Download Additional Application Application 4.1 – Cell formation problem The

Additional Application
Application 4.1 – Cell formation problem
The cell formation problem in manufacturing is as follows: Given a set of m machines and n
parts and their incidence data, the problem is to group the machines and parts into given
number of groups (called machine cells and part families respectively) such that all the
machines required to make a part (in the part family) are available in the corresponding
machine cell (group). If this is not feasible, the problem reduces to minimizing the number
of visits of parts to machines in other groups.
This problem can be modeled as an Integer Programming problem but has also been solved
as a k-median problem (Kusiak, 1987). The data on the visit of machines and parts is given in
the form of a binary incidence matrix where aij = 1 indicates that part j visits machine i and =
0 otherwise.
We consider each machine (row) of the incidence matrix as a point in n dimensional zeroone space, we compute a distance matrix among machines (rows) using the equation 𝑑𝑖𝑘 =
∑𝑛𝑗=1|𝑎𝑖𝑘 − 𝑎𝑗𝑘 |. We solve a k-median problem to get the machine groups.
We consider each part (column) of the incidence matrix as a point in m dimensional zeroone space, we compute a distance matrix among parts (columns) using the equation 𝑑𝑗𝑘 =
∑𝑚
𝑖=1|𝑎𝑖𝑗 − 𝑎𝑖𝑘 |. We solve a k-median problem to get the part groups. We allocate the
machine group to the associated part group and find the number of visits of parts to
machines in other cells.
Consider the machine component incidence matrix shown in Table 3.1. The distance matrix
among machines is given in Table 4.13
Table 4.13 – Machine component incidence matrix
1
2
3
4
5
6
7
1
1
1
1
1
2
1
1
1
3
1
1
1
4
1
1
1
5
1
1
1
1
6
1
1
1
1
1
8
1
1
1
Considering machines 1 and 2, We have
d12  1  0  0 1  0  0  1  0  1  0  1 1  0 1  0 1  6 . The distance matrix is
given in Table 4.14
Table 4.14 – Distance matrix
1
2
3
4
1
0
6
1
1
2
6
0
5
5
5
7
3
6
6
4
3
4
5
6
1
1
7
6
5
5
3
4
0
2
8
5
2
0
6
7
8
6
0
3
5
7
3
0
The p-median formulation has 36 variables and 37 constraints. For p = 2, the solution is X11 =
X31 = X41 = X55 = X25 = X65 = 1 with Z = 8. The machine groups are {1, 3, 4} and {5, 2, 6}.
For p = 3, the solution is X11 = X31 = X41 = X22 = X52 = X66 = 1 with Z = 5. The machine groups
are {1, 3, 4}, {2, 5} and {6}.
We solve the part families problem considering each part (column) to be a point in six
dimensional zero-one space. Considering parts 1 and 2, d12 = 1+1+1+1+1+0 = 5. The
distances are d12 = 5, d13 = 4, d14 = 1, d15 = 2, d16 = 2, d17 = 5, d18 = 5, d23 = 1, d24 = 4, d25 = 6,
d26 = 5, d27 = 0, d28 = 0, d34 = 3, d35 = 3, d36 = 6, d37 = 1, d38 =1, , d45 = 3, d46 = 3, d47 = 4, d48 =4,
d56 = 3, d57 = 4, d58 =4, d67 = 5, d68 =5 and d78 = 0. The optimum solution with k = 2 is given by
X11 = X22 = X41 = X51 = X61 = X32 = X72 = X82 with Z = 6. The two part families are {1, 4, 5, 6} and
{2, 3, 7, 8}.
We observe that the machine group {1, 3, 4} goes with the part group {1, 4, 5, 6} and the
machine group {2, 5, 6} goes with the part group {2, 3, 7, 8}. There are four intercell moves
where parts have to visit machines belonging to other cells.
