000/2)
Next, obtain a 96 well plate and take 1 sample of 190 μl from the blank and put it in well #A1. Then take 3 different 190 μl samples of concentration 0 and put it in Wells F2, G2, and H2. Repeat this step again by taking 3 more different 190 μl samples of concentra-tion 1 and putting it in wells F3, G3, H3. It should be noted that it is important to vortex each
Repeat step number seven 19 more times or until ten minutes are up. Do this
1.2 1.0 0.8 0.6 0.4 0.2 0.0 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 Concentration (mM) y = 0.1897x R2 = 0.99
10. I then repeated steps 1-9 using alcohol as my solvent, instead of the 1% salt solution.
The k value in run 7 is 75% of that in run 5. This will be directly related to the addition of KNO3 in run 5 and H2O in run 7. These will make different ionic concentrations, resulting in the change in rate constant
1. 10 drops of 1 M K2CrO4 was added to the solution and stirred for about 10 minutes.
8) Steps 1 - 8 were repeated using the wavelengths of 360 nm to 900
∏i = (Pe + 40 - 100) x 100 = 100 x Pe –6,000 -------------------- [1]
correct state (Right Status) to the correct location (Right Place) - "6R", and to minimize the total
This is used in main iteration of proposed algorithm for weight updation.The Fig 2 shows the architecture of iteration process defined in (9) for 2 sources .