NAME: RIVERA, MARY QUEEN S.
MANAGEMENT SCIENCE
CBET-603P
ASSIGNMENT 4
PROBLEM 1.
TABLE 1: INITIAL FEASIBLE SOLUTION
R;
K;
To
from
k1 = 40
k2 = 120
k 3= 200
A
B
C
Plant capacity
R1 = 0
Plant W
56
40
-40
80
-120
80
56
R2 = 120
Plant X
16
160
66
240
-40
160
82
R3 = 40
Plant Y
0
80
36
160
41
240
77
PR
72
102
41
Compute the values of R; and K;
R1 + K1 = 40
R2 + K2 = 240
0 + K1 = 40
120 + K2 =240
K1 = 40
K2 = 240-120
K2 = 120
R2 + K1 = 160
R2 + 40 = 160
R3 + K2 = 160
R2 = 160 - 40
R3 + 120 = 160
R2 = 120
R3 + 160 - 120
R3 = 40
R3 + K3 = 240
40 + K3 = 240
k3 = 240 - 40
k3 = 200
IMPROVEMENT INDEX :
C1, - R1, - K;
Unused cells:
W B: 80 - 0 - 120 = -40
W C: 80 - 0 - 200 = -120
X C: 160 - 120- 200 = -160
Y A: 80- 40 - 40 = 0
215
215
TABLE 2: SECOND IMPROVED SOLUTION
R;
K;
To
from
k1= 40
k2= 120
k3= 200
A
B
C
R1=0
Plant W
56
40
R2=120
Plant X
16
160
25
240
R3=40
Plant Y
0
80
77
160
PR
72
80
102
41
41
Plant capacity
80
56
160
82
240
77
215
215
R1 + K1 = 40
IMPROVEMENT INDEX :( C1, - R1, - K;)
0 + K1 = 40
Unused cells:
k1= 40
W B: 80 - 0 - 120 = -40
W C: 80 - 0 - 200 = -120
R2 + K1 = 160
X C : 80 - 40- 40 = 0
R2 + 40 = 160
Y A : 240- 40 - 40 = 160
R2 = 160 - 40
R2 = 120
R3 + K2 = 160
R3 + 120 = 160
R2 + K2 = 240
R3 + 160 - 120
120 + K2 = 240
R3 = 40
K2 = 240-120
K2 = 120
R2 + K3 = 160
120 + K3 = 160
K3 = 160 - 120
K3 = 40
TABLE 3: FINAL SOL
K;
R;
k1= 40
k2= 120
k3= 200
A
B
C
To
from
R1=0
Plant W
R2=120
Plant X
41
160
R3=40
Plant Y
31
80
PR
72
40
56
80
240
46
41
160
102
R3 + K2 = 160
0 +0 =0
80 + K2 = 160
K1 = 0
80
56
160
82
240
77
41
R1 + K1 = 0
k2 = 160 - 80
k2 = 80
R2 + K1 = 160
R2 + 0 = 160
R2 + K3 = 160
R2 = 160
160 + K3 = 160
k3 = 160 - 160
R1 + K2 = 80
k3 = 0
0 + K2 = 80
K2 = 80
R3 + K1 = 80
R3 + 0 = 80
R3 = 80
Unused cells: (C1, - R1, - K;)
W B : 40 - 0 - 0 = 40
W C : 80 - 0 - 0 = 80
X B : 240 - 160- 80 = 0
Y C : 240- 80 - 0 = 160
OPTIMAL TRANSPORTAION COST
Shipping
W, B
80 x 56
4,480
X, A
160 x 41
6,560
X, C
160 x 41
6,560
Y, A
80 x 31
2,480
Y, B
160 x 46
7,360
Transport Cost
27,440
Plant capacity
215
215