The optimum solution with k = 3 is given by X11 = X22 = X41 = X51 = X61 = X32 = X77 = X87 with Z =
2. The three part families are {1, 4, 5, 6}, {2, 3} and {7, 8}. The machine group {1, 3, 4} goes
with the part group {1, 4, 5, 6}. The machine group {2, 5} goes with the part group {7, 8} and
the machine group {6} goes with the part group {2, 3}. There are eleven intercell moves
where parts have to visit machines belonging to other cells.
We observe that though the objective function value for the p median problem is lesser for
p = 3 compared to p = 2, the number of intercell moves is more. This creates some
difficulties in determining the correct value of p. This also leads us to consider other
formulations to obtain the best set of machine cells and part families. A heuristic way to
assign parts to a given set of machine groups is as follows:
1. Allot a part to the machine group where it visits maximum number of machines.
2. In case of a tie choose the smallest sized cell.
3. Further ties are broken arbitrarily.
If we apply this rule, we get the same solution for k=2. We need not solve the k-median
problem for parts. For p = 3, the machine groups are {1, 3, 4}, {2, 5} and {6}. The
corresponding part families are {1, 4, 5, 6}, {2, 7, 8} and {3} with 8 intercell moves.
Application 4.2 (Upgrading medical facilities)
There are five villages and five primary health care centers (PHC)that take care of maternity
related health issues. There is a full-fledged hospital that takes care of emergency
operations. There are three types of demand for each village. These are (100, 20, 10), (80,
40, 15), (200, 60, 25), (180, 50, 20) and (240, 70, 30). The first village has 100 patients who
should have regular check up out of whom 20 have planned C-section operation. Ten would
have further complications expected to result in emergency surgeries.
At present, the five health care centers take care of all the regular check-ups while the
second and third demand types are handled by the hospital. Each primary health care
center has a capacity of 200 while the hospital has a capacity to handle 200 planned Csection surgeries which includes 80 emergencies.
Since the capacity in the hospital is not enough, it is decided to upgrade two of the PHCs to
UPHC that can handle the planned C-section operations up to 100. This is done by a
combination of shifting some of the resources from the hospital and more importantly
creating new facilities in the UPHCs. The cost of upgrading the 5 PHCs is 1000, 1200, 900,
800 and 1100. The upgraded PHCs will now have enough resources to take care of the
normal deliveries that will be referred to the hospital only in case of emergency (10% of the
patients). There is a budget restriction of 2000. Since some resources are moved from the
hospital, the capacity to handle C-section operations reduces to 100. The emergency
handling increases to 200 due to reallocation of some existing resources.
The distance matrix among the existing PHCs is given in Table 4.15
Table 4.15 – Distance matrix
1
9
10 12 9
12 2
13 11 7
14 11 2
15 10
16 15 14 1
9
11 8
18 16 2
The distance between the villages and the existing PHCs is 10, 12, 8, 9 and 10. The distance
between the hospital and the PHC/UPHC is 15, 12, 18, 20 and 16.
Let i = 1 to 5 represent the villages and j=1 to 5 represent the PHCs. Let Y j = 1 if PHC j is
upgraded. Let Xij be the number of patients (check up) allotted from village i to PHC/UPHC j.
Let Rij be the number of normal delivery patients from village i allotted to UPHC j. Let Z ij be
number of planned C-section patients from village i to UPHC j. Let Pj be the number of Csection patients referred to hospital. We assume that 10% of the normal delivery patients
and 40% of C-section patients have emergency operations that are carried out in the
hospital. Let fj be the fixed cost to upgrade PHC j and B is the budget restriction. Let Di1 and
Di2 represent the demand for normal delivery and planned C-section in village i.
The dij be the distance between village i and PHC/UPHC j. Let bi be the distance between
hospital and village i and let aj be the distance between PHC/UPHC j and the hospital.
