Week 1
Excercise 2.a
Variables
X1= 100 Juice glasses
X2= 100 Cocktail glasses
Objective function
Z (maximized profit) =500X1+450X2
Constraints
Production capacity=6X1+5X2≤60
Storage capacity= (10*100) X1 +(20*100) X2≤15000= X1+2X2≤15
Demand constraint =X1≤8 (geen demand constraint op X2, in dit geval)
X1,X2 ≥ 0 (niet vergeten)((anders is het ook mogelijk om negatieve aantallen te produceren
wat niet kan))
,Excercise 2.b
Production Capacity Storage Capacity
6X1+5X2≤60 X1+2X2≤15
X1=0 X1=0
X2= 60/5= 12 X2=2/15=7,5
6X1+5X2≤60 X1+2X2≤15
X2=0 X2=0
X1= 60/6= 10 X1=15
Z =5X1+4,5X2
*Objective kan je ook tekenen met rechte lijnen om dan je intersect met je CPF solution vinden.
A (0,0)
B (0;7,5): Z = 5*0+4,5*7,5= 33,75
C X1+2X2≤15
X1=(15-2X2)
6(15-2X2) +5X2≤60
90-12X2+5X2 ≤ 60
90-7X2 ≤ 60
7X2 ≤ 30
X2 ≤ (30/7); 4+(2/7)
X1=15-2*(30/7)
X1= 6+(3/7)
(6+(3/7);4+(2/7))
Z =5(6+(3/7))+4,5(4+(2/7)) = 32+(1/7) + 19 (2/7)= 51+(3/7)
D X1=8
6*(8) +5X2≤60
5X2=12
X2=2+(2/5)
(8;2+(2/5))
Z =5(8)+4,5(2+(2/5))= 50,8
E (8,0)= Z =5X1+4,5X2= 40
Optimum is point C. This is the most feasible solution. To maximize profit 6(3/7) batches of juice
glasses should be sold and 4(2/7) batches of cocktail glasses.
Excercise 2.d
,Not a linear function anymore but a quadratic function after solving the brackets.
Excercise 3.a
Variables
X1= number of 60-inch televisions
X2= number of 42-inch televisions
Objective function
Z (maximized profit) =120X1+80X2
Constraints
workhour capacity=20X1+10X2 ≤ 500
Demand constraint =X1 ≤ 40; x2 ≤ 10
X1,X2 ≥ 0
Excercise 3.b
10X1+20X2 ≤ 500
X2=25; X1=50
A (0,0): Z=0
B (0;10): Z = 120*0+80*10= 800
C (20; 10)
20X1+10X2 ≤ 500
X2=10
20X1+100 ≤ 500
20X1 ≤ 400
X1 ≤ 20
120*20+80*10= 3200
D (25;0)= 120*25+80*0= 3000
,
Excercise 2.a
Variables
X1= 100 Juice glasses
X2= 100 Cocktail glasses
Objective function
Z (maximized profit) =500X1+450X2
Constraints
Production capacity=6X1+5X2≤60
Storage capacity= (10*100) X1 +(20*100) X2≤15000= X1+2X2≤15
Demand constraint =X1≤8 (geen demand constraint op X2, in dit geval)
X1,X2 ≥ 0 (niet vergeten)((anders is het ook mogelijk om negatieve aantallen te produceren
wat niet kan))
,Excercise 2.b
Production Capacity Storage Capacity
6X1+5X2≤60 X1+2X2≤15
X1=0 X1=0
X2= 60/5= 12 X2=2/15=7,5
6X1+5X2≤60 X1+2X2≤15
X2=0 X2=0
X1= 60/6= 10 X1=15
Z =5X1+4,5X2
*Objective kan je ook tekenen met rechte lijnen om dan je intersect met je CPF solution vinden.
A (0,0)
B (0;7,5): Z = 5*0+4,5*7,5= 33,75
C X1+2X2≤15
X1=(15-2X2)
6(15-2X2) +5X2≤60
90-12X2+5X2 ≤ 60
90-7X2 ≤ 60
7X2 ≤ 30
X2 ≤ (30/7); 4+(2/7)
X1=15-2*(30/7)
X1= 6+(3/7)
(6+(3/7);4+(2/7))
Z =5(6+(3/7))+4,5(4+(2/7)) = 32+(1/7) + 19 (2/7)= 51+(3/7)
D X1=8
6*(8) +5X2≤60
5X2=12
X2=2+(2/5)
(8;2+(2/5))
Z =5(8)+4,5(2+(2/5))= 50,8
E (8,0)= Z =5X1+4,5X2= 40
Optimum is point C. This is the most feasible solution. To maximize profit 6(3/7) batches of juice
glasses should be sold and 4(2/7) batches of cocktail glasses.
Excercise 2.d
,Not a linear function anymore but a quadratic function after solving the brackets.
Excercise 3.a
Variables
X1= number of 60-inch televisions
X2= number of 42-inch televisions
Objective function
Z (maximized profit) =120X1+80X2
Constraints
workhour capacity=20X1+10X2 ≤ 500
Demand constraint =X1 ≤ 40; x2 ≤ 10
X1,X2 ≥ 0
Excercise 3.b
10X1+20X2 ≤ 500
X2=25; X1=50
A (0,0): Z=0
B (0;10): Z = 120*0+80*10= 800
C (20; 10)
20X1+10X2 ≤ 500
X2=10
20X1+100 ≤ 500
20X1 ≤ 400
X1 ≤ 20
120*20+80*10= 3200
D (25;0)= 120*25+80*0= 3000
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