Delft University ME 44205 Assignment: Steel Recycling Q2 2018

This assignment solution addresses a steel recycling optimization problem where the objective is to minimize the cost of producing five batches of stainless steel with specific chromium, nickel, iron, and copper proportions using ten different heaps of scrap iron and pure chromium. The solution includes a mathematical formulation with indices, sets, parameters, decision variables, an objective function to minimize cost, and functional constraints related to heap quantity, batch weight, and metal balance. The implementation of the model involves defining variables for the amount of scrap iron from each heap used in each batch, pure chromium addition, and copper removal, along with constraints for weight, heap availability, and metal composition. The provided mathematical model aims to find the optimal mix of scrap iron and pure chromium to meet the required steel specifications at the lowest cost, though the assignment notes that no feasible solution exists.

ME 44205 -
ASSIGNMENT Q2 2018
Steel Recycling
DECEMBER 31, 2018

Model Description:
Key Concepts:
 There are 10 heaps of scrap iron named A to J. Each heap contains a consistent mixture of metals
like Chromium, Nickel, Copper and Iron as shown in the table 1.
 The amount of each heap available is mentioned in tons.
 The cost in euro/ton for each heap is also provided.
Table 1
heap A B C D E F G H I J
Chromium 10 25 25 14 0 18 10 2 25 10
Nickel 0 15 10 16 15 1 12 8 19 2
Copper 4 0 2 5 3 0 1 3 0 15
Iron 86 60 63 65 82 81 77 87 56 73
Amount (ton) 30 90 50 70 20 30 10 20 5 100
Cost (Euro/ton) 50 100 90 105 85 115 90 80 70 5
 Any proportion from each heap can be mixed to produce a batch of 18/10 stainless steel. Moreover,
pure Chromium can also be added in the batch at the cost of 125 Euro/ton.
 The main objective is to produce five batches of steel with at-least 18% Chromium, 10% Nickel, 65%
of Iron and variable proportions of Copper as mentioned in table 2. Also, the weight required for
each batch is also given in table 2.
Table 2
Batch 1 2 3 4 5
Weight (tons) 125 50 25 80 20
Max Copper 5 2.5 1 0.5 0
Indices and Sets
 i for Heaps I= [A, B, C, D, E, F, G, H, I, J]
 j for batches J= [Batch1, Batch2, Batch3, Batch4, Batch5]
Parameters
 per heap: maximum quantity (tons), cost (euro/ton)
 per batch: actual quantity (tons)
 per metal: proportion (%)
 all numerical values mentioned in the problem
Variables
 per heap quantity (real), per batch type
 often multiple types of variables are needed for modelling
 detail of variables depends upon the objective and constraints
Objective
 cost: sum of all the batches, per heap
 minimize
 often not all the variables are needed in the objective function

Functional constraints
1. total quantity of heaps available, sum over all heaps, each batch
2. total weights of each batch required, sum over all the proportions from each heap
3. balance of maximum proportion of metals in each batch
a. chromium
b. nickel
c. iron
d. copper
4. non-negativity constraint
Ques 1: Mathematical Formulation
Indices and Sets
i Index of heap i ∈ I [A, B, C, D, E, F, G, H, I, J]
j Index of batch j ∈ J [P, Q, R, S, T]
Parameters
mi Maximum quantity of heap Tons
nj Quantity of batch Tons
ai Proportion of chromium %
bi Proportion of nickel %
di Proportion of iron %
ei Proportion of copper %
fj Maximum copper per batch %
yi Cost of heap Euro/ton
Decision Variables
Xij : proportion of heap I in batch j in tons
Objective Function
 Minimize cost
M nⅈ ∑
i=1
10
∑
j=1
5
yi xij
Functional Constraints
1. Heap quantity
For i ∈ I ∑
j ∈ J
xij ≤ mj
2. Batch weight
For j ∈ J ∑
i ∈ I
xij=n j
3. Chromium balance
∑
i ∈ I
ai xij ≥ 1 8
4. Nickel balance

∑
i ∈ I
bi xij ≥ 1 0
5. Iron balance
∑
i ∈ I
di xij ≥ 6 5
6. Copper balance
∑
i ∈ I
e xij ≥ fj
7. Non-negativity constraint
xij ≥ 0
8. Real-value constraint
xij ∈ R
Ques 2: Implementation of the model
Let the amount of scrap iron chosen for Batch 1 from Heaps A-J be PA, PB, PC, PD, PE,
PF, PG, PH, PI and PJ tons respectively.
Let the amount of scrap iron chosen for Batch 2 from Heaps A-J be QA, QB, QC, QD, QE,
QF, QG, QH, QI, and QJ tons respectively.
Let the amount of scrap iron chosen for Batch 3 from Heaps A-J be RA, RB, RC, RD, RE,
RF, RG, RH, RI, and RJ tons respectively.
Let the amount of scrap iron chosen for Batch 4 from Heaps A-J be SA, SB, SC, SD, SE,
SF, SG, SH, SI, and SJ tons respectively.
Let the amount of scrap iron chosen for Batch 5 from Heaps A-J be TA, TB, TC, TD, TE,
TF, TG, TH, TI, and TJ tons respectively.
Let the Pure Chromium added to the batches 1 to 5 be Cr1, Cr2, Cr3, Cr4 and Cr5 tons
respectively.
Let the amount of Copper removed by electrolysis from each batch be C1, C2, C3, C4
and C5 kg respectively.
Cost Function
Min(z)= 50 (PA+QA+RA+SA+TA) +
100 (PB+QB+RB+SB+TB) +
90 (PC+QC+RC+SC+TC) +
105 (PD+QD+RD+SD+TD) +
85 (PE+QE+RE+SE+TE) +

