Summary Modelling and simulation 4 notes and excercises

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

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