Problem 1: Process Analysis
You manage an assembly line consisting of 6 work stations (referred to as Work Station 1, Work Station 2, ... , Work Station 6) in series, each manned by a single worker (Alex, Barbara, Chuck, Donna, Ed, and Fran). Each unit of product is processed first at Work Station 1, then at Work Station 2, and onward through the six work stations in numerical order. For reasons not completely understood, the amount of time required to complete the work at each work station depends on which worker is assigned to that station. The following table gives the number of minutes each worker requires to complete the tasks associated with each work station.
|
Workers
|
Work Stations
|
|
1
|
2
|
3
|
4
|
5
|
6
|
|
Alex
|
10
|
35
|
41
|
45
|
15
|
23
|
|
Barbara
|
33
|
40
|
24
|
19
|
49
|
13
|
|
Chuck
|
16
|
20
|
39
|
32
|
18
|
48
|
|
Donna
|
47
|
12
|
14
|
26
|
44
|
33
|
|
Ed
|
25
|
43
|
37
|
31
|
27
|
21
|
|
Fran
|
50
|
28
|
22
|
11
|
38
|
42
|
Therefore, if Alex is assigned to Work Station 1, he will require a total of 10 minutes to complete the tasks at that work station for every unit of product. In contrast, he will require 45 minutes if assigned to Work Station 4. Similarly, Donna will require 47 minutes if assigned to Work Station 1, but only 12 minutes if assigned to Work Station 2. Clearly, one worker must be assigned to each work station, and these assignments are considered permanent for the foreseeable future.
Assume that the system is always busy (i.e., there is always demand for this product), and there is no transient state (so, e.g., workers can stop work on a unit of product at the end of a shift and resume work at precisely the same point the next day). Also assume that no inventory is allowed between work stations; instead, workers who complete their tasks faster than the bottleneck work station accumulate idle time.
a. To which work station should each worker be assigned in order to minimize the flow time of the assembly line? What is the associated flow time?
Alex _10____ Barbara 19_ Chuck 18_ Donna 12_ Ed _21 Fran _22
Flow Time (minutes) _81
Problem 2:
You order two products from the same supplier. The annual demand for Product 1 is 10,000 units and the annual demand for Product 2 is 20,000 units. Note that demand for both products is constant throughout the year. The holding cost is the same for both products, $1 per unit per year. However, you incur a fixed cost of $200 each time you order Product 1, and a fixed cost of $100 each time you order Product 2. These fixed order costs are independent of the size of the order.
a. How many units of Product 1 and Product 2 should be ordered at a time in order to minimize total holding + order cost?
Product 1 :
EOQ = Sqrt (2*K*R/H) where
R = Annual demand = 10,000
K = Order cost = $200
H = Holding cost factor = $1
EOQ (P1) = Sqrt (2*K*R/H) = SQRT(2*200*10000/1) = 2,000 Nos
Number of Units of Product 1 to Order 2000 Nos
EOQ (P2) = Sqrt (2*K*R/H) = SQRT(2*100*20000/1) = 2,000 Nos
Number of Units of Product 2 to Order 2000 Nos
b. Suppose that the supplier insists that orders for Product 1 and Product 2 be coordinated so that they can be shipped at the same time (still incurring the fixed cost of $200 and $100, respectively). Given this requirement, how many units of Product 1 and Product 2 (to the nearest integer) should be ordered at a time in order to minimize total holding + order cost? (Hint: the products must be ordered the same number of times per year)
We use Different values of K to calculate TC1 + TC2 = TC as shown below.