115 (PF+QF+RF+SF+TF) +
90 (PG+QG+RG+SG+TG) +
80 (PH+QH+RH+SH+TH) +
70 (PI+QI+RI+SI+TI) +
5 (PJ+QJ+RJ+SJ+TJ) +
125 (Cr1+Cr2+Cr3+Cr4+Cr5) +
5*1000 +
0.50 *1000 (C1+C2+C3+C4+C5)
Constraints:
Weight of Batch constraints:
PA+PB+PC+PD+PE+PF+PG+PH+PI+PJ-C1 =125
QA+QB+QC+QD+QE+QF+QG+QH+QI+QJ-C2 =50
RA+RB+RC+RD+RE+RF+RG+RH+RI+RJ-C3 =25
SA+SB+SC+SD+SE+SF+SG+SH+SI+SJ-C4 =80
TA+TB+TC+TD+TE+TF+TG+TH+TI+TJ-C5 =20
Heap Weight Constraints:
PA+QA+RA+SA+TA<=30
PB+QB+RB+SB+TB<=90
PC+QC+RC+SC+TC<=50
PD+QD+RD+SD+TD<=70
PE+QE+RE+SE+TE<=20
PF+QF+RF+SF+TF<=30
PG+QG+RG+SG+TG<=10
PH+QH+RH+SH+TH<=20
PI+QI+RI+SI+TI<=5

PJ+QJ+RJ+SJ+TJ<=100
Metal Constraints:
Chromium
0.01PA+0.25PB+0.25PC+0.14PD+0PE+0.18PF+0.10PG+0.02PH+0.25PI+0.10PJ>=18
0.01QA+0.25QB+0.25QC+0.14QD+0QE+0.18QF+0.10QG+0.02QH+0.25QI+0.10QJ>=18
0.01RA+0.25RB+0.25RC+0.14RD+0RE+0.18RF+0.10RG+0.02RH+0.25RI+0.10RJ>=18
0.01SA+0.25SB+0.25SC+0.14SD+0SE+0.18SF+0.10SG+0.02SH+0.25SI+0.10SJ>=18
0.01TA+0.25TB+0.25TC+0.14TD+0TE+0.18TF+0.10TG+0.02TH+0.25TI+0.10TJ>=18
Nickel
0PA+0.15PB+0.10PC+0.16PD+0.15PE+0.01PF+0.12PG+0.08PH+0.19PI+0.02PJ>=10
0QA+0.15QB+0.10QC+0.16QD+0.15QE+0.01QF+0.12QG+0.08QH+0.19QI+0.02QJ>=10
0RA+0.15RB+0.10RC+0.16RD+0.15RE+0.01RF+0.12RG+0.08RH+0.19RI+0.02RJ>=10
0SA+0.15SB+0.10SC+0.16SD+0.15SE+0.01SF+0.12SG+0.08SH+0.19SI+0.02SJ>=10
0TA+0.15TB+0.10TC+0.16TD+0.15TE+0.01TF+0.12TG+0.08TH+0.19TI+0.02TJ>=10
Iron
0.86PA+0.60PB+0.63PC+0.65PD+0.82PE+0.81PF+0.77PG+0.87PH+0.56PI+0.73PJ>=65
0.86QA+0.60QB+0.63QC+0.65QD+0.82QE+0.81QF+0.77QG+0.87QH+0.56QI+0.73QJ>=65
0.86RA+0.60RB+0.63RC+0.65RD+0.82RE+0.81RF+0.77RG+0.87RH+0.56RI+0.73RJ>=65
0.86SA+0.60SB+0.63SC+0.65SD+0.82SE+0.81SF+0.77SG+0.87SH+0.56SI+0.73SJ>=65
0.86TA+0.60TB+0.63TC+0.65TD+0.82TE+0.81TF+0.77TG+0.87TH+0.56TI+0.73TJ>=65
Copper
0.04PA+0PB+0.02PC+0.05PD+0.03PE+0.00PF+0.01PG+0.03PH+0.00PI+0.15PJ>=5
0.04QA+0QB+0.02QC+0.05QD+0.03QE+0.00QF+0.01QG+0.03QH+0.00QI+0.15QJ>=2.5
0.04RA+0RB+0.02RC+0.05RD+0.03RE+0.00RF+0.01RG+0.03RH+0.00RI+0.15RJ>=1
0.04SA+0SB+0.02SC+0.05SD+0.03SE+0.00SF+0.01SG+0.03SH+0.00SI+0.15SJ>=0.5
0.04TA+0TB+0.02TC+0.05TD+0.03TE+0.00TF+0.01TG+0.03TH+0.00TI+0.15TJ>=0

Non-negativity Constraints
All variables>=0
Ques 3,4:
When implementing this complex mathematical model, there exists no feasible
solution for such problem.