Question

The demand for a perishable item over the next four months is 400, 300, 420, and 380 tons, respectively. The supply capacities for the same months are 500,600, 200, and 300 tons The purchase price per ton varies from month to month and is estimated at $100, $140, $120, and $150, respectively. Because the item is perishable, a current month's supply must be consumed within 3 months (starting with current month). The storage cost per ton per month is $3. The nature of the item does not allow back-ordering. Solve the problem as a transportation model, and determine the optimum delivery schedule for the item over the next 4 months.

A) Formulate the problem as a transportation problem by hand

B) Using the initial solution with the lowest cost use the method of multiplier to fond the optimal solution

Answer 1

(a)

The transportation table is as follows:

TOTAL number of supply constraints : 4
TOTAL number of demand constraints : 4
Problem Table is

	Month1	Month2	Month3	Month4		Supply
S1	150	120	140	100		500
S2	150	120	140	100		600
S3	150	120	140	100		200
S4	150	120	140	100		300
Demand	380	420	300	400

(b)

Solution with the lowest cost method is as follows:

Here Total Demand = 1500 is less than Total Supply = 1600. So We add a dummy demand constraint with 0 unit cost and with allocation 100.
Now, The modified table is

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100	0		500
S2	150	120	140	100	0		600
S3	150	120	140	100	0		200
S4	150	120	140	100	0		300
Demand	380	420	300	400	100

The smallest transportation cost is 0 in cell S1Ddummy

The allocation to this cell is min(500,100) = 100.
This satisfies the entire demand of Ddummy and leaves 500 - 100 = 400 units with S1

Table-1

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100	0(100)		400
S2	150	120	140	100	0		600
S3	150	120	140	100	0		200
S4	150	120	140	100	0		300
Demand	380	420	300	400	0

The smallest transportation cost is 100 in cell S1Month4

The allocation to this cell is min(400,400) = 400.
Table-2

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100(400)	0(100)		0
S2	150	120	140	100	0		600
S3	150	120	140	100	0		200
S4	150	120	140	100	0		300
Demand	380	420	300	0	0

The smallest transportation cost is 100 in cell S4Month4

The allocation to this cell is min(300,0) = 0.
This satisfies the entire demand of Month4 and leaves 300 - 0 = 300 units with S4

Table-3

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100(400)	0(100)		0
S2	150	120	140	100	0		600
S3	150	120	140	100	0		200
S4	150	120	140	100	0		300
Demand	380	420	300	0	0

The smallest transportation cost is 120 in cell S2Month2

The allocation to this cell is min(600,420) = 420.
This satisfies the entire demand of Month2 and leaves 600 - 420 = 180 units with S2

Table-4

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100(400)	0(100)		0
S2	150	120(420)	140	100	0		180
S3	150	120	140	100	0		200
S4	150	120	140	100	0		300
Demand	380	0	300	0	0

The smallest transportation cost is 140 in cell S4Month3

The allocation to this cell is min(300,300) = 300.
Table-5

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100(400)	0(100)		0
S2	150	120(420)	140	100	0		180
S3	150	120	140	100	0		200
S4	150	120	140(300)	100	0		0
Demand	380	0	0	0	0

The smallest transportation cost is 140 in cell S3Month3

The allocation to this cell is min(200,0) = 0.
This satisfies the entire demand of Month3 and leaves 200 - 0 = 200 units with S3

Table-6

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100(400)	0(100)		0
S2	150	120(420)	140	100	0		180
S3	150	120	140	100	0		200
S4	150	120	140(300)	100	0		0
Demand	380	0	0	0	0

The smallest transportation cost is 150 in cell S3Month1

The allocation to this cell is min(200,380) = 200.
This exhausts the capacity of S3 and leaves 380 - 200 = 180 units with Month1

Table-7

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100(400)	0(100)		0
S2	150	120(420)	140	100	0		180
S3	150(200)	120	140	100	0		0
S4	150	120	140(300)	100	0		0
Demand	180	0	0	0	0

The smallest transportation cost is 150 in cell S2Month1

The allocation to this cell is min(180,180) = 180.
Table-8

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100(400)	0(100)		0
S2	150(180)	120(420)	140	100	0		0
S3	150(200)	120	140	100	0		0
S4	150	120	140(300)	100	0		0
Demand	0	0	0	0	0

Initial feasible solution is

	Month1	Month2	Month3	Month4	Ddummy		Supply
S1	150	120	140	100 (400)	0 (100)		500
S2	150 (180)	120 (420)	140	100	0		600
S3	150 (200)	120	140	100	0		200
S4	150	120	140 (300)	100	0		300
Demand	380	420	300	400	100

The minimum total transportation cost =100×400+0×100+150×180+120×420+150×200+140×300=189400

Here, the number of allocated cells = 6, which is two less than to m + n - 1 = 4 + 5 - 1 = 8
∴ This solution is degenerate

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