Question

Create a worksheet called "Transport a. Enter the following transportation model into excel following the linear programming layout. Hint: this is unbalanced Make sure to include your 1) decision variables, 2) objective function, 3) constraints, 4) inequalities, 5) confounds, 6) and solution b. c. Once you run your model, select "keep solver solution". Name the sheet that's created "solution_basic" DESTINATIONS 13 20 20 25 10 12 30 16 35 20 30 25

Answer 1

Here Total Demand = 75 is greater than Total Supply = 65. So We add a dummy supply constraint with 0 unit cost and with allocation 10.
Now, The modified table is

	A	B	C	Supply
1	10	25	13	20
2	20	25	8	10
3	12	30	16	35
Sdummy	0	0	0	10
Demand	20	30	25

The rim values for 1=20 and A=20 are compared.
The smaller of the two i.e. min(20,20) = 20 is assigned to 1 A
This exhausts the capacity of 1 and leaves 20 - 20 = 0 units with A
Table-1

	A	B	C	Supply
1	10(20)	25	13	0 0=20-20
2	20	25	8	10
3	12	30	16	35
Sdummy	0	0	0	10
Demand	0 0=20-20	30	25

The rim values for 2=10 and A=0 are compared.
The smaller of the two i.e. min(10,0) = 0 is assigned to 2 A
This meets the complete demand of A and leaves 10 - 0 = 10 units with 2
Table-2

	A	B	C	Supply
1	10(20)	25	13	0
2	20	25	8	10 10=10-0
3	12	30	16	35
Sdummy	0	0	0	10
Demand	0 0=0-0	30	25

The rim values for 2=10 and B=30 are compared.
The smaller of the two i.e. min(10,30) = 10 is assigned to 2 B
This exhausts the capacity of 2 and leaves 30 - 10 = 20 units with B
Table-3

	A	B	C	Supply
1	10(20)	25	13	0
2	20	25(10)	8	0 0=10-10
3	12	30	16	35
Sdummy	0	0	0	10
Demand	0	20 20=30-10	25

The rim values for 3=35 and B=20 are compared.
The smaller of the two i.e. min(35,20) = 20 is assigned to 3 B
This meets the complete demand of B and leaves 35 - 20 = 15 units with 3
Table-4

	A	B	C	Supply
1	10(20)	25	13	0
2	20	25(10)	8	0
3	12	30(20)	16	15 15=35-20
Sdummy	0	0	0	10
Demand	0	0 0=20-20	25

The rim values for 3=15 and C=25 are compared.
The smaller of the two i.e. min(15,25) = 15 is assigned to 3 C
This exhausts the capacity of 3 and leaves 25 - 15 = 10 units with C
Table-5

	A	B	C	Supply
1	10(20)	25	13	0
2	20	25(10)	8	0
3	12	30(20)	16(15)	0 0=15-15
Sdummy	0	0	0	10
Demand	0	0	10 10=25-15

The rim values for Sdummy=10 and C=10 are compared.
The smaller of the two i.e. min(10,10) = 10 is assigned to Sdummy C
Table-6

	A	B	C	Supply
1	10(20)	25	13	0
2	20	25(10)	8	0
3	12	30(20)	16(15)	0
Sdummy	0	0	0(10)	0 0=10-10
Demand	0	0	0 0=10-10

The initial feasible solution is

	A	B	C	Supply
1	10 (20)	25	13	20
2	20	25 (10)	8	10
3	12	30 (20)	16 (15)	35
Sdummy	0	0	0 (10)	10
Demand	20	30	25

The minimum total transportation cost =10×20+25×10+30×20+16×15+0×10=1290
Here, the number of allocated cells = 5, which is one less than to m + n - 1 = 4 + 3 - 1 = 6

This solution is degenerate