I. Problem Description The Darby Company is re-evaluating its current production and distribution system in order to determine whether it is cost-effective or if a different approach should be considered. The company produces meters that measure the consumption of electrical power. Currently, they produce these meters are two locations – El Paso, Texas and San Bernardino, California. The San Bernardino plant is newer, and therefore the technology is more effective, meaning that their cost per unit is $10.00, while the El Paso plant produces at $10.50. However, the El Paso plant has a higher capacity at 30,000 to San Bernardino’s 20,000. Once manufactured, the meters are sent to one of three distribution centers – Ft. Worth, Texas, Santa …show more content…
Without distribution limitations: Min 3.2x1 + 2.2x2 + 4.2x3 + 3.9x4 + 1.2x5 + 0.3x6 + 2.1x7 + 3.1x8 + 4.4x9 + 2.7x10 + 4.7x11 + 3.4x12 + 2.1x13 + 2.5x14 + 6.0x15 + 5.2x16 + 5.4x17 + 4.5x18 + 6.0x19 + 3.3x20 + 2.7x21 + 5.4x22 + 3.3x23 + 2.4x24 S.T. (subject to): 1. x1 + x2 + x3 ≤ 30,000 2. x4 + x5 ≤ 20,000 3. -x1 + x6 + x7 + x8 + x9 + x15 = 0 4. -x2 - x4 + x10 + x11 + x12 + x16 + x17 + x18 + x19 + x20 + x21 = 0 5. –x3 – x5 + x13 + x14 + x22 + x23 + x24 = 0 6. x6 + x16 =3600 7. x7 + x17 = 4880 8. x8 + x18 = 2130 9. x9 + x19 = 1210 10. x10 + x15 + x22 = 6120 11. x11 + x23 = 4830 12. x12 + x24 = 2750 13. x13 + x20 =8580 14. x14 + x21 = 4460 Without distribution limitations and supplying direct to Los Angeles, San Diego and San Antonio: Min 3.2x1 + 2.2x2 + 4.2x3 + 3.9x4 + 1.2x5 + 0.3x6 + 2.1x7 + 3.1x8 + 4.4x9 + 2.7x10 + 4.7x11 + 3.4x12 + 2.1x13 + 2.5x14 + 6.0x15 + 5.2x16 + 5.4x17 + 4.5x18 + 6.0x19 + 3.3x20 + 2.7x21 + 5.4x22 + 3.3x23 + 2.4x24 + 0.3x25 +0.7x26 + 3.5x27 s.t.: 1. x1 + x2 + x3 + x27 ≤ 30,000 2. x4 + x5 + x25 + x26 ≤ 20,000 3. -x1 + x6 + x7 + x8 + x9 + x15 = 0 4. -x2 - x4 + x10 + x11 + x12 + x16 + x17 + x18 + x19 + x20 + x21 = 0 5. –x3 – x5 + x13 + x14 + x22 + x23 + x24 = 0 6. x6 + x16 =3600 7. x7 + x17 + x27 = 4880 8. x8 + x18 = 2130 9. x9 + x19 = 1210 10. x10 + x15 + x22 = 6120 11. x11 + x23 = 4830 12. x12 + x24 = 2750 13. x13 + x20 + x25 =8580 14. x14 + x21 + x26 = 4460
|0 |$ 60 |$ 0 |$ 60 |$ 0 |$ 0 |$ 0 |$ 0 |
[191844 + 44100 + 100489 + 99225 + 53361 + 5776 + 13689 + 51076]/7
Passage Date: AR2001 = 5 GW2000 + 6 NI2000 + 7 REV2000 + 8 NA2000 + 2
Once you subtract two with the ten, you’d have to reverse the sign because the negative is with the ‘ x’, which would make the problem x is greater than 4.
for(index = positionToSort + 1; index < length; index++){ ///c3 * sum(from position + 1 to length + 1)
1 21.3 ± 2.5 20.9 ± 2.6 20.4 ± 2.3 20.5 ± 1.2 22.5 ± 3.9 21.8 ± 3.9 21.2 ± 2.1 23.4 ± 3.2 26.1 ± 4.5* 25.1 ± 4.6* 25.1 ± 3.7* 25.1 ± 3.5*
665 0.34 174 216 537 234 122 1.2 0.2 60.4 9.6 0.22 65.0 2,149 554 66.6 2,483 80,300 492 14 23 6 21 15 874 524 12,216 30
2(X4 + X5 + X6 + X7 + X10 + X11 + X16 + X17 + X18 + X27 + X28) – (X1 + X2 +X3
22 x C-H (412) + 8 x C-C (348) + 2 x C-O (360) + 2 x O-H (463) + 15 x O=O (498) 20 x C=O (805) + 24 x O=H (463)
sin2 x + 4 sin x + 3 3 + sin x 12. = 2x cos 1 sin x 13. cos x 1 sin x tan x = sec x 1 + sin x cos2 x
3.375(1 + r)-1 + 3.375 (1 + r)-2 + 3.375 (1 + r)-3 + ...…+ 3.375 (1 + r)-40+100(1 + r)-40 = 95.6
correct state (Right Status) to the correct location (Right Place) - "6R", and to minimize the total
u = 0,1, , M − 1; v = 0,1, , N − 1; w = 0,1,
( 26 ) n Cr - 1 = 36 n C r = 84 and n Cr + 1 = 126 , then r is (a) 1 (b) 2 (c) 3 ( d ) none of these [ IIT 1979 ]
y(2) = (1)(1) + ( 2)(3) + (3)( 2) + 0 + 0 + 0 = 13