The objective is to Minimize ∑𝑖 ∑𝑗 𝑑𝑖𝑗 (𝑋𝑖𝑗 + 𝑍𝑖𝑗 ) + ∑𝑖 𝑏𝑖 𝑃𝑖 + ∑𝑖 ∑𝑗 0.4𝑎𝑗 (𝑋𝑖𝑗 + 𝑍𝑖𝑗 )
Subject to
∑ 𝑌𝑗 = 2
𝑗
∑ 𝑓𝑗 𝑌𝑗 ≤ 𝐵
𝑗
∑ 𝑋𝑖𝑗 ≤ 600
𝑖
∑ 𝑍𝑖𝑗 ≤ 100𝑌𝑗
𝑖
∑ 𝑃𝑖 ≤ 100
𝑖
∑ 𝑋𝑖𝑗 = 𝐷𝑖1
𝑗
∑ 𝑍𝑖𝑗 + 𝑃𝑖 = 𝐷𝑖2
𝑗
The formulation considering the data would be to Minimize 7X11 + 13.8X12 + 17.2X13 + 20X14
+ 15.4X15 + 18X21 + 6.8X22 + 20.2X23 + 19X24 + 13.4X25 + 20X31 + 15.8X32 + 9.2X33 + 23X34 +
16.4X35 + 22X41 +19.8X42 + 21.2X43 + 9X44 + 15.4X45 + 17X51 + 12.8X52 + 25.2X53 + 246X54 +
8.4X55 + 7Z11 + 13.8Z12 + 17.2Z13 + 20Z14 + 15.4Z15 + 18Z21 + 6.8Z22 + 20.2Z23 + 19Z24 + 13.4Z25
+ 20Z31 + 15.8Z32 + 9.2Z33 + 23Z34 + 16.4Z35 + 22Z41 + 19.8Z42+ 21.2Z43 + 9Z44 + 15.4Z45 + 17Z51
+ 12.8Z52 + 25.2Z53 + 24Z54 + 8.4Z55 + 10P1 + 12P2 + 8P3 + 9P4 + 10P5
subject to
Y1 + Y2 + Y3 + Y4 + Y5 = 2
10Y1 + 12Y2 + 9Y3 + 8Y4 + 11Y5 ≤ 20
P1 + P2 + P3 + P4 + P5 ≤ 100
X11 + X21 + X31 + X41 + X51 - 400Y1 ≤ 0
X12 + X22 + X32 + X42 + X52 - 400Y2 ≤ 0
X13 + X23 + X33 + X43 + X53 - 400Y3 ≤ 0
X14 + X24 + X34 + X44 + X54 - 400Y4 ≤ 0
X15 + X25 + X35 + X45 + X55 - 400Y5 ≤ 0
X11 + X12 + X13 + X14 + X15 = 100
X21 + X22 + X23 + X24 + X25 = 80
X31 + X32 + X33 + X34 + X35 = 200
X41 + X42 + X43 + X44 + X45 = 180
X51 + X52 + X53 + X54 + X55 = 240
Z11 + Z21 + Z31 + Z41 + Z51 - 100Y1 ≤ 0
Z12 + Z22 + Z32 + Z42 + Z52 - 100Y2 ≤ 0
Z13 + Z23 + Z33 + Z43 + Z53 - 100Y3 ≤ 0
Z14 + Z24 + Z34 + Z44 + Z54 - 100Y4 ≤ 0
Z15 + Z25 + Z35 + Z45 + Z55 - 100Y5 ≤ 0
Z11 + Z12 + Z13 + Z14 + Z15 + P1 = 20
Z21 + Z22 + Z23 + Z24 + Z25 + P2 = 40
Z31 + Z32 + Z33 + Z34 + Z35 + P3 = 60
Z41 + Z42 + Z43 + Z44 + Z45 + P4 = 50
Z51 + Z52 + Z53 + Z54 + Z55 + P5 = 70
Xij, Zij, Pj ≥ 0, Yj = 0,1.
The optimum solution to the MILP is given by Y3 = Y5 = 1; P1 = 20, P2 = 30, P4 = 50, X13 = 100,
X25 = 80, X33 = 200, X43 = 100, X45 = 80, x55 = 240, Z25 = 10, Z33 = 60, Z55 = 70 with objective
function value of 12284.
