C = t(c(-1,4))
A = matrix(c(-10,20,5,10,1,0),nrow = 3, byrow = T)
dir = rep("<=",3)
b = c(22,49,5)
sol = lp(direction = 'max',
objective.in = C,
const.mat = A,
const.dir = dir,
const.rhs = b,
int.vec = c(1,2))
print (sol$solution)## [1] 2 2print (sol$objval)## [1] 6As shown in the graph, there are five branches of which only three are feasible.
Let \(F_A,F_D,W_A,W_D\) represent number of factories/warehouses to be built in Austin/Dallas.
\(max\ 9F_A + 5F_D + 6W_A + 4W_D\)
\(s.t.\)
\(6F_A + 3F_D + 5W_A + 2W_D <= 11\)
\(W_A + W_D <=1\)
\(F_A,F_D,W_A,W_D\ are\ binary\)
C2 = t(c(9,5,6,4))
A2 = matrix(c(6,3,5,2,
0,0,1,1,
1,1,0,0), nrow = 3, byrow = T)
dir2 = c("<=","<=",">=")
b2 = c(11,1,1)
sol2 = lp(direction = 'max',
objective.in = C2,
const.mat = A2,
const.dir = dir2,
const.rhs = b2,
binary.vec = c(1,2,3,4))
print (sol2$solution)## [1] 1 1 0 1print (sol2$objval)## [1] 18## city num
## 1 ATL 1
## 2 BOS 2
## 3 CHI 3
## 4 DEN 4
## 5 HOU 5
## 6 LAX 6
## 7 NO 7
## 8 NY 8
## 9 PIT 9
## 10 SLC 10
## 11 SF 11
## 12 SEA 12Let \(x_1,\ x_2,\ x_3,\ x_4,\ x_5,\ x_6,\ x_7,\ x_8,\ x_9,\ x_10,\ x_11,\ x_12\) be binary variables, so \(x_i\ =\ 1\) means a hub will be built in place \(i\). The problem can be formulated as:
\(min\ x_1+x_2+x_3+x_4+x_5+x_6+x_7+x_8+x_9+x_10+x_11+x_12\)
\(s.t.\)
\(x_1+x_3+x_5+x_7+x_8+x_9\ >=\ 1\ (Force\ ATL\ to\ be\ covered,\ same\ for\ follows)\)
\(x_2+x_8+x_9\ >=\ 1\)
\(x_1+x_3+x_7+x_8+x_9\ >=\ 1\)
\(x_4+x_10\ >=\ 1\)
\(x_1+x_5+x_7\ >=\ 1\)
\(x_6+x_10+x_11\ >=\ 1\)
\(x_1+x_3+x_5+x_7\ >=\ 1\)
\(x_1+x_3+x_5+x_8+x_9\ >=\ 1\ (Same\ for\ NY\ and\ PIT)\)
\(x_4+x_6+x_10+x_11+x_12\ >=\ 1\)
\(x_6+x_10+x_11+x_12\ >=\ 1\)
\(x_10+x_11+x_12\ >=\ 1\)
C3 = t(rep(1,12))
A3 = matrix(c(1,0,1,0,1,0,1,1,1,0,0,0,
0,1,0,0,0,0,0,1,1,0,0,0,
1,0,1,0,0,0,1,1,1,0,0,0,
0,0,0,1,0,0,0,0,0,1,0,0,
1,0,0,0,1,0,1,0,0,0,0,0,
0,0,0,0,0,1,0,0,0,1,1,0,
1,0,1,0,1,0,1,0,0,0,0,0,
1,1,1,0,0,0,0,1,1,0,0,0,
1,1,1,0,0,0,0,1,1,0,0,0,
0,0,0,1,0,1,0,0,0,1,1,1,
0,0,0,0,0,1,0,0,0,1,1,1,
0,0,0,0,0,0,0,0,0,1,1,1),
nrow = 12, byrow = T)
dir3 = rep(">=",12)
b3 = rep(1,12)
sol3 = lp("min",
objective.in = C3,
const.mat = A3,
const.dir = dir3,
const.rhs = b3,
binary.vec = 1:12)
print (sol3$objval)## [1] 3print (sol3$solution)## [1] 1 0 0 0 0 0 0 1 0 1 0 0Therefore, a minimal of 3 hubs should be built at ATL, NY, SLC so as to cover all 12 hubs.
