tutor tues apr 2, 2-4 pm since no classes apr 1 set 6 due apr 9 set 7 due april 11 c-3 due april 18
TRANSCRIPT
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Tutor Tues Apr 2, 2-4 PM since no classes Apr 1
Set 6 due Apr 9
Set 7 due April 11
C-3 due April 18
![Page 2: Tutor Tues Apr 2, 2-4 PM since no classes Apr 1 Set 6 due Apr 9 Set 7 due April 11 C-3 due April 18](https://reader030.vdocuments.mx/reader030/viewer/2022032516/56649c765503460f9492a0f3/html5/thumbnails/2.jpg)
Typo: Summer Schedule
• Page 27, MSCI 306M
• Session 1: June 3-July 12 (6 weeks)
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Answers to Practice Problems II
Problem (1)
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(1) 4X1 + 3X2 >96
X1 X2
0 96/3=32
96/4=24 0
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X1
X2
0,32
24,0
(1)
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(2) X1 + X2 <50
X1 X2
0 50
50 0
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X1
X2
0,32
24,0
0,50
50,0
(1)(2)
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(3) X1 +3X2 >60
X1 X2
0 60/3=20
60 0
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X1
X2
0,32
24,0
0,50
50,0
(1)(2)
0,20
60,0
(3)
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Corner Point: (1),(3)
• (1) 4X1 + 3X2 = 96
• (3) X1 + 3X2 = 60
• (1)-(3) 3X1 = 36
• X1 = 12
• (3) 12 + 3X2 = 60, so X2 = 16
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Corner Point: (2),(3)
• (2) X1 + X2 = 50
• (3) X1 + 3X2 = 60
• (3)-(2) 2X2= 10
• X2 = 5
• (2) X1 + 5 = 50
• X1 = 45
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X1
X2
0,32
24,0
0,50
50,0
(1)(2)
0,20
60,0
(3)
Corner points
Not feasible
Notfeasible 12,16 45,5
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MINIMUM COST
X1 X2 2X1+7X2
12 16 136
45 5 125=MIN
0 32 224
0 50 350
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EXAM FORMAT
MAKE 45 UNITS OF ALASKA AND 5 UNITS ARABIAN FOR
$ 125 COST
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ANSWER: (2)
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SIMULTANEOUS EQUATIONS
• (1) 2X1 + X2 = 12
• (2) X1 – X2 = -3
• (1)+(2) 3X1 = 9
• X1 =3
• 3-X2 = -3
• X2 = 6
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0,12
6,0
0,3
3,6
INFEASIBLE
INFEASIBLE
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MINIMIZE COST
X1 X2 COST=
4X1+3X2
0 12 36
3 6 30=MIN
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ORIGINAL PROBLEM
• MAKE BOTH X1 AND X2
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NEW PROBLEM
• CHANGE (1) 2X1+X2 >8
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0,12
6,0
0,3
3,6
INFEASIBLE
0,8
4,0
1.7,4.7
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MIN COST
X1 X2 COST=
4X1+3X2
0 8 24
1.7 4.7 20=MIN
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OUTPUT INSENSITIVE
• NEW: MAKE BOTH X1 AND X2
• ALSO INPUT INSENSITIVE SINCE BOTH ORIGINAL PROBLEM AND NEW PROBLEM HAVE NO SLACK
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(3) ALGEBRA
STEP 1: OBJECTIVE FUNCTION
OLD C1 = 5
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Z= C1X1 +12X2
• 12X2 = Z – 5X1
• X2 = (1/12)Z – (C1/12)X1
• COEFFICIENT OF X1 IS –C1/12
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STEP 2: CONSTRAINT (1)
• 2X1 + X2 = 200
• X2 = 200 –2X1
• COEF OF X1 IS -2
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STEP 3: STEP 1 = STEP 2
• -C1/12 = -2
• C1 = 24
• OLD C1 = 5
• C1 < 24
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STEP 4: CONSTRAINT (2)
• X1 + 2X2 = 300
• 2X2 = 300 –X1
• X2 = 150 –0.5 X1
• COEF OF X1 IS –0.5
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STEP 5: STEP 1 = STEP 4
• -C1/12 = -0.5
• C1 = 6
• OLD C1 = 5
• C1 < 6
• STEP 3: C1 < 24
• ANSWER: C1 < 6
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(4) MAX X1+3X2+2X3+4X4
• X1=AM PHONE, X2=AM RIDE, X3=AFT PHONE, X4=AFT RIDE
• CONSTRAINTS
• AM: X1+20X2 < 12(60)=720
• AFT: 2X3+30X4 < 14(60) = 840
• GAS: X2+X4 < 100
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(5)MIN 3X1+4X2+2X3+X4, X1=FISH,X2=BEEF,
X3=LETTUCE, X4=BEANS
• CAL: 1000<400X1+700X2+40X3+500X4<1800
• FAT: 15<4X1+80X2+2X4<40
• PROT: 90X1+100X2+2X3+20X4>25
• CHOL: 100X1+300X2<19
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(6)MIN 100000X1+50000X2, X1=TV, X2= RADIO
• X2 X1 > 2
• 1000000X1+ 400000X2 > 3000000
• .6(1000000)X1+(1-.7)(40000)X2 1000000X1+400000X2 > .4
• X1 < 5