We observe that for No of Order = 7, Q1 = 1429 & Q2 = 2857 giving us lowest total cost of $4243.
|
K1
|
$ 200
|
|
K2
|
$ 100
|
|
H
|
$ 1
|
|
Q1
|
10000
|
|
Q2
|
20000
|
|
K
|
Q1
|
Q2
|
TC1
|
TC2
|
TC
|
|
1
|
10,000
|
20,000
|
$ 5,200
|
$ 10,100
|
$ 15,300
|
|
2
|
5,000
|
10,000
|
$ 2,900
|
$ 5,200
|
$ 8,100
|
|
3
|
3,333
|
6,667
|
$ 2,267
|
$ 3,633
|
$ 5,900
|
|
4
|
2,500
|
5,000
|
$ 2,050
|
$ 2,900
|
$ 4,950
|
|
5
|
2,000
|
4,000
|
$ 2,000
|
$ 2,500
|
$ 4,500
|
|
6
|
1,667
|
3,333
|
$ 2,033
|
$ 2,267
|
$ 4,300
|
|
7
|
1,429
|
2,857
|
$ 2,114
|
$ 2,129
|
$ 4,243
|
|
8
|
1,250
|
2,500
|
$ 2,225
|
$ 2,050
|
$ 4,275
|
|
9
|
1,111
|
2,222
|
$ 2,356
|
$ 2,011
|
$ 4,367
|
|
10
|
1,000
|
2,000
|
$ 2,500
|
$ 2,000
|
$ 4,500
|
|
11
|
909
|
1,818
|
$ 2,655
|
$ 2,009
|
$ 4,664
|
|
12
|
833
|
1,667
|
$ 2,817
|
$ 2,033
|
$ 4,850
|
|
13
|
769
|
1,538
|
$ 2,985
|
$ 2,069
|
$ 5,054
|
|
14
|
714
|
1,429
|
$ 3,157
|
$ 2,114
|
$ 5,271
|
|
15
|
667
|
1,333
|
$ 3,333
|
$ 2,167
|
$ 5,500
|
Problem 3:
A professor holds online office hours all day Saturday and Sunday the weekend prior to final exams. He is available from 8AM to 8PM both days to answer student questions. Students who arrive while the professor is busy with another student simply wait until their turn comes up. Students are processed in the order they arrive, and all students tolerate whatever wait is necessary to get their questions answered. The professor notes from past experience that students arrive randomly with questions - the average time between arrivals is 36 minutes and the coefficient of variation of interarrival times is 1. Similarly, the time required to answer student questions is randomly distributed with an average of 24 minutes and a coefficient of variation of 1.
a. On average how long does a student have to wait to get in to see the professor?
Coeff of variation CVa = Std Dev of Interarrival time / Mean of interarrival time
As CVa = 1, It is exponential distribution.
Coeff of variation CVp = Std Dev of Processing time / Mean of Processing time = 1
Utilization U = Activity time/ Interarrival time = 24/36 = 0.67
So 1-U = 1-0.67 = 0.33
So Avge waiting time = Activity time *(U/(1-U) * (CVa^2 + CVp^2)/2
= 24*(0.67/0.33)*(1^2+1^2)/2
= 48.73 Mins
Average waiting time (minutes) 48.73 Mins
b. Suppose the professor would prefer the average waiting time to be no more than 40 minutes. By how much would the average interarrival time have to grow in order to meet this standard?
Increase in Average Interarrival Time (minutes) __________
Avge waiting time = Activity time *(U/(1-U) * (CVa^2 + CVp^2)/2
We also know that CVa & CVp are not affected by arrival or processing time.
So (CVa^2 + CVp^2)/2 = (1^1+1^1)/2 = 1 will remain same
So Avge waiting time = Activity time *(U/(1-U)*1
Or (U/(1-U) = Avge waiting time / Activity time = 40/24 = 1.67
Now Utilization U = Activity time/ Interarrival time = a/p
So 1-U = 1 - a/p = (p-a)/p
So U/(1-U) = (a/p) / [(p-a)/p] = a / (p-a)
So p/(p-a) = 1.67
Solving for p, we get p = 1.67*(p-a)
Or 0.67p = 1.67*a = 1.67*24
So p = 1.67*24/0.67 = 38.4 mins
So If interarrivakl time is 38.4 mins, Avge waiting time will be no more than 40 mins.
