1. GEOP 4355 Distribution Problems
Outline
Pooling effect problem
Distrbution problem
Sources/references used in the preparation of this presentation are listed in the Introduction presentation

2. Distribution models Variety of problems in distribution
Models have been developed to support strategic, tactical, and operational distribution decision making.
Pooling effect problem: tradeoff between inventory and transportation
Distribution/ transportation problem: How to move material effectively given customer requirements and source limitations.

3. Pooling effect problem
month
dc1
dc2
dc3
dc4 1
141
300
373
385 2
207
280
314
112 3
199
194
354
259 4
273
201
206
360 5
166
135
276
125 6
293
339 7
109
303
133
118 8
115
250
291 9
157
333
223
378 10
102
244
139 11
235
352
114
343 12
278
265
327
288
Total
2,354
3,076
3,048
3,133
Monthly demand for product X
Serves by 4 DCs

4. Pooling effect problem
month
dc1
dc2
dc3
dc4
total 1
141
300
373
385
1199 2
207
280
314
112
913 3
199
194
354
259
1006 4
273
201
206
360
1040 5
166
135
276
125
702 6
293
339
1121 7
109
303
133
118
663 8
115
250
291
850 9
157
333
223
378
1091 10
102
244
139
824 11
235
352
114
343
1044 12
278
265
327
288
1158
Total
2,354
3,076
3,048
3,133
11,611
mean
196.2
256.3
254.0
261.1
967.6
std dev
64.4
80.4
91.2
108.8
176.3
Column to the left shows the total demand per month
Rows at the bottom show
average demand per month (d)
standard deviation of demand (sd)

5. Pooling effect problem
month
dc1
dc2
dc3
dc4
total 1
141
300
373
385
1199 2
207
280
314
112
913 3
199
194
354
259
1006 4
273
201
206
360
1040 5
166
135
276
125
702 6
293
339
1121 7
109
303
133
118
663 8
115
250
291
850 9
157
333
223
378
1091 10
102
244
139
824 11
235
352
114
343
1044 12
278
265
327
288
1158
Total
2,354
3,076
3,048
3,133
11,611
mean
196.2
256.3
254.0
261.1
967.6
std dev
64.4
80.4
91.2
108.8
176.3
Assuming monthly replenishments to DC
Replenishment lead time to each DC is
L = 0.5 months,
sL = 0
sddlt = 𝐿 × 𝑠 𝑑 2 + 𝑑 2 × 𝑠 𝐿 2
sddlt 46 57 65 77
125

6. Pooling effect problem
month
dc1
dc2
dc3
dc4
total 1
141
300
373
385
1199 2
207
280
314
112
913 3
199
194
354
259
1006 4
273
201
206
360
1040 5
166
135
276
125
702 6
293
339
1121 7
109
303
133
118
663 8
115
250
291
850 9
157
333
223
378
1091 10
102
244
139
824 11
235
352
114
343
1044 12
278
265
327
288
1158
Total
2,354
3,076
3,048
3,133
11,611
mean
196.2
256.3
254.0
261.1
967.6
std dev
64.4
80.4
91.2
108.8
176.3
Desired TSL = 99% Z = 2.33 SS = SS  sddlt Aggregate Safety stock needed with 4 DCs = ( ) SS = 567 If all orders were to be served by a single DC. SS = 290
sddlt 46 57 65 77
125 SS
106
132
150
179
290

7. Pooling effect problem
Assume each unit costs $1,000. h = 20%
ASC = SS  h  cu = $113,400 with 4DCs,
ASC = $58,000 with one DC
dtc/unit
units/yr
transp c
dc1 $
2,354
$ ,593
dc2 $
3,076
$ ,688
dc3 $
3,048
$ ,802
dc4 $
3,133
$ ,652
Delivery Annual Transportation costs = $49,735
Delivery Annual Transportation costs = $80,116
one DC $
11,611
$ 80,116
TRC (4DCs) = $163,135
TRC(oneDC) = $138,116

8. Distribution problem
How many units to ship from each warehouse to each of the customers based on availability, needs and transportation costs.
$5/u
Store 1
Need = 60
How many units
would you ship
in each of the
four routes?
DC1
Available = 80
$3/u
$11/u
Store 2
Need = 40
$6/u
DC2
Available = 25

9. Distribution problem Could you optimally solve a larger problem?
Store 1
Need = 190
DC1
Available = 356
Could you optimally solve a larger problem?
Store 2
Need = 98
DC2
Available = 408
Store 3
Need = 141
DC3
Available = 230
Store 4
Need = 330
Total Available = 994
Total Needs = 759
Store1
Store 2
Store 3
Store 4
DC1
7.2
8.1
6.4
8.4
DC2
6.0
5.0
7.3
9.4
DC3
5.4
3.6
6.3
3.1

