ISYE 3104 Summer 2003 Homework 6 Solution
Chapter 9 Layout Strategy
PROBLEM SET
Question 1: Your third-party warehouse has space available for 10,000 pallets and you have 20 forklift operators per 8-hour day for 250 working days a year. If the average trip from receiving to storage to shipping is 10 minutes, how many inventory turns a year could you support for a full warehouse?
Assuming that one forklift operator can carry only one pallet at a time.
Maximum pallets flow per hour = (20 operators)*(60 min/hr) / (10 min/trip)
= 120 pallets per hour
Maximum pallets flow per year = (120 pallets/hr)*(8 hr/day)*(250 day/yr)
= 240,000 pallets per year
Thus,
Maximum inventory turns = (240,000 pallets/yr) / (10,000 pallets)
= 240 times per year
Question 2: Your third-party warehouse is bidding for a contract to store widgets as they are manufactured. Widgets are perishable and should be turned on average 6 times per year. The manufacturer produces at an average rate of 32 pallets per day. How many pallet positions should you devote to widgets to ensure that widgets turn as required?
Assuming that there are 250 working days per year.
Maximum pallets flow per year = (32 pallets/day)*(250 day/yr)
= 8,000 pallets per year
Thus,
Maximum pallets position = (8,000 pallets/yr) / (6 times/yr)
= 1,334 pallets.
2. Suppose that you have 10 cubic feet available in flow rack, which is restocked from pallet rack, and you have just added three SKU’s with projected activity as follows:
a. Suppose that you have decided to put all these three SKU’s in the flow rack. How much space should be allocated to each SKU?
See table below.
b. How many restock trips per month for each SKU will result from the space allocation computed in part a?
First, compute the monthly flows of each SKU using
ƒi = [(units/month) / (units/case)] x [ƒt3 / case] = ƒt3 / month :
ƒA = (2000 / 200) x 2 = 20
ƒB = (1200 / 15) x 7 = 560
ƒC = (2000 / 10) x 1 = 200
We have = 4.47, = 23.66, and = 14.14.
SKU i should occupy ui = *V and the number of restocks is
The results obtained using the above formulas are tabulated below:
SKU / Volume stored (ui) / Restocks / month (ƒi / ui)A
B
C / 1.06
5.60
3.34 / 20/1.06 = 18.87
560/5.60 = 100
200/3.34 = 59.88
Total / 10.0 / 178.75
c. What would be the average restock trips per month for each SKU if the available space was shared equally among all the three SKU’s?
Equal space allocations: ui = V / 3 and the number of restocks =
SKU / Volume stored (ui) / Restocks / month (ƒi / ui)A
B
C / 3.33
3.33
3.33 / 20/3.33 = 6
560/3.33 = 168
200/3.33 = 60
Total / 10.0 / 234
d. What is the total net benefit achieved under (i) the optimal space allocation implied by part (a), and (ii) the space allocation suggested in part c? Assume that the savings associated with every pick from the fast-pick area amount to $1.5, while each replenishment trip costs about $10 in labor.
For optimal space allocation, the net benefit achieved is the total pick savings – the replenishment costs = $1.5 * 1050 picks/month - $10 * 179 restock/month = -$215. For equal space allocation, the benefit is $1.5 * 1050 picks/month - $10 * 234 restock/month = -$765.
Note: Since the net benefits are negative for both cases, this implies that the replenishment costs are greater than the pick saving and thus we should not put all SKUs in the fast pick area.
e. If you can select the SKU’s to enter to the fast pick area, which ones would you choose in order to maximize the resulting net benefit? How much space should be allocated to the each of the selected SKU’s?
In order to determine which SKU’s are to be included in the fast pick area, first we calculate the viscosity pi /Ö fi (where pi = # of picks per month) of each SKU, and rank them in decreasing viscosity:
PA = 500 / Ö20 = 111.80 => Ranking (1)
PB = 300 / Ö560 = 12.68 => Ranking (3)
PC = 250 / Ö200 = 17.68 => Ranking (2)
Then, we evaluate the resulting net benefit, using the same procedure as above, in case that the fast pick area contains only the most viscous SKU, which is SKU A.
SKU / Volume stored (ui) / Restocks / month (ƒi / ui)A / 10.0 / 20/10 = 2
Total / 10.0 / 2
Net benefit is $1.5*500 - $10*2 = $730.
Next, we compute the optimal allocation of the fast pick area and the resulting net benefit, if the fast pick area contains only the two most viscous SKU’s, which are SKU A and C.
SKU / Volume stored (ui) / Restocks / month (ƒi / ui)A
C / 2.40
7.60 / 20/2.40 = 8.33
200/7.60 = 26.32
Total / 10.0 / 31.95
Net benefit is $1.5*750 - $10*31.95 = $805.5.
Since the obtained net benefit is greater than that obtained from choosing only one SKU and from choosing all SKUs, we conclude we should bring both SKU A and C in to the fast pick area. It is obvious that SKU B is not a good choice not be in the fast pick area since it is bulky with lesser pick and thus low viscosity.
Note: As we explained in class, the above procedure does not guarantee optimality, but it provides a very good, near-optimal solution. For this small example, one can search explicitly for the optimal solution, by generating all possible combinations of SKUs to be included in the fast-pick area and compute the associated net benefits; such an approach is known as exhaustive search. However, something like this would be impossible (the computational cost would be prohibitive) in a more realistic situation with tens, hundreds or even thousands of SKU’s; on the other hand, the viscosity ranking procedure is a viable alternative for these larger problems, since for N SKU’s you have to evaluate at most N+1 alternatives (i.e., the most viscous SKU, the two most viscous SKU’s, …., the N-1 most viscous SKU’s, all SKU’s, plus the (baseline) case of no SKU at all.)