So Increase in Interarrivale time = 38.4-36 = 3.4 mins
c. Assume again an average interarrival time of 36 minutes and suppose the professor is considering reducing student waiting time by answering questions faster. How much faster would the professor have to answer questions in order to reduce the average waiting time to 40 minutes?
Decrease in the Average Processing Time (minutes) __________
Avge waiting time = Activity time *(U/(1-U) * (CVa^2 + CVp^2)/2
We also know that CVa & CVp are not affected by arrival or processing time.
So (CVa^2 + CVp^2)/2 = (1^1+1^1)/2 = 1 will remain same
So Avge waiting time = Activity time *(U/(1-U)*1= 40 mins .....(i)
Now Utilization U = Activity time/ Interarrival time = a/p = a/36
So 1-U = 1 - a/p = (p-a)/p
So U/(1-U) = (a/p) / [(p-a)/p] = a / (p-a) = a/(36-a)
So (i) become a*U/(1-U) = 40
Or a*a/(36-a) = 40
Or a^2 = 40*(36-a) = 1440 - 40*a
Or a^2 + 40*a - 1440 = 0
Solving this, we get [-40 +Sqrt(40^2 - 4*1*(-1440))]/(2*a) = 22.90 Mins
Or [-40 - Sqrt(40^2 - 4*1*(-1440))]/(2*a) = -62.90 Mins
As Activity time can't be negative, Activity time a = 22.90 Mins
Current Activity time is 24 mins
So Decreas in processing time = 24-22.90 = 1.10 mins
d. Again assume an average interarrival time of 36 minutes and an average processing time of 24 minutes. What would be the average waiting time if the professor could clone himself (thereby creating a system with two servers and a single queue)?
For no of servers = m, U = (1/Interarrival time) / (m/activity time) = p/(a*m)
So 1-U = 1 - p/(a*m) = (a*m - p)/(a*m)
So U/(1-U) = [p/(a*m)] / [(a*m - p)/(a*m)] = p/(a*m - p)
Avge waiting time = (Activity time/m) *(U/(1-U) * (CVa^2 + CVp^2)/2
= (24/2)*(24/(2*36-24)*1
= 6 mins
Average Waiting Time (minutes) 6
Problem 4:
Susan sells snow cones from a pushcart. Snow cones come in two flavors, Redeye Raspberry (RR) and Boozy Banana (BB). Susan's cost for each cone is the same, $0.50/unit, and she charges $2.00/unit for cones of either flavor. From experience, Susan knows that the daily demand for RR cones is normally distributed with mean 100 and standard deviation 30, while demand for BB cones is normally distributed with mean 120 and standard deviation 60.
Assume that the demand for RR cones is independent of the demand for BB cones (and vice versa), and that demand in excess of supply is lost (no substitutions). Leftover snow cones are discarded at the end of the day.
a. How many RR and BB cones should be stocked at the beginning of the day to maximize Susan's expected profit? What is the expected profit of this policy?
If Susan order too few snow cones of RR, RR shortage cost Cu = Sale price - Purchase price = 2 - 0.50 = $1.50
If Susan Orders too many RR, RR overage cost Co = 0.50
So Critical ratio F(Q) = ?Cu/(Co+Cu) = 1.50 / (1.50+0.50) = 0.75
We find from Normal table that for Probability = 0.75, z = NORM.S.INV(0.75) = 0.6745
So Optimal Qty of RR cones Susan should stock Q = Mean + z*Std Dev = 100 + 0.6745*30 = 121 Cones
Similarly Optimal Qty of BB cones Susan should stock Q = Mean + z*Std Dev = 120 + 0.6745*60 = 161 Cones
Number of RR Cones 121 Cones
Number of BB Cones 161 cones
Expected Profit = Cu*(RR+BB) = 1.50*(121+161) = $423
b. If Susan can stock no more than 250 snow cones, how many RR and BB cones should be stocked at the beginning of the day to maximize Susan's expected profit? What is the expected profit of this policy (Hint: the maximum order size makes the cost of overstocking cones of either flavor increase)?