Facilities 3 and 5 are upgraded and the entire budget is consumed. The allocation to the
upgraded PHCs for normal deliveries and planned C-sections are given through the Xij and Zij
values. 40% of the Zij and Xij come to the hospital for emergency surgeries.
Case 4.1 – TAKEHOME Industries
TAKEHOME Industries is in the business of making and distributing notebooks. They make
two types of notebooks and wish to set up factories out of three potential sites. The fixed
cost of setting up a factory in the three sites is 2000, 5000 and 2500 respectively. The
capacities are 450, 500 and 400 each month. The factory in site 1 if set up can produce only
type 1 notebook while the factory in 3 can produce only type 2 notebook. The factory in site
2 can produce both notebooks and the capacity is a total of 500 notebooks including both
the types each month. The fixed cost of setting up the factories is 2000, 5000 and 2500
respectively.
The notebooks produced in the factories reach the customer through warehouses. Again
the company is considering three sites to locate warehouses. The fixed costs of setting up
the warehouses are 1800, 1600 and 900 respectively and the capacities are 800 each for the
warehouses. The unit cost of transportation between the potential factory sites and
potential warehouse sites as well as between the warehouse sites and the customers are
given in Table 4.16
Table 4.16 – Cost of transportation
F-W
Cost
F-W
Cost
1-1
4
2-3
4
1-2
8
3-1
6
1-3
6
3-2
8
2-1
5
3-3
3
2-2
7
W-C
1-1
1-2
1-3
2-1
2-2
Cost
10
12
14
12
9
W-C
2-3
2-4
3-2
3-3
3-4
Cost
10
13
16
14
13
The demand for the two items at the four customers for four months is given in Table 4.17
Table 4.17 – Demand data
Customer M1P1 M1P2
1
100
80
2
180
120
3
60
50
4
100
120
M2P1
60
140
40
120
M2P2
90
100
70
140
M3P1
70
90
30
110
M3P2
50
120
80
140
M4P1
80
80
60
150
M4P2
60
150
60
100
The data is shown in Figure 4.4
100, 60,70,80
2000, 450
1
80
10
4
1
100, 60,70,80
1
12
6
5000, 500
8
5
2
14
800
7
2
13
120, 100,120,150
10
3
8
3
3
60, 40,30,60
50, 70,80,60
16
14
3
2500, 400
180, 140,90,80
9
4
6
12
2
800
13
100, 120,110,150
4
120, 140,140,100
Figure 4.4 – Data for TAKEHOME Industries
Solution
The problem at hand is a location, allocation problem where we have to locate facilities
(factories and warehouses) and allocate the production quantities to the customers through
the warehouses to meet the demand for four months.
We define Fi = 1 if a factory is opened at site i and Zj = 1 if a warehouse is opened at site j.
We define Xijlt as the quantity of item l sent during month t between factory i and
warehouse j. We define Yjklt as the quantity of item l sent during month t between
warehouse j and customer k.
The known values are as follows:
fi = Fixed cost of setting up a factory at site i
zj = Foxed cost of setting up a warehouse at site j
Ail = production capacity of factory (at site i) for product l.
Bj = storage capacity of warehouse j.
Dklt = Demand for item l at customer k for period t.
The objective function is to minimize the location cost and allocation cost.
Minimize ∑𝑖 𝑓𝑖 𝐹𝑖 + ∑𝑗 𝑧𝑗 𝑍𝑗 + ∑𝑖 ∑𝑗 𝐶𝑖𝑗 ∑𝑙 ∑𝑡 𝑋𝑖𝑗𝑙𝑡 + ∑𝑗 ∑𝑘 𝐶𝑗𝑘 ∑𝑙 ∑𝑡 𝑌𝑗𝑘𝑙𝑡
The constraints are
Factory i can produce a maximum of Ail for item l in month t. This is given by
∑ 𝑋𝑖𝑗𝑙𝑡 ≤ 𝐴𝑖𝑙
∀𝑖,𝑙,𝑡
𝑗
We have 12 constraints considering three facilities and four time periods. We include the
details that factory 1 can produce item 1 only and factory 3 can produce item 2 only. Factory
2 produces both the items.