There are six combiniations with least possible waste:
## Unit Combos Waste
## 1 x1 4 * W25 20
## 2 x2 3 * W25 + 1 * W37 8
## 3 x3 2 * W25 + 1 * W54 16
## 4 x4 1 * W25 + 1 * W37 + 1 * W54 4
## 5 x5 3 * W37 9
## 6 x6 2 * W54 12
## 7 x7 2 * W37 + 1 * W25 21Let \(x_1,\ x_2,\ x_3,\ x_4,\ x_5,\ x_6\) represent number of each combos. The objective is to minimize waste while satisfying quantity demands.The problem, therefore, can be formulated as follows:
\(min\ 20x_1 + 8x_2 + 16x_3 + 4x_4 + 9x_5 + 12x_6 + 21x_7\)
\(s.t.\)
\(4x_1 + 3x_2 + 2x_3 + x_4 + x_7\ >=\ 233\)
\(x_2 + x_4 + 3x_5 + 2x_7\ >=\ 148\)
\(x_3 + x_4 + 2x_6\ >=\ 106\)
\(x_1,\ x_2,\ x_3,\ x_4,\ x_5,\ x_6\ are\ integers\)
C4 = t(c(20,8,16,4,9,12,21))
A4 = matrix(c(4,3,2,1,0,0,1,
0,1,0,1,3,0,2,
0,0,1,1,0,2,0), nrow = 3, byrow = T)
dir4 = rep(">=",3)
b4 = c(233,148,106)
sol4 = lp("min",
objective.in = C4,
const.mat = A4,
const.dir = dir4,
const.rhs = b4,
int.vec = c(1:7))
print (sol4$objval)## [1] 764print (sol4$solution)## [1] 0 42 0 107 0 0 0The solution says we should produce 42 sets of Combo 2 and 107 sets of Combo 4.
## Type Mon Tue Wed Thu Fri Sat Sun Cost
## 1 x1 1 1 1 1 1 0 0 300
## 2 x2 0 1 1 1 1 1 0 330
## 3 x3 0 0 1 1 1 1 1 360
## 4 x4 1 0 0 1 1 1 1 360
## 5 x5 1 1 0 0 1 1 1 360
## 6 x6 1 1 1 0 0 1 1 360
## 7 x7 1 1 1 1 0 0 1 330Let \(x_1,\ x_2,\ x_3,\ x_4,\ x_5,\ x_6\) represent number of 6 types of workers, so the question can be formulated as follows:
\(min\ 300x_1 + 330x_2 + 360x_3 + 360x_4 + 360x_5 + 360x_6\)
\(s.t.\)
\(x_1 + x_4 + x_5 + x_6 + x_7\ >=\ 13\)
\(x_1 + x_2 + x_5 + x_6 + x_7\ >=\ 12\)
\(x_1 + x_2 + x_3 + x_6 + x_7\ >=\ 10\)
\(x_1 + x_2 + x_3 + x_4 + x_7\ >=\ 14\)
\(x_1 + x_2 + x_3 + x_4 + x_5\ >= 8\)
\(x_2 + x_3 + x_4 + x_5 + x_6\ >= 6\)
\(x_3 + x_4 + x_5 + x_6 + x_7\ >= 5\)
\(x_i\ are\ integers\)
C5 = t(Cost)
A5 = matrix(c(Mon,Tue,Wed,Thu,Fri,Sat,Sun),
nrow = 7, byrow = T)
dir5 = rep(">=",7)
b5 = c(13,12,10,14,8,6,5)
sol5 = lp("min",
objective.in = C5,
const.mat = A5,
const.dir = dir5,
const.rhs = b5,
int.vec = 1:7)
print (sol5$objval)## [1] 4830print (sol5$solution)## [1] 8 2 0 3 1 0 1Therefore, we need 8 workers of Type 1, 2 of Type 2, 3 of Type 4, 1 of Type 5, and 1 of Type 7.
Type 1 workers who work from Mon to Fri are most popular. This makes sense because Mon - Fri are higher in workder demand and lower in cost comparing to Sat and Sun.