Number of RR Cones ____________
Number of BB Cones ____________
Expected Profit ____________
Problem 5:
Purchasing road salt for towns in the Northeast is a challenging task. The town of Homer, New York has calculated a forecast of their annual salt needs using historical data. The forecast is summarized in the table below (Q is the quantity needed):
For example, there is a 60.6% chance they will need 50,000 tons or fewer, there is a 3.3% chance they will need exactly 100,000 tons and there is a very small chance they will need more than 200,000 tons. Suppose Homer wants to minimize the amount of inventory it purchases while at the same time having no more than a 6% change of running out of salt (which would force it to purchase salt on the spot market for a premium).
a. At the start of the season, how much salt (in tons) should Homer have available in its storage sheds? Assume salt must be purchased in increments of 10,000.
|
Q (ton)
(a)
|
Distn Fn
|
Density Fn
(b)
|
c=a*b
|
|
0
|
0.0%
|
0.0%
|
0
|
|
10000
|
9.7%
|
9.7%
|
970
|
|
20000
|
23.9%
|
14.2%
|
2840
|
|
30000
|
37.8%
|
13.9%
|
4170
|
|
40000
|
50.2%
|
12.4%
|
4960
|
|
50000
|
60.6%
|
10.4%
|
5200
|
|
60000
|
69.1%
|
8.5%
|
5100
|
|
70000
|
76.0%
|
6.9%
|
4830
|
|
80000
|
81.4%
|
5.4%
|
4320
|
|
90000
|
85.7%
|
4.3%
|
3870
|
|
100000
|
89.0%
|
3.3%
|
3300
|
|
110000
|
91.6%
|
2.6%
|
2860
|
|
120000
|
93.6%
|
2.0%
|
2400
|
|
130000
|
95.1%
|
1.5%
|
1950
|
|
140000
|
96.3%
|
1.2%
|
1680
|
|
150000
|
97.2%
|
0.9%
|
1350
|
|
160000
|
97.9%
|
0.7%
|
1120
|
|
170000
|
98.4%
|
0.5%
|
850
|
|
180000
|
98.8%
|
0.4%
|
720
|
|
190000
|
99.1%
|
0.3%
|
570
|
|
200000
|
99.3%
|
0.2%
|
400
|
|
|
|
Mean
|
53460
|
|
|
|
Std Dev
|
1749
|
Stockout Probailbiy = 6% = 0.06
So Instock Prob = 1-0.06 = 0.94
Corresponding z-stat is NORM.S.INV(0.94) = 1.5548
Z = (S-Mean)/Std Dev
So S = Mean + z*Std Dev = 53460 + 1.5548*1749 = 56179
As Order is in multiple of 10,000, So Order Qty will be 60,000
Order Quantity (tons) 60,000
b. Now suppose Homer has been offered the following deal from American Salt Mine (ASM). ASM will sell Homer salt options for $30 per option with an exercise price of $40 for each option Homer purchases in advance of the season, Homer can "exercise" an option during the season to receive one ton of salt during the season. For example, if Homer purchases 100,000 options before the season starts, then it pays ASM $30 x 100,000 = $3,000,000 for those options. As Homer needs salt during the season, ASM will deliver up to 100,000 tons for a price of $40 per ton. Options are good only for this season - any unexercised options at the end of the season have no value. Finally, if Homer exercises all of its options and still needs more salt, then it will have to purchase salt in the spot market, for an estimated $80 per ton. Given this deal, how many options should Homer purchase from ASM? Assume options must be purchased in increments of 10,000 tons.