We show one of the constraints for each i.
X1111+X1211+X1311 ≤ 400Y1
X2112+X2212+X2312+X2122+X2222+X2322 ≤ 400Y2
X3124+X3224+X3324 ≤ 400Y3
The first constraint takes care of the condition that factory 1 does not produce item 2. The
corresponding variables are not defined. Similarly factory 3 does not produce item 1 and the
corresponding variables are not defined. Factory 2 produces both the items and therefore
has six terms. The constraints also ensure that items can be transported (and produced)
only when the facility is located.
The next sets of constraints are the intermediate constraints that limit the storage in the
chosen warehouse. Each warehouse has a capacity and the amount of material coming into
the warehouse should not exceed the capacity. There are three warehouses and four
months and we have twelve constraints for each warehouse for each month. A sample
constraint for the second warehouse and third month is given below:
X1213+X2213+X2223+X3223 ≤ 800Z2. Factory 1 can send only item 1 while factory 3 can send only
item 3.
The next set of constraints relate the total number of items entering and leaving each
warehouse in each month. There are 3 warehouses, 2 items and 4 months resulting in 24
constraints. A sample constraint for warehouse 3, item 2 and month 4 is given below:
X2324+X2324-Y3224-Y3324-Y3424 ≥ 0. Item 2 is delivered only from factories 2 and 3.
The next set of constraints is the demand constraints. The total number of items reaching a
customer should meet the demand. We have 4 customers, 2 items and 4 months resulting in
32 constraints. A sample constraint for customer 2, item 1 and month 3 is given below:
Y1213+Y2213+Y3213 ≥ 90
The formulation has 48 Xijlt variables and 80 Yjklt variables. There are three Yi and three Zj
variables resulting in 134 variables out of which six are binary. There are 80 constraints. The
optimum solution is given by F1 = F2 = Z1 = Z3 = 1 with total cost = 59250. The allocations are
X1111 = 340, X1112 = 240, X1113 = 190, X1114 = 220, X1312 = 20, X1314 = 20, X2121 = 200, X2122 = 190,
X2123 = 170, X2124 = 210, X2311 = 100, X2312 = 100, X2313 = 110, X2314 = 130, X2321 = 170, X2322 =
210, X2323 = 220, X2324 = 160, Y1111 = 100, Y1112 = 60, Y1113 = 70, Y1114 = 80, Y1121 = 80, Y1122 = 90,
Y1123 = 50, Y1124 = 60, Y1211 = 180, Y1212 = 140, Y1213 = 90, Y1214 = 80, Y1221 = 120, Y1222 = 100,
Y1223 = 120, Y1224 = 150, Y1311 = 60, Y1312 = 40, Y1313 = 30, Y1314 = 60, Y3321 = 50, Y3322 = 70, Y3323
= 80, Y3324 = 60, Y3411 = 100, Y3412 = 120, Y3413 = 110, Y3414 = 150, Y3421 = 120, Y3422 = 140, Y3423
= 140, y3424 = 100.
The solution is given in Figure 4.5
2000,
1
800
1
1
2
3
2
3
2500, 400
800
4
Figure 4.5 – Solution for TAKEHOME Industries
The numbers shown under the arrow are for item 2 while the numbers shown above are for
Item 1. The four numbers represent the quantity transported in the four periods. It is
observed that two factories out of the possible three are created and two warehouses out
of the possible three are created.
We consider a different demand scenario. This is given in Table 4.18
Table 4.18 – Demand data
Customer M1P1 M1P2
1
50
80
2
180
120
3
60
50
4
100
120
M2P1
60
140
40
120
M2P2
90
100
70
140
M3P1
70
90
80
110
M3P2
50
120
80
140
M4P1
230
180
60
200
M4P2
160
250
110
150
The total demand for month 4 is 670 + 670 = 1340. The total production capacity of the two
chosen factories in month 4 is 450 + 500 = 950. This will necessitate creation of a factory at
site 3. We also observe that the total demand over 4 months is 3600 and the total
production capacity for four months is 950 x 4 = 3800. It is also possible that we can produce
more during the earlier months and store the excess for use in the later months. We wish to
find out whether it is economical to create three factories or use the excess inventory.
We now define Ijlt as the inventory of item l stored in warehouse k at the end of period t.
When we did not consider inventory storage, the corresponding constraint set was
∑𝑖 𝑋𝑖𝑗𝑙𝑡 = ∑𝑘 𝑌𝑗𝑘𝑙𝑡 . This set is modified to 𝐼𝑗𝑙𝑡−1 + ∑𝑖 𝑋𝑖𝑗𝑙𝑡 = ∑𝑘 𝑌𝑗𝑘𝑙𝑡 + 𝐼𝑗𝑙𝑡 . Since we have
3 warehouses, 2 items and 4 periods, 24 constraints get modified. The storage constraints in
the warehouses also get modified. ∑𝑙 𝐼𝑗𝑙𝑡−1 + ∑𝑖 ∑𝑙 𝑋𝑖𝑗𝑙𝑡 ≤ 𝐵𝑗 . There is a cost of storing
the excess inventory. We keep this as Re 1 per unit/period. The objective function has
additional terms, representing the inventory cost, given by 𝑢 ∑𝑗 ∑𝑙 ∑𝑡 𝐼𝑗𝑙𝑡 where u is the
cost of storing one unit of the item.
We added 3 x 2 x 3 = 18 variables to the formulation and 36 constraints have been modified.
The optimum solution is given by F1 = F2 = Z1 = Z3 = 1 with total cost = 70190. The allocations
are X1111 = 290, X1112 = 360, X1113 = 140, X1114 = 450, X1312 = 50, X1313 = 310, X2121 = 230, X2122 =
220, X2123 = 170, X2124 = 350, X2311 = 100, X2312 = 70, X2321 = 170, X2322 = 210, X2323 = 330, X2324
= 150, Y1111 = 50, Y1112 = 60, Y1113 = 70, Y1114 = 230, Y1121 = 80, Y1122 = 90, Y1123 = 50, Y1124 =
160, Y1211 = 180, Y1212 = 140, Y1213 = 90, Y1214 = 180, Y1221 = 120, Y1222 = 100, Y1223 = 120, Y1224 =
250, Y1311 = 60, Y1312 = 40, Y1313 = 80, Y1314 = 60, Y3321 = 50, Y3322 = 70, Y3323 = 80, Y3324 = 110,
Y3411 = 100, Y3412 = 120, Y3413 = 110, Y3414 = 200, Y3421 = 120, Y3422 = 140, Y3423 = 140, y3424 =
150.I112 = 120, I113 = 20, I121 = 30, I122 = 60, I123 = 60, I313 = 200, I323 = 110.
The solution is given in Figure 4.6.
2000,
I112 = 120, I113 = 20,
I121 = 30, I122 = 60, I123 = 60
1
1
1
2
3
2
3
2500, 400
I313 = 200,
I323 = 110
Figure 4.6 – Solution to TAKEHOME Industries
Some more variations can be considered:
1. Assume that whenever facility 1 produces, the quantity is 450. Facility produces
exactly 250 of each and facility 3 produces 400 of item 2. Incorporate this into the
formulation? (Note that this may result in production more than demand and ending
inventory at the end of period 4).
2. Assume that trucks carry the items from the factories to warehouses and from
warehouses to customers. We have trucks of two capacities 80 and 60. The costs are
1000 and 900. Replace the unit transportation costs with the truck cost and
formulate an optimization problem?
4