1. WL2. An ambulance company receives a request for service about every 45 minutes. The service time, which includes time for the ambulance to get to the customer, processing, and “delivery” of the customer to his/her destination, is 55 minutes (exponentially distributed). The cost of each ambulance is $75,000 per year (lease, maintenance, fuel, ….) and the cost per ambulance “crew” is about $50/hour. Assume 24/7 (365 days a year).
How many ambulances should be in service if the target average wait time is 5 minutes (for an ambulance to become available)? What about if 15 minutes? 45 minutes?
If this is a service for non emergency care and the cost of waiting is $100/hour, what is the least cost solution?
How do the answer in b) changes if the cost of waiting is $10/hour?

2. How to solve this problem
First Step: ID problem parameters
What are the entities /customers: patients requesting transportation
What are the resources: ambulances
Number of servers (how many ambulances): S;
Arrival rate (how many patients request transportation): l;
Service rate (how many patients can be transported by a single ambulance): m;

3. S = is undefined at the start of the problem definition
Number of Servers
There is the possibility of one or more ambulances.
S = is undefined at the start of the problem definition

4. ID the Arrival Rate Interarrival time = 45 min/ ts
The average time between requests for transportation service is 45 minutes.
Interarrival time = 45 min/ ts
ts = transportation service
Arrival Rate (l ) is the inverse of the interarrival time.
l = 1 𝑡𝑠 45 𝑚𝑖𝑛 = ts/min
convert to hours or days to make more intuitive. we convert to ts/ day as it is an integer number.
l = 1 𝑡𝑠 45 𝑚𝑖𝑛 × 60 𝑚𝑖𝑛 ℎ𝑜𝑢𝑟 = ts/hour
l = 1 𝑡𝑠 45 𝑚𝑖𝑛 × 60 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 ℎ𝑜𝑢𝑟 × 24 ℎ𝑜𝑢𝑟𝑠 𝑑𝑎𝑦 = 32 ts/day

5. ID the Service Rate Service time = 55 mins/ts
The average time to provide the transportation service is 55 minutes.
Service time = 55 mins/ts
Service Rate (m) is the inverse of the service time.
m = 1 𝑡𝑠 55 𝑚𝑖𝑛𝑠 = ts/min
we will use the ts/days conversion as l and m need to be in the same time scale
m = 1 𝑡𝑠 55 𝑚𝑖𝑛 × 60 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 ℎ𝑜𝑢𝑟 × 24 ℎ𝑜𝑢𝑟𝑠 𝑑𝑎𝑦 = 26.2 ts/day

6. Number of Servers
We can now determine the minimum number of servers that can be considered.
For a problem to be in steady state this must be true: l < Sm
Assuming S = 1 then
Is 32 ts/day < 26.2 ts/day true? No. Therefore S cannot be 1.
Logic: One ambulance that can only provide 26.2 services per day is unable to handle a demand of 32 patients requesting transportation services per day.
Therefore this system would never be in steady state (line will increase forever).

7. Number of Servers We can increase S to 2.
For a problem to be in steady state this must be true: l < Sm
Is 32ts/day < 2 × 26.2 ts/day true? Yes. Therefore S can be 2 (or larger).
Logic: Two ambulances can provide 52.4 (2 × 26.2) services, and therefore they are able to handle a demand of 32 patients requesting transportation services.
Therefore this system can be in steady state (line will vary in size).

8. a) How many ambulances should be in service if the target average wait time is 5 minutes (for an ambulance to become available)? What about if 15 minutes? 45 minutes?
To answer this question we need to calculate Wq for different levels of S.
We will use the Lq table
Note this is wait time to get an ambulance assigned
to pickup the patient, not the wait time before
it arrives to pick up the patient.

9. Lq Table

10. Lq Table Based on the ratio l / m l / m = 32/26.2 = 1.22
Given is not in the table we use the next number = 1.3

11. Develop a table the gives the average waiting time per number of ambulances
These are the
values from the table
Wq = Lq / l S
Lq (ts) Wq 2
0.95
0.0297d
× 24 ℎ𝑜𝑢𝑟𝑠 𝑑𝑎𝑦 × 60 𝑚𝑖𝑛𝑠 ℎ𝑜𝑢𝑟 = mins 3
0.13 d
× 24 ℎ𝑜𝑢𝑟𝑠 𝑑𝑎𝑦 × 60 𝑚𝑖𝑛𝑠 ℎ𝑜𝑢𝑟 = 5.85 mins 4
Results.
If two ambulances the average wait time is 42.8 minutes.
If three ambulances the average wait time is 5.9 minutes
If four ambulances the average wait time approximates 0

