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.
H 1
aC1 
L
b
j 1 l 1
jl

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.
C 1
bH L 
L
a
i 1 k 1
ik

Lastly assume the problem is in heat balance
overall.
C 1
aC1 
a
i 1 k 1

H 1
L
ik
 bH L  
L
b
jl
j 1 l 1
The above simply say, choose both aC1 and
bHL to be large numbers. Then adjust them so
the entire problem is heat balanced.


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)
2)
Start in the upper left (northwest) corner.
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)
6)
Cross out the row or column which hast its heating or
cooling requirement aik or bjl reduced to zero.
Repeat from step 2 until all rows and columns are
deleted.
∞
b jl
Hot
∞
a ik
Cold
1050
C15
2550
C14
1170
C24
520
C23
221
C22
2189
C
580
410
370
353
-∞
γ jl
590
420
380
363
-∞
600
1700
3400
400
800
100
200
500
H15
H14
H24
H13
H23
H12
H22
H
0
M
600
I
0
M
I
0
0
1700
0
0
850
0
0
0
0
0
0
0
1
1
1
1
M
M
0
1
1
0
I
0
1
0
I
M
0
0
M
M
I
1
450
I
I
0
0
0
M
M
0
M
I
I
I
0
0
221
M
M
0
M
I
I
I
0
0
0
M
0
520
M
I
I
1170
0
0
M
I
1
0
0
0
1
1
0
1
0
0
0
639
400
800
100
200
50
0
2
2
2
2
2
2
1
Pinch
ρ ik
0
Pinch
-2
-2
-2
-2
-1
1)
2)
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.
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.
∞
590
b jl
Hot
∞
a ik
Cold
1050
C15
2550
C14
1170
C24
520
C23
221
C22
2189
C
580
410
370
353
-∞
γ jl
420
380
363
-∞
600
1700
3400
400
800
100
200
500
H15
H14
H24
H13
H23
H12
H22
H
0
0
2
600
-2
0
0
0
-2
0
0
0
1
0
1
0
1
0
1
1
-1
M
0
-1
0
1
1
0
-1
0
1
1
0
I
0
1
0
M
0
0
1
-1
I
M
1
450
M
0
I
0
1
I
M
0
0
1
0
I
0
M
I
M
0
0
0
2
I
M
0
0
1
0
I
0
M
I
M
0
0
0
2
I
M
0
221
1
0
I
0
M
I
M
0
520
0
2
I
0
0
0
1
0
1170
0
M
I
0
0
0
0
2
850
0
0
0
-1
0
0
0
M
I
0
0
0
-2
2
1700
0
-2
M
I
1
0
1
0
0
0
0
639
400
800
100
200
50
0
2
2
2
2
2
2
1
Pinch
ρ ik
0
Pinch
-2
-2
-2
-2
-1

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.