3. The warehouse depicted below will be used for the storage of six product families. The warehouse consists of storage bays of size 20ft x 20ft. Dock 1 has been designated as the receiving dock, while dock 2 is used as the shipping dock. The area requirement and monthly load rate for each product family are as follows:
Assuming that a dedicated storage policy will be adopted, assign the warehouse bays to the various product families.
Hint: In your calculations of the travel distance, assume that traveling between a dock and a bay takes place along the grid depicted above, and that a trip to/from a bay ends/starts at the center of the bay. Bays can be accessed from all four sides.
Solution:
We have 6 product families and 40 storage bays that can be used for storing product families. For further reference, we number each location as follows:
1 / 2 / 3 / 4 / 5 / 6 / 7 / 89 / 10 / 11 / 12 / 13 / 14 / 15 / 16
17 / 18 / 19 / 20 / 21 / 22 / 23 / 24
25 / 26 / 27 / 28 / 29 / 30 / 31 / 32
33 / 34 / 35 / 36 / 37 / 38 / 39 / 40
Then the problem can be solved as follows:
Assumptions:
· We can consider single-command type of operation, i.e., each trip is either a storage trip or a retrieval trip (there is no mixing of storage and retrieval activity).
· Distance is computed to/from the center of each location.
· The activity associated with each location is distributed evenly between receiving and shipping (otherwise, we would have accumulation of material in the warehouse or “systematic” shortages).
· Distance is measured according to the rectilinear metric.
Parameters:
· I = the index set of SKU’s: I = {1,2, …, 6}
· J = the index set of locations: J = {1, 2, …, 40}
· THi = number of units of SKU i handled per unit of time
· Ni = number of storage locations allocated to SKU i
· di = expected travel distance per unit load stored in location j
Decision variables:
· Xij = 1 if location j is allocated to SKU i; 0 otherwise for all i Î I and j Î J.
From the problem statement, we have the following data with respect to each product family:
ProductFamily / Area / Load Rate(THi) / Ni / THi/Ni / Rank of THi/Ni
1 / 2,400 / 500 / 6 / 83.33 / 3
2 / 3,200 / 250 / 8 / 31.25 / 6
3 / 2,000 / 650 / 5 / 130.00 / 2
4 / 2,800 / 450 / 7 / 64.29 / 4
5 / 4,000 / 375 / 10 / 37.50 / 5
6 / 1,600 / 750 / 4 / 187.50 / 1
where Ni = storage locations required by SKU I = Area to be occupied by SKU I / (20 ft * 20 ft) and THi = throughput rate of SKU I.
We can get the optimal solution of the above problem using the following matching procedure (presented in class):
· Rank all the available storage locations in increasing distance dj
· Rank all SKU’s in decreasing turns of THi/Ni (this quantity expresses the per location monthly traveling activity generated by each SKU)
· Move down the two lists, assigning the next most highly ranked SKU i to the next Ni locations
Now we can compute the expected distance dj for each location j as follows:
j / 1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / 11 / 12 / 13 / 14 / 15 / 16 / 17 / 18 / 19 / 20Lin,j / 40 / 60 / 80 / 100 / 120 / 140 / 160 / 180 / 20 / 40 / 60 / 80 / 100 / 120 / 140 / 160 / 20 / 40 / 60 / 80
Lj,out / 180 / 160 / 140 / 120 / 100 / 100 / 120 / 140 / 160 / 140 / 120 / 100 / 80 / 80 / 100 / 120 / 140 / 120 / 100 / 80
dj / 220 / 220 / 220 / 220 / 220 / 240 / 280 / 320 / 180 / 180 / 180 / 180 / 180 / 200 / 240 / 280 / 160 / 160 / 160 / 160
Rank / 4 / 4 / 4 / 4 / 4 / 5 / 7 / 8 / 2 / 2 / 2 / 2 / 2 / 3 / 5 / 7 / 1 / 1 / 1 / 1
j / 21 / 22 / 23 / 24 / 25 / 26 / 27 / 28 / 29 / 30 / 31 / 32 / 33 / 34 / 35 / 36 / 37 / 38 / 39 / 40
Lin,j / 100 / 120 / 140 / 160 / 40 / 60 / 80 / 100 / 120 / 140 / 160 / 180 / 60 / 80 / 100 / 120 / 140 / 160 / 180 / 200
Lj,out / 60 / 60 / 80 / 100 / 120 / 100 / 80 / 60 / 40 / 40 / 60 / 80 / 100 / 80 / 60 / 40 / 20 / 20 / 40 / 60
dj / 160 / 180 / 220 / 260 / 160 / 160 / 160 / 160 / 160 / 180 / 220 / 260 / 160 / 160 / 160 / 160 / 160 / 180 / 220 / 260
Rank / 1 / 2 / 4 / 6 / 1 / 1 / 1 / 1 / 1 / 2 / 4 / 6 / 1 / 1 / 1 / 1 / 1 / 2 / 4 / 6
where, Lin,j and Lj,out represent the distance from the location j to the receiving and shipping dock respectively and dj = Lin,j + Lj,out. Using the actual layout of the 40 locations, we get the following ranking profile for them:
4 / 4 / 4 / 4 / 4 / 5 / 7 / 82 / 2 / 2 / 2 / 2 / 3 / 5 / 7
1 / 1 / 1 / 1 / 1 / 2 / 4 / 6
1 / 1 / 1 / 1 / 1 / 2 / 4 / 6
1 / 1 / 1 / 1 / 1 / 2 / 4 / 6
The rank of each SKU has been already computed in the first table above. So, following the matching procedure outlined above, an optimal allocation is as follows: