a transportation problem

T

H

Let cik be cold stream i in interval k and hjl be hot stream j in the interval l.

Define aik as the heat needed by cik, which can be readily calculated after partitioning.

Similarly define bjl as the heat available from stream hjl.

Let qik,jl be the heat transferred from hjl to cik. The qik,jl are to be calculated.

Assume there are L intervals.

Let there be C-1 cold process streams and H-1 hot process streams in our problem.

Then the cold utility will be the Cth cold stream and the hot, the Hth stream.

Assume the heat needed by the cold utility is at the lowest level in the problem.

Also assume it is sufficient quantity to satisfy all the hot process stream cooling needs.

1

11 1

H L

C jlj l

a b

Assume similarly that the hot utility is available at the highest level and is in sufficient quantity to satisfy by itself all the cold stream heating requirements.

1

1 1

C L

HL iki k

b a

Lastly assume the problem is in heat balance overall.

The above simply say, choose both aC1 and bHL to be large numbers. Then adjust them so the entire problem is heat balanced.

1 1

11 1 1 1

C L H L

C ik HL jli k j l

a a b b

The heat required by cold stream i in interval k must be satisfied by transferring heat somewhere among the hot streams.

Similar statement for hot stream j in interval l - it must give up its heat somewhere to other streams.

All heats transferred must be nonnegative, that is no heat can flow from a cold stream to a hot one.

The objective function is minimized, with cost coefficients defined as below.

Each entry is a “cell” which can contain 3 numbers.

The upper right is the cost coefficient, Cik,jl.

The bottom number is the assigned qik,jl for the match.

The upper left is the value for optimality justification ξik,jl.

For each row aik is given to the far left and for each column bij to the very top.

We place the hot utility column (labeled H) to the far right and the cold utility column (labeled C) to the bottom.

Cells have been marked “I” if they are thermodynamically infeasible, i.e. if k > l for entry qik,jl.

1) Start in the upper left (northwest) corner.

2) Move from left to right in the uppermost row to the first column having a cost less than M, finding the cell corresponding to row cik, column hjl.

3) Assign qik,jl = Min(aik,bjl) to the cell.

4) Decrement both aik and bij by qik,jl.

5) Cross out the row or column which hast its heating or cooling requirement aik or bjl reduced to zero.

6) Repeat from step 2 until all rows and columns are deleted.

b jl

Hota ik Cold

0 M M M M M M 1

0 0 0 M M M M 1

0 0 0 M M M M 1

0 0 0 0 0 M M 1

0 0 0 0 0 0 0 1

1 1 1 1 1 1 1 0

∞ 590

1050∞

580

H15

C15

2550 C14

600

600

C-∞

γ jl

0

0

0

0

0

0

C241170410

520

221

C23

C22370

3532189

639

2

H24

3400

I

H14

1700

I

850

1170

2

0

0

0

0

1700

1

H22

I

I

I

I

0

800

2

H12

I

I

I

I

0

100

2

H23

I

I

I

0

0

Pinch

ρ ik

-1400

2

400 800 100 200

H13

I

I

I

0

0

200

2

H

450

0

0

0

0

50

Pinch0

-2

-2

-2

-2

500420 380 363 -∞

520

221

1) We must first establish for each row a “row cost”, ρik, and for each column a “column cost”, γjl. We show row and column costs along the right side and bottom of the tableau. Start with the top row and assign it a row cost of zero.

2) For any row cik for which a row cost is already assigned, find an active cell (qik,jl > 0) in that row. Assign a column cost γjl for the column corresponding to the active cell such that

γjl + ρik = Cik,jl

3) Repeat step 2 for assigned columns to set row costs.

4) Repeat steps 2 and 3 as needed until all row and column costs are set.

5) For every cell (or at least every inactive cell) write

ξik,jl = γjl + ρik

6) If no cell exists where ξik,jl > Cik,jl exist. The current tableau is optimal.

b jl

Hota ik Cold

0 0 2 M 2 M 2 M 2 M 2 M 2 M 1 1

-2 0 0 0 0 0 0 M 0 M 0 M 0 M -1 1

-2 0 0 0 0 0 0 M 0 M 0 M 0 M -1 1

-2 0 0 0 0 0 0 0 0 0 0 M 0 M -1 1

-2 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 1

-1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0

∞ 590

1050∞

580

H15

C15

2550 C14

600

600

C-∞

γ jl

0

0

0

0

0

0

C241170410

520

221

C23

C22370

3532189

639

2

H24

3400

I

H14

1700

I

850

1170

2

0

0

0

0

1700

1

H22

I

I

I

I

0

800

2

H12

I

I

I

I

0

100

2

H23

I

I

I

0

0

Pinch

ρ ik

-1400

2

400 800 100 200

H13

I

I

I

0

0

200

2

H

450

0

0

0

0

50

Pinch0

-2

-2

-2

-2

500420 380 363 -∞

520

221

The minimum utility usage for the given problem are◦ Steam: 450 kW◦ Cooling water: 2139 kW

for ΔTmin = 10o.

There are no heat crosses the pinch point so that we can locate the pinch point easily by noting the pattern.

If only thermodynamic constraints are involved and if ΔTmin is the same for all matches, the initial solution is always optimal for such problem.