Co = Overage Cost = Option Cost = 30
Cu = Underage cost = Mkt Price - Option excerise Cost = 80-40 = 40
So Critical Ratio F(Q) = Cu/(Co+Cu) = 40/(30+40) = 0.5714
Corresponding z-stat is NORM.S.INV(0.5714) = 0.1799
So Optimal Order Qty Q = Mean + Std Dev*z = 53460 + 1749*0.1799 = 53775 ton
So Homer should buy 60,000 Options
Problem 6: Risk Pooling
The UGA Bookstore stocks two types of cashmere sweaters. The two sweaters are identical in every way except on the first sweater is stitched UGA FOOTBALL (FF) while on the second is stitched DAWGS FOOTBALL (RF) (we'll refer to these two types as UGA sweaters and DAWGS sweaters). Both sweaters retail for $100 apiece and cost the Bookstore $40 to procure. Because the procurement lead time is long relative to the length of the football season, the Bookstore places a single order to cover anticipated sales for the entire season. Any sweaters left over at the end of the season are shipped to a reseller for $20 apiece. The demand for UGA sweaters is normally distributed with a mean of 1000 and a standard deviation of 400. The demand for DAWGS sweaters is normally distributed with a mean of 800 and a standard deviation of 300. It's been noted that in previous years when the demand for one type of sweater is high, the demand for the other type of sweater is low, leading the Bookstore to estimate the correlation between the two sweaters at -0.40.
a. How many UGA sweaters should the Bookstore order for the season to maximize expected profit? What is the expected profit?
Co = Overage Cost = Cost - Salvage = 40-20 = 20
Cu = Underage cost = Price - Cost = 100-40 = 60
So Critical Ratio F(Q) = Cu/(Co+Cu) = 60/(60+20) = 0.75
Corresponding z-stat is NORM.S.INV(0.75) = 0.6745
Expected Pooled demand = Mean(FF) + Mean(RF) = 1000+800 = 1800
Std Dev = Sqrt ((400^2 + 300^2)*(1+ Corrreation))
= Sqrt((400^2 + 300^2)*(1-0.40))
= 387.30
So Optimal Order Qty Q = Mean + Std Dev*z = 1800 + 387.30*0.6745 = 2062
Number of UGA Sweaters to Order (units) 2062
If All sweaters are Sold, Expected Profit = Cu*Q = 60*2062 = $123,720
b. How many DAWGS sweaters should the Bookstore order for the season to maximize expected profit? What is the expected profit?
Problem 7: Supply Chain Coordination
A supplier manufactures a product at cost m = $40 and sells it to a retailer at wholesale price c = $100. The retailer sells this product to consumers at retail price p = $200. The replenishment lead time for this product is long relative to the selling season, so the retailer only has one opportunity to stock the product. Demand for the product is normally distributed with a mean of 1000 units and a standard deviation of 400 units. Units unsold at the end of the season are salvaged at price v = $15.
a. How many units of the product should the retailer order to maximize its expected profit? What is the associated retailer's expected profit and the supplier's expected profit?
Number of Units to Order ____________
Retailer's Expected Profit ____________
Supplier's Expected Profit ____________
b. how many units of the product should the retailer order to maximize total supply chain expected profit? What is the associated retailer's expected profit and the supplier's expected profit?
Number of Units to Order __________
Retailer's Expected Profit ______________
Supplier's Expected Profit _____________
c. To motivate the retailer to buy more product, the supplier offers the following deal: the wholesale price is reduced to c = $80, but the retailer must pay the supplier an additional $40 for every unit that it sells. The supplier is also willing to buy back units that the retailer is unable to sell at buyback price b. These units are then salvaged by the supplier. What value of b should the supplier offer to maximize total supply chain expected profit? What is the associated retailer's expected profit and the supplier's expected profit?
Optimal Buyback Price ___________
Retailer's Expected Profit _______________
Supplier's Expected Profit _