10. Distribution problem
Mathematical modelling: use of equations to represent/model/solve a complex business problem.
Three components
Decision variables: what must be determined
Objective functions: the key metric of the problem
Constraints: the system characteristics representing its limitations and operations

11. Distribution problem Decision variables
The number of units to ship from each warehouse to each customer
Let U indicate units and Us,d = Units shipped from source s to destination d
Thus UW1,C2 means the number of units shipped from warehouse 1 to customer 2.
There are 12 decision variables (3 × 4)
We use sets to represent all possible warehouses (S) and customers (D)
S = {1…3}, D = {1…4}

12. Distribution problem Objective function
Let T represent the transportation cost per unit and Ts,d = transportation cost per unit shipped from warehouse s to customer d.
This assumes a fixed rate per unit, rate is independent of the total number of units shipped.
Therefore TW1,C1 = $7.1, TW1,C2 = $8.1, … C1 C2 C3 C4 W1
7.2
8.1
6.4
8.4 W2
6.0
5.0
7.3
9.4 W3
5.4
3.6
6.3
3.1

13. Distribution problem Objective Function
Transportation costs for each route equals
U × T
If UW1,C1 = 100 and TW1,C1 = $7.1 then transportation costs for that route are $710.
Total system transportation costs are then
UW1,C1 × TW1,C1 + UW1,C2 × TW1,C2 + UW1,C3 × TW1,C3 + …. + UW3,C4 × TW3,C4
General form:  sS, dD Us,d × Ts,d
Objective of the model is to minimize total system transportation costs

14. Distribution problem Constraints
Supply side constraints establishes the limits on what can be shipped from each source. It’s a maximum, not all units must be shipped. W1
Available = 356
UW1,C1 + UW1,C2 + UW1,C3 + UW1,C4 ≤ 356
UW2,C1 + UW2,C2 + UW2,C3 + UW2,C4 ≤ 408 W2
Available = 408 W3
Available = 230
UW3,C1 + UW3,C2 + UW3,C3 + UW3,C4 ≤ 230

15. Distribution problem Constraints
Destination side constraints establish the amount that must be received at each destination.
UW1,C1 + UW2,C1 + UW3,C1 = 190
C1, Need = 190
UW1,C2 + UW2,C2 + UW3,C2 = 98
C2, Need = 98
UW1,C3 + UW2,C3 + UW3,C3 = 141
C3, Need = 141
UW1,C4 + UW2,C4 + UW3,C4 = 330
C4, Need = 330
Non negativity constraints establishes that the number of units shipped from any source to any destination must not be a negative number.
Us,d ≥ 0 for all sources and destinations.

16. Distribution problem Complete mathematical formulation
Decision Variables
Us,d = Units shipped from warehouse s to customer d.
s = 1…3 and d =1…4. define sets : S = {1…3}, D = {1…4}
Objective Function
Minimize total transportation costs
Ts,d = transportation cost per unit from warehouse s to customer d.
 sS, dD Ts,d ×Us,d
Constraints
 s, dD Us,d ≤ Units available at warehouse s.  s  S.
 sS,d Us,d = Units needed at customer d.  d  D.
Us,d ≥ 0  s  S,  d  D

17. Distribution problem OF Constraints
= Minimize (7.2Uw1,c Uw2,c1 + ……
Constraints
W1: Uw1,c1 + Uw1,c2 + Uw1,c3 + Uw1,c4 ≤ 356
W2: Uw2,c1 + Uw2,c2 + Uw2,c3 + Uw2,c4 ≤ 408
W3: Uw3,c1 + Uw3,c2 + Uw3,c3 + Uw3,c4 ≤ 230
C1: Uw1,c1 + Uw2,c1 + Uw3,c1 = 190
C2: Uw1,c2 + Uw2,c2 + Uw3,c2 = 98
C3: Uw1,c3 + Uw2,c3 + Uw3,c3 = 141
C4: Uw1,c4 + Uw2,c4 + Uw3,c4 = 330
Us,d ≥ 0 for all s in S and d in D

18. Distribution problem
Mathematical models are implemented/solved using special software applications.
Excel can be used to solve small/medium sized problems.
Finding the optimal solution is the first step.
Second step is to perform sensitivity analysis.
Analysis of how changes to some parameters change the decisions and the objective function.

19. Distribution problem Extensions
Route constraints: Us,d ≤ Cs,d where Cs,d is the maximum that can be shipped in route s,d during the decision timeframe.
Transshipment problem
Transshipment nodes between warehouses and customers with a capacity
Missing requirements problem
Units available from the warehouses cannot satisfy the demand. Let md be the unmet demand at d.
Unmet requirements have a cost per unit per customer.
UW1,C4 + UW2,C4 + UW3,C4 = 330 – mC4