MATH 3170 Assignment #2 Solution (20 mark)
Questions marked: 3.8.10, 3.9.12, 3.10.9, 3.11.9
3.6.4
Let Xi = fraction undertaken of investment I (I = 1, 2, …, 9).
The LP formulation is:
max z = 14X1 + 17X2 + 17X3 + 15X4 + 40X5 + 12X6 + 14X7 + 10X8 + 12X9
s.t.
12X1 + 54X2 + 6X3 + 6X4 + 30X5 + 6X6 + 48X7 + 36X8 + 18X9<=50 (Year 1)
3X1 + 7X2 +6X3 + 2X4 + 35X5 + 6X6 + 4X7 + 3X8 + 3X9<=20 (Year 2)
All variables>=0.
Note that the question does not mention that the cash not used in Year 1
can be used in Year 2.
3.7.1
Graphically, we find the optimal solution to be x1 = 50, x2 = 100 with
optimal value z = 2500.
3.8.10 (5 marks)
Let Mi = Tons of coal shipped from Mine i and Xij = Tons of coal shipped
from Mine i to Customer j. Here is appropriate formulation in LINDO:
MIN 50 M1 +
X22 + 7 X23 +
SUBJECT TO
2)
3)
4)
5)
6)
7)
8)
9)
10)
11)
12)
13)
END
55 M2 + 62 M3 + 4 X11 + 6 X12 + 8 X13 + 12 X14 + 9 X21 + 6
11 X24 + 8 X31 + 12 X32 + 3 X33 + 5 X34
M1 <=
120
M2 <=
100
M3 <=
140
X11 + X21 + X31 =
80
X12 + X22 + X32 =
70
X13 + X23 + X33 =
60
X14 + X24 + X34 =
90
0.08 M1 + 0.06 M2 + 0.04 M3 <=
0.05 M1 + 0.04 M2 + 0.03 M3 <=
M1 - X11 - X12 - X13 - X14 =
M2 - X21 - X22 - X23 - X24 =
M3 - X31 - X32 - X33 - X34 =
15
12
0
0
0
Of course, all variables are nonnegative.
Constraint (9) comes from 0.08M1+0.06M2+0.04M3 <= 0.05(M1+M2+M3) but
since we know M1+M2+M3 is simply the sum of all the Xij’s which is
80+70+60+90=300, the RHS is imply 0.05(300)=15. Establishing constraint
(10) is similar.
You can do the above formulation without defining M1, M2 and M3 by simply
replacing each Mi with the corresponding expression given in (11) to
(13).
3.9.12 (5 marks)
Let A = hundreds of liters of A purchased and processed
B = hundreds of liters of B produced from A and processed
C = hundreds of liters of C from A
D = hundreds of liters of D from processed C
iS = hundreds of liters of i sold (i=B,C or D)
CP = hundreds of liters of C processed
iL = hundreds of liters of product i left unsold (i=B,C or D)
The correct LP formulation in LINDO is as follows:
MAX
12 BS
SUBJECT TO
2) 3) 4) 5)
6)
7)
8) -
+ 16 CS + 26 DS - 9 A - CP
CS - CP + C
BS + B - BL
DS + D - DL
BS <=
30
CS <=
60
DS <=
40
0.6 A - 0.4
- CL =
=
0
=
0
CP + B =
0
0
9) - 0.4 A + C =
10) - 0.6 CP + D =
11)
3 A + CP <=
0
0
200
END
Of course, all variables are nonnegative. Note that there are many
different ways to model this problem as an LP problem. The above
formulation is just one of them.
3.10.9 (5 marks)
Let
xi = number of workers who get quarter i off
it = inventory of mixers at end of quarter t
mt = mixers produced during quarter t
The LP formulation is:
min z = 30(i1 + i2 + i3 + i4) + 30,000(x1 + x2 + x3 + x4)
s.t.
i1 = 600 + m1 - 4000, i2 = i1 + m2 - 2000
i3 = i2 + m3 - 3000, i4 = i3 + m4 - 10,000
m1500(x2 + x3 + x4), m2500(x1 + x3 + x4),
m3500(x1 + x2 + x4), m4500(x1 + x2 + x3)
All variables 0
3.11.9 (5 marks)
Let Pi = Fraction of project I invested in, Bt = Money borrowed at time
t, Ct= ending cash position at time t
max z = 5.5P1 +6P3-P2+1.03C2.5-1.035B2.5
s.t.
C0 = 2-3P1-2P2-2P3+B0
C.5=1.03C0-1.035B0-P1-.5P2-2P3+B.5
C1=1.03C.5-1.035B.5+1.8P1+1.5P2-1.8P3+B1
C1.5=1.03C1-1.035B1+1.4P1+1.5P2+P3+B1.5
C2=1.03C1.5-1.035B1.5+1.8P1+1.5P2+P3+B2
C2.5=1.03C2-1.035B2+1.8P1+.2P2+P3+B2.5
All Bt<=2
ALL VARIABLES >=0
The above formulation assumes that all debts with interest must be
repaid in exactly 6 months.