12. For 5 minutes = 4 ambulances For 15 minutes = 3 ambulances
a) How many ambulances should be in service if the target average wait time is 5 minutes (for an ambulance to become available)? What about if 15 minutes? 45 minutes? S
Lq (ts) Wq 2
0.95
42.75 mins 3
0.13
5.86 mins 4
0 mins
For 5 minutes = 4 ambulances
For 15 minutes = 3 ambulances
For 45 minutes = 2 ambulances

13. b) If this is a service for non emergency care and the cost of waiting is $100/hour, what is the least cost solution?
First we must calculate the waiting cost per entity: WCe WCe = Wq × wcr The wcr =$100/hour (given in the problem description) = $2,400/day For S = 2, then WCe = d × $2,400 = $71.25/ts For S = 3, then WCe = 5 min × $1.667/min = $9.75/ts S
Lq (ts) Wq
WCe
RCd
RCe
TCe 2
0.95
0.0297d
$71.25 3
0.13 d
$9.75 4 $0

14. b) If this is a service for non emergency care and the cost of waiting is $100/hour, what is the least cost solution?
Next we must calculate the resource cost per year: RCY The problem indicates that: The cost of each ambulance is $75,000 per year (lease, maintenance, fuel, ….) and the cost per ambulance “crew” is about $50/hour. Assume 24/7 (365 days a year). Thus the annual cost per ambulance is 75, × 24 × 365 = $513K/ambulance =$513K/yr / 365 = $1,405/day For S = 2, then RCd =2 ambulances × $1,405/day= $2,810 For S = 3, then RCd=3 ambulances × $1,405/day= $4,215 S
Lq (ts) Wq
WCe
RCd
RCe
TCe 2
0.95
0.0297d
$71
$2,810 3
0.13 d
$9.75
$4,215 4 $0
$5,620

15. b) If this is a service for non emergency care and the cost of waiting is $100/hour, what is the least cost solution?
Determine RCe
We need to “distribute”the resource costs across all customers.
RCe = RCd / demand per day = arrival rate per day = l
$2,810 / 32 = $87.81
Best solution is 3
ambulances S
Lq (ts) Wq
WCe
RCd
RCe
TCe 2
0.95
0.0297d
$71
$2,810
$87.81
$158.81 3
0.13 d
$9.75
$4,215
$131.72
$141.47 4 $0
$5,620
$175.63

16. c) How do the answer in b) changes if the cost of waiting is $10/hour?
All the WCe are 10% of the original
WCe
$71
$9.75 $0
original S
Lq (ts) Wq
WCe
RCd
RCe
TCe 2
0.95
0.0297d
$7.1
$2,810
$87.81
$94.91 3
0.13 d
$0.98
$4,215
$131.72
$132.70 4 $0
$5,620
$175.63

17. c) How do the answer in b) changes if the cost of waiting is $10/hour?
All the WCe are 10% of the original
WCe
$71
$9.75 $0
Best solution is 2
ambulances S
Lq (ts) Wq
WCe
RCd
RCe
TCe 2
0.95
0.0297d
$7.1
$2,810
$87.81
$94.91 3
0.13 d
$0.98
$4,215
$131.72
$132.70 4 $0
$5,620
$175.63

18. Extra: What is the best solution if the cost of waiting is $1,000/hour
Extra: What is the best solution if the cost of waiting is $1,000/hour? (wcr = $1,000/hr)
All the WCe are 10 x of the original
WCe
$71
$9.75 $0
original S
Lq (ts) Wq
WCe
RCd
RCe
TCe 2
0.95
0.0297d
$710
$2,810
$87.81
$797.81 3
0.13 d
$97.5
$4,215
$131.72
$229.22 4 $0
$5,620
$175.63

19. Extra: What is the best solution if the cost of waiting is $1,000/hour
Extra: What is the best solution if the cost of waiting is $1,000/hour? (wcr = $1,000/hr)
All the WCe are 10 x of the original
WCe
$71
$9.75 $0
Best solution is 4
ambulances S
Lq (ts) Wq
WCe
RCd
RCe
TCe 2
0.95
0.0297d
$710
$2,810
$87.81
$797.81 3
0.13 d
$97.5
$4,215
$131.72
$229.22 4 $0
$5,620
$175.63

20. Problem insights
The problem serves to illustrate how waiting and resource costs change as the resource capacity changes.
The optimal solution is highly dependent on the waiting cost ratio (wcr).
As the wcr decreases the optimal solution (lowest cost) will use fewer resources
Inversely, as wcr increases, the optimal solution (lowest cost) will use more resources.